Answer:
natural capital
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____________________ syndrome occurs when inflamed and swollen tendons are caught in the narrow space between the bones of the shoulder joint.
Answer:
Tendonitis
Explanation:
When a part of the body becomes too overused it can cause the tendons(fiborous connective tissue) to swell and fluid will accumulate around the tendon which connects muscle to bone. Along with this the lining of the tendon can become inflamed which causes Tenosynovitis.
Which of the following is not a function of the digestive system? Select one: A. mechanical processing B. absorption C. secretion D. ingestion E. filtration
Answer: option E.
Filtration.
Explanation:
Digestive system consist of the gastrointestinal tracts, organs that help to breakdown large complex consumed food into to small pieces which can easily be absorbed by the body.
Filtration is not a function of digestive system because it play a major role in circulatory system which help in movement of solutes out of the capillary into interstitial fluid by separating the bigger particles from the smaller particles, filtering out nitrogenous waste and regulating salt concentration .
A response to an unconditioned stimulus that occurs naturally without learning is known as a(n) __________ response.
A. neutral
B. conditioned
C. reconditioned
D. unconditioned
Answer:
it is going o be B. conditioned
Answer:
D
Explanation:
edge
Some people use the terms "cell cycle" and "mitosis" as if they were the same thing. Which of these best describes why they are not the same?
a. the cell cycle is just a stage of mitosis
b. mitosis and the cell cycle are different types of cell division that occur in different of cells in a multicellular organism
c. mitosis is a sequence of events within the cell cycle that separates the genetic material
d. mitosis is a type of cell division that occurs in eukaryots and the cell cycle takes place in prokaryots
Answer: C. Cells in meiosis have unique genetic information
Explanation: Meiosis is the cell division that forms four daughter cells from one parent cell. It includes two successive divisions called as meiosis I and meiosis II. Crossing over during prophase I of meiosis I imparts new gene combinations to the daughter cells of meiosis. Hence, daughter cells formed by the end of meiosis have some new gene combinations, that is, unique genetic information.
The client has experienced a stroke, and her speech is now garbled. The client may have experienced damage to which of the following?
A. Brainstem
B. Medulla oblongata
C. Frontal lobe
D. Cerebellum
Answer:
C
Explanation:
A person with garbled speech may have experienced damage to the frontal lobe portion of the brain.
This is because the left portion of the frontal lobe of the brain is an area that has been touted to be responsible for the control of speech in humans. Research in the past has shown that damage to that region of the brain affects the formation of spontaneous speech as well as motor speech control.
Hence, garbled speech in someone that has experienced stroke can be attributed to damage to the frontal lobe.
Correct option: C.
Type the correct answer in the box. Spell all words correctly. Henry went on a field trip to a national park. He not only observed the aquatic life in the pond, but also the trees, vegetation, and animal’s species on land. Which type of diversity did Henry come across in the national park? Henry came across the______ diversity.
Answerflora
Explanation:
Every interaction with the outside world is because of A) epithelial tissue B) connective tissue C) nervous tissue D) muscular tissue
Answer:
A cutaneous membrane is a multi-layered membrane composed of epithelial and connective tissues. The apical surface of this membrane exposed to the external environment and is covered with dead, keratinized cells that help protect the body from desiccation and pathogens....
If the Gulf Stream stopped running, how would the climates of North America and Europe be affected?
Answer: Land would become infertile, habitat loss, & environmental changes
Explanation:
Because there wouldn't be alot of water to keep the soil moist, therefore the ground would begin to harden and get dry. :)
If the Gulf Stream stopped, the climates of North America and Europe would be colder and drier.
The gulf stream is a huge current of warm and fast-moving ocean water. This ocean current plays a key role in regulating temperatures across the globe and has an impact on the frequency and intensity of storms and rainfall. The collapse and stoppage of this stream will cause an array of problems, including but not limited to:
Reduced rainfallLower temperaturesFewer stormsAreas that would be particularly affected by this are North America and Europe. This is due to the direction that the gulf stream flows, carrying warm water along these boundaries.
The vast amount of warm water carried by this stream explains how it is able to affect the average temperatures of these coastal areas, with the disappearance of the stream, temperatures in North America and Europe would initially decrease.
The warm water carried by this current is also responsible for increased rainfall. Warm water on the surface of the ocean is heated by the sun and evaporates, creating clouds, which then provide rain. The absence of this current would dramatically affect cloud formation and reduce the amount of rain severely. This can lead to droughts and reduced agriculture.
Though Fewer storms may seem like a positive impact, it is not entirely. Storms play a crucial part (together with ocean currents) in regulating global temperatures, as well as provide a source of rainfall to end droughts and promote biodiversity in areas that depend on water and humid climates.
The gulf stream is important for maintaining climates and rainfall. If it were to stop, climates in North America and Europe would be colder and drier.
For more information on ocean currents visit:
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A mutation that hides the effect of another mutation at a site that is distinct from the site of the original mutation, but with in the same gene. This mutation is best described as a
Answer:
The correct answer is intragenic suppressor mutation.
Explanation:
A suppressor mutation partially or completely converses the effects of a different mutation. A suppressor mutation is of two kinds, that is, an intragenic suppressor and an intergenic suppressor. An intragenic suppressor converses the influence of a mutation at a distinct site in a similar gene.
On the other hand, an intergenic suppressor converses the influence of a mutation at a distinct locus of the gene. It is the intragenic suppressor mutation, which takes place in a similar gene where the occurrence of the first mutation had taken place. Therefore, the mutation, which hides the influence of another mutation at a location, which is different from the location of the original mutation, but taking place in a similar gene can be illustrated as the intragenic suppressor mutation.
When the blood glucose concentration is low, the pancreas secretes glucagon to signal the liver to produce glucose and glucose-1-phosphate. Which metabolic steps are involved in this biosignaling pathway
Answer:
See below (Correct Answer: I and IV)
Explanation:
Inactivation of glycogen synthase via phosphorylation by protein kinase A
Activation of glycogen synthase via phosphorylation by protein kinase A
Activation of phosphofructokinase-2 via phosphorylation by protein kinase A
Activation of fructose-2,6-bisphosphatase via phosphorylation by protein kinase A
Activation of phosphofructokinase-1 by fructose-2,6-bisphosphate
Inhibition of fructose-1,6-bisphosphatase by fructose-2,6-bisphosphate
Inactivation of pyruvate kinase via phosphorylation by protein kinase A
II
I and IV
I, III, V, VI, and VII
I, IV, and VII
Correct Answer: (b) I and IV
Glycogenolysis: It is the process by which glycogen stored in liver and muscle cells, break down to form glucose.
which of the following MOST directly influences a measurable outcome in an experiment?
Answer:
Hello. You did not enter the answer options, but the factor that most directly influences a measurable result in an experiment is the manipulation of the variables.
Explanation:
In an experiment, the manipulation of variables becomes highly important so that it is possible to measure, that is, evaluate the result. This is because it is the variables that express values that represent the characteristics that are being analyzed and studied within the experiment. Therefore, the manipulation between them must be done in a very rational and balanced way so as not to modify the values shown by them, changing the data and generating false or immeasurable data.
Activity 12-5: A Dihybrid Punnett Square. Consider your answers in the previous question. In a separate piece of paper, solve the Punnett square for a cross between two heterozygous individuals (BbRr x BbRr); write the genotypes of the gametes along the top and the side of the square and fill in the squares.
Answer:
See the answer below
Explanation:
For each heterozygous parent with the genotype BbRr, the possible gametes are: BR, Br, bR, and br. A set of these gametes from one of the parents will be lined up along the top of the Punnet's square while another set from the other parent will be lined up along the side of the square.
The result is shown in the attached image.
A cell biologist examines a diploid cell from a barley plant during prometaphase of mitosis and determines that 28 chromatids are present. The role of meiosis in this plant is to (Check all that apply)
Answer:
A cell biologist examines a diploid cell from a barley plant during prometaphase of mitosis and determines that 28 chromatids are present. The role of meiosis in this plant is to reduce the number of chromosomes per cell from 14 to 7.
Explanation:
Meiosis is a process that happens in two different phases, and there are many more steps in every phase. This procedure reduces the number from diploid to haploid, that's why the number of chromosomes per cell reduces from 14 to 7. Thanks to this process we have diversity in every organism that has sexual reproduction.
Which statement best describes the rock shown?
O The grains of this rock are jagged.
O The grains in this rock are tiny.
O This rock has a non-banded pattern.
O The color of this rock is determined by its texture.
Answer:
this rock has a non-banded pattern
Answer:
Confirming that the other person is correct
Explanation:
As observed in Protein Workshop, the guanidinobenzoyl group of the substrate is bound to the domain of the protein that ___________ contain the L1 and L2 regions.
Questions
41 Explain why you needed to use a water plant for this experiment
I suspect you are referring to a specific experiment, which measures the rate of photosynthesis at differing intensities of light. This experiment could certainly be done with land plants, but using a water plant makes it easier to have a simple way to measure the rate of photosynthesis--the counting of bubbles which come off a specific area of the plant. A water plant such as elodea (the type you commonly see in aquariums) does the process of photosynthesis just as plants that grow in the air do, but you can actually see the bubbles that indicate photosynthesis is occurring--they form on the leaves before they float to the surface, enabling an easy measurement of bubbles per minute.
Plants have chloroplasts, which in addition to giving plants their green color, perform photosynthesis, the process on which life on earth is dependent. They use the energy of the sun to use carbon dioxide and water to produce sugars and oxygen. When you are counting the bubbles on the underwater leaf, you are counting bubbles of oxygen formed during photosynthesis.
A graduate student studying biology at the University of Nebraska has identified a new species of spider found only in Eastern Nebraska around Omaha. The graduate student determines that the spider has six homologous pairs of chromosomes. How many chromosomes would a cell in that spider have during metaphase of mitosis?
Answer:
12 chromosomes
Explanation:
Mitosis is a type of cell division that involves the formation of two genetically identical daughter cells. The two daughter cells are genetically identical in the sense that they contain the same number of chromosomes as the parent cell. Mitosis involves four stages namely: Prophase, Metaphase, Anaphase and Telophase.
In the metaphase stage as stated in this question, homologous chromosomes align at the equator of the cell called cell plate, before each chromatids are pulled apart by microtubules at the Anaphase stage.
According to the question, the spider being worked on has 6 pairs of chromosomes, which will align at the cell's equator during metaphase stage of mitosis. Since the replicated chromosomes (chromatids) are yet to separate to opposite poles of the cell, the cell will still contain 12 chromosomes at the metaphase stage.
N.B: Each chromosome contains 2 chromatids or replicated chromosome, which will be separated at the Anaphase stage. Each chromatid will be an individual chromosome after cytokinesis.
Choose a strategy to clone genes known only by a mutant phenotype in organisms with highly efficient transformation procedures and relatively small genomes.
Answer:
By combining traditional genetic mapping with whole-genome sequencing (next-generation sequencing)
Explanation:
Map-based cloning, also named positional cloning, is a technique used to identify genetic sequences linked to markers whose exact physical localization in the genome is known. Map-based cloning has shown to be useful to clone genes, however, this approach is limited to the existence of mutant phenotypes in different organisms/accessions/strains. In recent years, map-based cloning has been combined with next-generation sequencing methodologies in order to clone genes in background of single mutants without information on multiple organisms. This type of strategy has the potential for identifying causal mutations and thus clone genes in species with small genomes and high transformation efficiencies such as, for example, Arabidopsis thaliana.
1. Question: How does temperature affect the growth of bacteria?
Procedures: A scientist places three different petri dishes in 20C, 35C,
and 60C. After 2 days, the number of bacteria colonies were counted.
Answer:
It directly affects
Explanation:
Bacteria are conditioned by external environmental factors such as partial pressure of oxygen, humidity, temperature, metabolites, radiation present, living with other types of colonies, partial pressures of carbon dioxide, etc.
The growth of a colony will always be better or worse as long as the ideal environmental conditions are met for that bacterium to nurture and develop widely.
In general, the ideal temperature is around 37 degrees for most bacteria, although it also depends a lot on the presence of the substrate they need to live and give their necessary metabolites.
This food web reveals that, as matter flows
through trophic levels,
A. matter from consumers, such as the green lynx
spider, is eventually recycled by decomposers,
such as a fungus.
B.matter from producers, such as the cutleaf daisy,
is immediately recycled by decomposers, such
as a bacterium.
C. matters from all organisms, such as the Bald
eagle, is dissipated at each level and none of it
is recycled.
D. matter from the Sun is recycled after many years
by consumers, such as a Bachman's sparrow.
Answer:
There is no food web attached to this question but it can be answered because matter flows uniformly through trophic levels in all food webs.
The correct answer is;
A. matter from consumers, such as the green lynx spider, is eventually recycled by decomposers, such as a fungus.
Explanation:
A food web is an interconnected series of food chain showing how organisms feed on one another to obtain energy. A food web always start with a unique set of autotrophic organisms i.e. capable of producing their own food called PRODUCERS. Producers such as the cutleaf daisy in this question create the organic matter via the process of photosynthesis, which gets transferred to organisms that feed on them called CONSUMERS e.g green lynx spider
Consumers are heterotrophic i.e rely on other organisms for source of energy. As they feed on other organisms and get fed on, the matter gets transferred until the last consumer dies. The dead matter in the dead organism gets decomposed by organisms called decomposers e.g fungi and it gets recycled back into the soil as nutrients for use by plants.
Therefore, dead matter from consumers, such as the green lynx spider, is eventually recycled by decomposers, such as a fungus.
The origin and the insertion are made up of __________.
Answer:tendon
I think so..
Why photosynthesis can't take place in such an environment , where there is no oxygen..? And exactly why trees need oxygen for photosynthesis ?
Answer:
Photosynthesis is a chemical process that leads to the release of oxygen as a by-product and is necessary for cellular respiration
Explanation:
The plants produce or make oxygen form the wastes of starch, sunlight carbon dioxide, and water. In order to burn the sugar that is stored in the plants, it needs to have sunlight which requires oxygen. As the break down of sugar C6 H12 O6 to CO2 and H2O needs O2 to do it. Which is called cellular respiration. This is also stated by the first law of thermodynamics that energy cannot be created and neither destroyed but transforms itself. Photosynthesis is the chemical process that is used to convert this chemical energy which makes carbon and water into glucose and oxygen is released as a by-product."If an organism were to die in an environment that had no saprotrophs or detritivores, what would you expect to see if you observed the area 2 weeks later?"
Answer: The options are not given, here are the options.
A. There will be no difference as compared to an area that has saprotrophs and detritivores. •
B. There would be nothing left, it will have completely decomposed. •
C. The body would remain, and be just as it was when it died except for weathering. •
D. The body would be decomposing at about half the normal speed.
The correct option is C.
The body would remain, and be just as it was when it died except for weathering.
Explanation:
"If an organism were to die in an environment that had no saprotrophs or detritivores, The body would remain, and be just as it was when it died except for weathering because detrivores and feed on organic matter and excrete the nutrients, without detrivores and saprotroph the the organic matter in the dead organisms will still remain the same.
Detrivores are organisms that feed on dead organic matter and releasing nutrients to the ecosystem through it's wastes.
Saprotroph are decomposers that survive on or in dead organic matter secreting enzymes and releasing nutrients to the ecosystem. Examples is fungi.
Without detrivore and and saprotroph the dead organisms will remain the same except from weathering be side weathering is carried out through the activities of air, water or some agents.
You continue this approach by designing two separate experiments. In the first experiment, you use glucose where "Coccupies position 1. In the second experiments, 14C occupies position 2 and 6 in the glucose molecules. For each experiment, you use 0.2 moles of radiolabeled glucose and you assume that all the pyruvate formed is converted to acetyl-CoA. What following statements are correct? (select 3)
a) When glucose is labeled on carbon #1, 0.2 mole of acetyl-CoA is radiolabeled
b) When glucose is labeled on carbon #1, 0.1 mole of acetyl-CoA is radiolabeled
c) When glucose is labeled on carbons #2 and #6, 0.2 mole of acetyl-CoA is radiolabeled
d) When glucose is labeled on carbons #2 and 6, 0.1 mole of acetyl-CoA is radiolabeled
e) When glucose is labeled on carbons #2 and 6, the carbonyl groups of half of the acetyl-CoA molecules are radiolabeled
f) When glucose is labeled on carbons #2 and 6, the methyl groups of half of the acetyl-CoA molecules are radiolabeled
Answer:
b) When glucose is labeled on carbon #1, 0.1 mole of acetyl-CoA is radiolabeled
c) When glucose is labeled on carbons #2 and #6, 0.2 mole of acetyl-CoA is radiolabeled
e) When glucose is labeled on carbons #2 and 6, the carbonyl groups of half of the acetyl-CoA molecules are radiolabeled
f) When glucose is labeled on carbons #2 and 6, the methyl groups of half of the acetyl-CoA molecules are radiolabeled
Explanation:
When glucose undergoes glycolysis, it is converted into two molecules of pyruvate with carbon 1 to 3 forming the first molecule of pyruvate and carbon 4 to 6 the second molecule of pyruvate. The C-1 and C-6 of the glucose molecule becomes the methyl groups of each of the two molecules of pyruvate. The C-2 and C-5 of the glucose molecule forms the carbonyl carbon of each of the two pyruvate molecules. Each of the two pyruvate molecules undergoes further oxidation to yield acetyl-CoA with the carbonyl and methyl groups of pyruvate retained in the acetyl-CoA molecules.
Thus when 0.2 moles of glucose are labelled at C-1 and then C-2 and C-6 in each of the two experiments the following results are obtained:
When glucose is labeled on carbon-1, 0.1 mole of acetyl-CoA is radiolabeled since half of the two pyruvate molecules are obtained from C-1
When glucose is labeled on carbon-2 and carbon-6, 0.2 mole of acetyl-CoA is radiolabeled since the C-2 and C-6 of the glucose molecules forms a part of one of each of the two pyruvate molecules.
When glucose is labeled on carbons #2 and 6, the carbonyl groups of half of the acetyl-CoA molecules are radiolabeled since one of the two carbonyl groups of the two pyruvate molecules is formed from C-2 of glucose.
When glucose is labeled on carbon-2 and carbon-6, the methyl groups of half of the acetyl-CoA molecules are radiolabeled since one of the two methyl groups of the two pyruvate molecules is formed from C-6 of glucose.
Help me please!!! I need this
Answer:
The correct answer is - option A. 1 (ovary).
Explanation:
Oocytes or egg are produced in the ovaries during the process of the female gametogenesis in female reproductive system. Ovaries are located on each side of the uterus that are oval and small in shape and size and located lower abdomen.
Among other female reproductive organs these are located above others. The ovaries produce oocytes and hormones It is the site at which primordial germ cell (PGC), become primary oocytes.
Thus, the correct answer is - option A. 1 (ovary).
You believe that the cytosolic protein p007 is involved in mediating growth hormone-stimulated proliferation in cells. You wish to establish a cell model using your cultured fibroblasts. Provide clear explanations to the following, each worth 2.5 pts each. a) Describe how you would transfect the growth hormone receptor DNA into your cells to create stable cell lines, and include a description of the benefits of using stable cell lines versus transiently transfected cell line. b) Describe how you would measure specific receptor binding of GH to your stable cell line. c) Describe how you would measure growth hormone-induced proliferation in your stable cell line. Make sure to include controls. d) Describe how you would use RNAi to demonstrate the requirement for p007 in growth-hormone stimulated proliferation. Make sure to include controls.
Answer:
Following are the answer to this question:
Explanation:
In option a:
There are many ways, which can be defined as follows:
It keeping cells "qualified," use heat shock and handle ions like Ca and ice incubating.It uses Micro-injection with DNA. It uses the biolistics or even the process of gene-gun.The stable cell charts indicate simple and definitive outcomes when their relationship to specific results. Moreover, relative to temporary cell lines, findings are more reliable and error-free. These lines also easily break and much less.
In option b:
Radio tests including linking tests could be used
The Ligand Binding Assays is an analysis, that is based on the binding of ligands to antibodies, or other macromolecules. Its existence and distance of formed ligand-receptor complexes are defined using a system of measurement that is usually electrical or fluorescent.
In option c:
The method for measuring the levels of cell growth is accessible and the Another way is to measure the metabolism within the same cell. So many color pigments were also accessible that can permeate an organism as well as react with some proteins as well as other factors, and that can effectively identify a color final product.
Throughout relation to the amount of metabolism exercise, cell proliferation could be evaluated by researching one or even more cell markings. The BrdU limited liability assessment is a very well-published illustration.
Its specific antibody can detect BrdU and high BrdUincorporations levels are linked to high levels of cell proliferation. Command cell without DNA receptor growth hormone.
In option d:
The cell growth will stop after each use of RNA interference ( RNAi). The p007 formation as well as the effect on growth caused by the intrusion will be a goal. If development is delayed effectively, the theory will be tested successfully to a cell without any of the required DNA. The RNAi is the free cell
The ostrich is a bird. However, it has traits that are different from those of the birds we typically see flying around our neighborhoods. Using credible sources, investigate the environment and adaptations of ostriches to explain how they evolved to survive in their environment.
Answer:
According to National Geographic, ostriches are a part of a very small group of birds that cannot fly because unlike most birds, their small wings are not strong enough to carry their body for flight and their breastbone isn't balanced enough for flying. Birds that are unable to fly are called ratites.
A number of scientists namely Thomas Huxley, Richard Owen, and others have tried to show that these ratites are actually related to each other and eventually, it was discovered that they all had one thing in common, the way the bones at the roof of the mouth were arranged was similar to that of reptiles rather than other birds.
Richard Owen found and assembled the remains of an extinct ostrich skeleton which was an extinct moa and contrary to already held opinion, one ratite known as tinamous did not really fit with the profile of a ratite because it could fly, even though almost grudgingly and they possessed keeled sternum which suggests that they evolved from flying birds.
DNA tests showed that tinamous evolved within ratites and not necessarily as a separate entity. The tests also showed that moas and tinamous are related.
It was also speculated that the division of the supercontinent Pangaea southern side led to the separation of flightless ratite ancestors, causing each landlocked group to evolve and become the flightless birds we know today such as the ostrich, rheas, etc.
Answer:
An ostrich is an extremely large bird that lives in Africa. It can’t fly, but it has powerful legs that allow it to outrun most predators. Its flexible knees and two-toed feet enable it to run easily. An ostrich has strong leg muscles that it can use to kick large predators. It has claws on its wings to defend itself from predators and thick feathers to keep it warm at night. Ostriches have thick eyelids that protect their eyes from sandstorms.
Explanation:
The role of the ___________ branch of the autonomic nervous system mediates control of organ processes when the body is essentially ______.
Answer: parasympathetic; at rest
Explanation: The parasympathetic nervous system branch of the autonomic nervous system is largely responsible for the relaxation of the body especially at rest where it undoes the activities of the sympathetic branch by decreasing respiration, heart rate and then increasing the body's digestion rate. As such, this branch is also responsible for digestion response in a relaxed, resting, or feeding states. The role therefore of the parasympathetic branch of the autonomic nervous system mediates control of organ processes when the body is essentially at rest.
explain where the pericardium is found and what it does for the heart
Answer:
the pericardium is found in your heat Problems can occur when the pericardium becomes enflamed or fills with fluid.
can someone tell me what crabs these are.
Answer:
Loxorhynchus grandis
