Answer:
The osmolarity of mammalian urine may vary over time. The osmolarity of mammalian urine varies little between species. Mammalian urine is always hyperosmotic to blood. The osmolarity of mammalian urine may vary over time.
Explanation:
I hope it will help you
geosphere burying organisms
hydrosphere respiration of marine life
biosphere photosynthesis atmosphere burning things
Answer:
This question is incomplete, the question is asking to match each activity with the sphere to which it adds carbon or CO2.
ANSWERS:
Geosphere = burying organisms
hydrosphere = respiration of marine life
biosphere = photosynthesis atmosphere = burning things
Explanation:
This question describes the types of spheres we have. The respective activities that occur in each sphere that adds CO2 or carbon are as follows:
- Geosphere: This refers to the solid part of the Earth. Dead organisms are buried in the geosphere (soil) where their carbon remains can be deposited.
- Hydrosphere: This constitutes all of the water bodies (sea, river, stream, ocean) on the Earth. When marine organisms respire in water, carbon dioxide (CO2) is released.
- Biosphere: A sphere consisting of all living things (plant, animal, microbes etc.) on Earth. Plants as a living organisms photosynthesize and make use of carbon dioxide to produce glucose, which stores in the plant.
Atmosphere: The gaseous component of the Earth. Burning substances release carbon based gases into the atmosphere.
Which organism exhibits behavioral adaptation?
Answer:
Humans
Birds
Bears
Chameleons
Cicadas
Explanation:
A Behavioral Adaptation is something an animal does - how it acts - usually in response to some type of external stimulus.
Examples of some Behavioral Adaptions:
Migration, Hibernation, Dormancy, Camouflage.
Answer:
Behavioral adaptations are the actions that organisms take in order to stay alive. Behavioral adaptations include things like bird cries and migration. Evolution produces adaptations. Evolution is a long-term change in a species.
OAmalOHopeO:
is important to the transmission of nerve impulses.
a) Zinc
b) Calcium
c) Sodium
d) Iron
Please help thx
is important to the transmission of nerve impulses.
a) Zinc
b) Calcium
c) Sodium
d) Iron
[tex] \bold \pink{sodium}[/tex]
Answer:
B)Calcium
Explanation:
Have a nice day
An enzyme is a _____ molecule.
a. polysaccharide
b. nucleic acid
c. carbohydrate
d. protein
The locus B on the X chromosome of a malaria-carrying mosquito shows a 49% recombination rate with respect to the locus M. Since a recombination rate of 50% is essentially indistinguishable from independent assortment, you might be tempted to look for a locus that falls between B and M. Before you decide to do all that work, you run a chi square test to determine the P value of your experiment.
Required:
What range of P values would tell you that you should accept the conclusion that locus B and locus M are, indeed, 49mu apart and that another locus is not necessary?
Answer:
The correct answer is - P = 0.45
Explanation:
The Chi-square test is a test that determines the assorted genes independently. Here p-values used to make conclusions in significance testing.
If the p value is less than the significance level we expect or choose, then this null hypothesis is rejected in favor of the alternative hypothesis. It p value is greater than or equal to the significance level, then we fail to reject the null hypothesis.
So in this case, as the recombination rate is 49% so the p value which satisfies the conclusion will be around 0.45 and values near to it.
The large number of alveoli in lungs improve gas exchange by ________. increasing the surface area for diffusion decreasing the distance for diffusion changing the diffusion constant increasing the concentration difference
Answer:
Explanation:
The large number of alveoli in lungs improve gas exchange by . increasing the surface area for diffusion
All of the members of a particular species that live in one area are called a(an):
A) biotope
B) community
C) ecosystem
D) biosphere
E) population
Answer:
Population
Explanation:
population- all the members of one species in a particular area.
Different measures of disease are useful to evaluate and assess public health programs and needs in different situations. For each of the following questions, state which measure would best support your goal.
Measures of disease:
I = incidence rate (p. 126)
P = prevalence (p. 113)
L = lifetime prevalence (p. 114)
M = crude mortality rate (crude death rate) (p. 112)
R = sex ratio (counts) (p. 109)
a. To demonstrate the risks of car-train crashes at railroad crossings without warning signals.
b. To demonstrate the amount of children's exposure to secondhand cigarette smoke.
c. To estimate the number of persons who have had leukemia during their lifetimes.
d. To estimate the number of healthcare facilities needed to support patients with Alzheimer's disease.
e. To argue that mortality from HIV infection is a more serious public health problem in one region of the US than another.
Answer:
Different Measures of Disease
Question Measure of Disease
a. I = incidence rate
b. P = prevalence
c. L = lifetime prevalence
d. L = lifetime prevalence
e. M = crude mortality rate
Explanation:
a) Measures of disease:
I = incidence rate (p. 126): measures the frequency of disease occurrence in a population over time. It is measured as new episodes of illness in a period divided by the population.
P = prevalence (p. 113): ratio or proportion of persons in a population with a disease at a time or over time. It is based on a sample.
L = lifetime prevalence (p. 114): the proportion or ratio of a population who had a disease at some point in their life.
M = crude mortality rate (crude death rate) (p. 112): the ratio of deaths in a period among a geographical population.
R = sex ratio (counts) (p. 109): ratio of male and female population.
H. pylori cannot grow in other microenvironments of the human body because the conditions are unsuitable for its growth, but other species require different conditions.
a. True
b. False
Answer:
it should be an true statement
Unlike skeletal muscle, the muscle cells of the heart (called "cardiomyocytes") metabolize very little glucose, relying almost exclusively on metabolism of fatty acids for their energy. Which of the following statements are likely to be true of cardiomyocytes, but NOT skeletal myocytes? (select two answers)
a. low levels of glycogen storage, even when blood glucose is high
b. high levels of glycogen storage, even when blood glucose is low
c. low lactate production, even at high activity
d. high lactate production, even at low activity
e. many mitochondria compared to other cells
f. few mitochondria compared to other cells
Answer: A. low levels of glycogen storage, even when blood glucose is high.
C. low lactate production, even at high activity.
Explanation:
Cardiomyocytes are also known as the myocardiocytes. They are the cells which makes up make up the muscle if the heart. They're responsible for the generation of the contractile force in the heart which helps in the pumping if blood.
From the options given, the statements that are likely to be true of cardiomyocytes, but not of skeletal myocytes include:
• low levels of glycogen storage, even when blood glucose is high.
• low lactate production, even at high activity.
Answer:
The answer is A and C
Explanation:
Since cardiomyocytes are dependent on fatty acids, they rely on oxidative phosphorylation to generate energy, and consume very little glucose. Due to this, they have very low need for fermentation of pyruvate. In fact, they can absorb lactate generated by skeletal myocytes and erythrocytes and oxidize it for additional energy. Another consequence of this metabolism is that cardiomyocytes store little or no glucose as glycogen.
Skeletal myocytes and cardiomyocytes both have high energy needs and thus have more mitochondria than other cells. Cardiomyocytes have an even greater density of mitochondria than skeletal myocytes because they depend almost entirely on oxidative phosphorylation for their energy needs. This dependence on oxidative phosphorylation also explains the extraordinary sensitivity of cardiac muscle to hypoxic (low-oxygen) events. Without oxygen, cardiomyocytes rapidly run out of energy and can quickly die.
In the oceans on either side of the Isthmus of Panama are 30 species of snapping shrimp, 15 species on the Pacific side and 15 different species on the Atlantic side. Species live at different water depths. Morphological and genetic data show that Atlantic and Pacific species that live at similar depths are sister species. The sister species on each side of the isthmus cannot interbreed because the water in the canal is fresh water, not salt water, and provides a barrier to reproduction. A sea-level, salt-water canal between the two oceans has been proposed to make transport across the isthmus easier. Which of the following outcomes is the most likely result if such a canal were built?
A. greater percentage of difference in DNA sequences between sister species that inhabit deep water than between sister species that inhabt stalow water
B. greater percentage of difference in DNA sequences between sister species that inhabit shallow water than between sister species that inhabit deep water
C. similar percentages of difference in DNA sequences between all pairs of sister species
D. greater percentage of difference in DNA sequences between Atlantic species than between Pacific species
Answer:
The options of this question are wrong, you can find the correct options by navigating on the web. The options of this question are as follow:
1) The sister species will continue to diverge from each other.
2) None of the sister species will interbreed with each other.
3) The Atlantic and Pacific shrimp will continue to live in their respective oceans and not enter the new canal.
4) Shallow-water species from the two oceans that are sister species would be more likely to interbreed with each other than would be deep-water species.
Answer:
4) Shallow-water species from the two oceans that are sister species would be more likely to interbreed with each other than would be deep-water species
Explanation:
In evolutionary biology, sister species are defined as descendant species formed when one species splits during the course of evolution. Moreover, adaptation refers to the evolutionary process of adjustment of organisms to the environment, which is usually due to natural selection. During the course of evolution, organisms under different environments must change to adapt to their environments. In this case, it is expected that sister species that live in similar environmental conditions (i.e., shallow-water species) exhibit fewer phenotypic differences, being therefore more likely to interbreed with each other.
Which of the following statements is false? Choose one: A. Some Bcl2 family members promote apoptosis, whereas others inhibit apoptosis. B. Bax and Bak are death-promoting members of the Bcl2 family that induce the release of cytochrome c from mitochondria into the cytosol. C. Some death-inhibiting members of the Bcl2 family inhibit apoptosis by blocking cytochrome c release from mitochondria. D. The death-promoting members of the Bcl2 family include Bcl2 itself.
Answer: The statement that is FALSE is D (The death-promoting members of the Bcl2 family include Bcl2 itself.)
Explanation:
Apoptosis is a process found in multicellular organisms whereby cells follows a order of events which eventually leads to their death without releasing harmful substances into the surrounding area of the cells. Through this process, old cells, unnecessary cells, and unhealthy cells are eliminated from the body.
The process of apoptosis can be regulated by cell regulator proteins such as Bcl-2 family.
Bcl-2 stands for B-cell lymphoma 2 which are encoded in human cells by BCL2 gene. The Bcl-2 family are capable of regulating apoptosis by either inducing it or inhibiting it.
Bcl-2 and Bcl-XL, PREVENT apoptosis by preventing the release of mitochondrial apoptogenic factors such as cytochrome c and AIF (apoptosis-inducing factor) into the cytoplasm
Part of the Bcl-2 family is the Bax and Bak and they are the core regulators of the intrinsic pathway of apoptosis. They PROMOTE apoptosis by inducing the release of cytochrome c from mitochondria into the cytosol. Therefore, the death-promoting members of the Bcl2 family does not include Bcl2 itself.
The following types of tissue all are part of the category named muscle tissue:
Skeletal muscle
Smooth muscle
Cardiac muscle
3
Visceral muscle
A. B. and Care all correct.
Answer:
A, B and C are all correct.
Explanation:
There are three main muscle types:
Muscle skeleton — the specialized tissue that connects to the bones and enables movement. The musculoskeletal system is called skeleton muscles and bones together (also known as the locomotor system). Generally speaking, the biceps and triceps of the front or the back of the upper arm are arranged into opposing pairs. Skeletal muscles are controlled by humans, and so are also called voluntary muscles. Another term is streaked muscles, as the tissue looks streaked on a microscope. smooth muscle -located throughout several internal tissues such as the digestive tract, the uterus, and blood vessels such as the artery. The smooth muscle in layered sheets is placed along the construction length in waves. Another typical word is an involuntary muscle, because, without our knowledge, the motion of the smooth muscle takes place.The cardiac muscle - the heart muscle. Without our conscious awareness the heart contracts and relaxes.help me in my hw landslide can be prevented by terrace farming why
Answer:
Reducing the intensity of runoff.
Explanation:
Terrace farming prevent landslides and erosion to occur by reducing the intensity of runoff. This farming also reduce the risk of drought due to raising the moisture content in the soil and allowing water to slowly infiltrate into the soil. Due to terrace farming, the runoff water moves very slowly which causes less erosion of the soil as well as reduces the intensity of water flow which causes more infiltration.
Explain factors that impact North Carolina ecosystems. (Examples: acid rain effects in mountains, beach erosion, urban development in the Piedmont leading to habitat destruction and water runoff, waste lagoons on hog farms, Kudzu as an invasive plant, etc.)
Answer:
Temperature, moisture, topography, elevation, soils, hydrology and disturbance.
Explanation:
Temperature, moisture, topography, elevation, soils, hydrology and disturbance are the factors that impact North Carolina ecosystems. Pollution, urbanization and invasive species inversely affected the ecosystem of North Carolina. Industries produces harmful gases that causes pollution as well as make the rainfall acidic which adversely affected the infrastructure as well as plants and animals. The introduction of invasive species also has a negative impact on the plants and animals of that ecosystem.
You are studying an enzyme that is inactivated by phosphorylation and create a mutant in which the threonine that is normally phosphorylated is replaced with glutamate. Predict the impact of this change on the activity of this enzyme. Group of answer choices
Answer:
always active
Explanation:
Phosphorylation is a posttranslational modification that consists of the addition of phosphate groups to specific amino acids on the protein. Phosphorylation acts as a molecular switch for proteins that are phosphorylated (i.e., in some situations phosphorylation acts to activate protein function, whereas in other situations phosphorylation can inactivate protein function). Phosphorylation modifies the three-dimensional structure of the protein, thereby affecting, for example, the accessibility of the active site of a phosphorylated enzyme to its substrate. Phosphorylation can occur only at the side chains of three amino acids: Serine, Threonine and Tyrosine. In this case, the enzyme is inactivated by phosphorylation on the Threonine residue, so it is expected that the mutant enzyme cannot be phosphorylated, remaining in an active state.
For both clinical and cosmetic reasons, plastic surgeons inject substances into connective tissue underlying the skin epidermis. This plumps up areas deficient in soft tissue, and is used to reduce surgical scars or wrinkles, for example. Which normal connective tissue components are good candidates to be injected as fillers in this type of procedure
Answer:B.) collagen
C.) glycosaminoglycan
Explanation:
The MN blood group in humans is under the control of a pair of co-dominant alleles, M (we will call the frequency of M, p) and N (we will call the frequency of N, q). In a group of 556 individuals, the following numbers of individuals are found for each of the genotypes:
167 MM
280 MN
109 NN
a) What is the frequency of each allele? p= q=
b) What is the value of the Chi-square statistic test to find if in this particular case the genotypic frequencies conform to the Hardy-Weinberg distribution.
Chi square value =
c) What is the probability associated with you chi square statistic calculated above? Please complete the blanks below with the corresponding symbol, < OR > than the critical value, your conclusion with respect to the null hypothesis of Hardy Weinberg equilibrium, Retain or Reject.
P value __ than 0.05
Conclusion ___ the null hypothesis of HW equilibrium
Answer:
a) f(M) = p = 0.55
f(N) = q = 0.45
b) X² = 12.12
c) P₀.₀₅ = 5.991
d) P₀.₀₅ < X²
e) Reject the null hypothesis of HW equilibrium
Explanation:
Due to technical problems, you will find the complete explanation in the attached files
Hypoxia is known to cause a disruption of somitogenesis. If a mouse embryo were temporarily exposed to low levels of oxygen during the final stages of somitogenesis, what would you expect to see in the vertebral column of the newborn mouse
In mammals, hypoxia may cause defects during fetal development. In this case, it would be expected to observe an abnormal development of the caudal (tail) vertebrae
Somitogenesis refers to the process during embryo development in which somites form. These somites are cells that will give rise to structures associated with the vertebrae body plan. Somites form sequentially from the head to the tail, where each new somite forms on the caudal or tail region of the previous one.Somitogenesis represents the first sign of segmentation of the developing vertebrate embryo. During somitogenesis, the unsegmented paraxial or presomitic mesoderm in the trilaminar embryonic stage is segmented in order to form pairs of somites. Moreover, caudal vertebrae refer to the bones that form the tail of vertebrates, which derive from caudal somites.Embryo hypoxia refers to the condition in which the developing embryo does not receive sufficient oxygen (O2) supply. It has been shown that hypoxia during embryo development can increase the incidence of malformations. In this case, embryo hypoxia affects normal caudal somite segmentation, thereby leading to defects in the caudal (tail) vertebrae.Learn more about this topic here:
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8.the following one is considered as safety rule in laboratory
1 point
never eat or drink anything in the science lab
always clean up when you have finished the experiment
wear safety goggles while doing experiments
all the above
If you were rolling dice in a casino and observed results comparable to those observed in Part 2, would you assume the dice were faulty or rigged? Explain your answer using data from your chi-square analysis.
Answer:
Yes, I would guess that the dice are faulty or rigged.
Explanation:
Yes, I would guess that the dice are faulty or rigged.
This is because the chi-square analysis indicates that there is a less than 5% probability that the difference between the expected and observed results were due to pure chance.
The development of DNA technology is bringing profound changes to science, agriculture and healthcare. Provide one example of a DNA technology and provide at least one advantage and one example of a concern or problem associated with its use.
Answer:
The CRISPR-Cas9 genome editing system can be used to edit genes and correct mutations associated with inherited diseases. However, this technology also has the potential to edit genes in germline cells in order to irreversibly modify the human species and the natural evolution of life
Explanation:
The CRISPR-Cas9 (Clustered regularly interspaced short palindromic repeats and CRISPR-associated protein 9) system is a natural prokaryotic defense system used by bacteria to defend against invading DNA. In the laboratory, the CRISPR-Cas9 system has been repurposed to create a versatile genome-editing tool that allows us to modify the genome of mammalian cells in a targeted fashion. The CRISPR-Cas9 is a simple gene-editing tool that consists of a single guide RNA (sgRNA) that guides the Cas9 enzyme to the exact genomic location where Cas9 needs to make a cut, which is then repaired by different DNA repair mechanisms. During DNA repair, nucleotides can be replaced and/or deleted, thereby producing desired genomic modifications. The CRISPR-Cas9 has an enormous potential to repair mutations in genes associated with inherited genetic disorders and cancer (i.e., oncogenes might be reversed in vivo by using this technology). However, the CRISPR-Cas9 genome editing system is also a subject of concern due to its dual use. For example, this technology can be used to modify the genome of germline cells by inducing mutations that can be passed across generations, thereby irreversibly modifying human DNA and altering the normal course of evolution.
why does a rational producer reject the business in increasing stage
Answer:
In three stages of production to which are hold by the law of diminishing marginal returns the second stage is considered to as the rational stage of production as compared to the first and third stage this is due to the fact that this stage it comes out with the best assumptions for efficient and sustainable production, thereby the stage it simultaneously shows the positivity of the average products and marginal product hence, in the increase of labour the total production or outputs increased likewise with the same quantity of fixed factor, therefore this stage it gives direction to the producer on how the higher units of output can be produced.
Unlike the first stage to which it shows each additional variable could enhance more production and such this it signifies an increasing of marginal return but for this case it do not give out the direction in permanent production since labours are needed fewer and in the decrease in number of labour can brings about underproduction.
Also the third stage marginal return is frankly negative because in adding more variable inputs will further led into adding more number of labour hence the increase of labour will contributes to under production as it may be due to labour capacity and efficiency concern.
define structure and function
Answer:
(biology) An arrangement or organization of parts to form an organ, system, or living thing. (2) (ecology) A network or a hierarchy of interrelated parts of a system. ... (4) The formation or mode of construction of a body or system from distinct units.
Name one component of the digestive system that performs mechanical digestion and describe how
it contributes to acquiring nutrients from food.
Answer:
The stomach and the teeth both perform mechanical digestion, which is physically (as opposed to chemically) breaking the food into smaller components. This exposes a larger surface area for chemical digestion and release of nutrients. The teeth are vital to mastication, which breaks large bites of food down into smaller pieces that are easily swallowed. The stomach’s muscle contractions churn the food to expose all particles to the acid and digestive enzymes..
Schistosoma becomes infectious to humans during which of the following stages?
After cercariae migrate to the liver
After the eggs are released in urine
After it is released from the snail
After mating of male and female worms
Before the eggs hatch
Which of the following groups includes a protozoan that is transmitted sexually?
Chromalveolata
Excavata
Amoebazoa
Archaeplasteda
Rhizaria
Schistosoma becomes infectious to humans during which of the following stages?
After cercariae migrate to the liver
After the eggs are released in urine
After it is released from the snail
After mating of male and female worms
Before the eggs hatch
Which part of a mushroom would be eaten?
Thallus
Sporangium
Mycelium
Septum
Hyphae
Answer:
After it is released from the snailExcavata SporangiumExplanation:
Schistosoma life cycle:
1. Human excretes schistosomes eggs through feces and/or urine. These eggs eventually reach the water.
2. Once in the water, schistosomes eggs hatch and release immature larvae named miracides.
3. Miracides swim and penetrate a freshwater snail, which is an intermediary host.
4–5. Once inside the snail, miracides turn into sporocites and then into cercariae. Cercariae characterize by having a bifurcated tail. Snails release the cercariae in water, and they swim until they get in touch with a mammal. Once they reach the mammal, they penetrate the skin and get into the body. This is the stage in which the parasite becomes infectious to humans. Just a few cercariae are enough to infect a person.
6. When cercariae penetrate their host body, they lose their tails and turn into schistosomules. These new forms travel to the liver, where they end their maturation process.
7. Males and females mate in the liver and migrate to veins of the intestine or bladder. Once there, the females lay the eggs, which are then excreted by the host. And the cycle begins again.
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Excavata are asymmetrical protists. The group receives its name because of the appearance of the feeding groove. Excavata is composed of many protists. One of them is Parabasalid.
Parabasalids are parasitic protists, and one of them is Trichomonas vaginalis, which is transmitted sexually.
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Sporangium. This is the reproductive structure of the mushroom and the one that is usually eaten. Spores develop inside the sporangium. Its surface is covered by a protector cuticle. It can be eaten when it is closed or when it is already open. However, when open is tastier.
What are the student’s observations and inferences before he starts his investigation?
Answer:
Hypothesis. A Hypothesis is an estimation of what might happen and the student's observation before moving on to investigate.
Muscle cells that are worked often in exercise will require more mitochondria. If we use more mitochondria, it will require us to use more carbohydrates to build ATP’s. What if we run out of carbohydrates stored in our body? What macromolecule can we break down and use in cellular respiration?
Group of answer choices
A) sucrose
B) lactose
C) glycogen
D) lipids
Answer:
Lipids
Explanation:
Which energy containing molecule is readily used around the cell to power its processes?
A. Glucose
B. ATP
C. ADP
D. Lipids
Answer:
Adenosine 5'-triphosphate, or ATP, is the most abundant energy carrier molecule in cells
ATP is an energy-containing molecule that is readily used around the cell to power its processes. Thus, the correct option is B.
What is ATP?ATP stands for Adenosine triphosphate. ATP may be defined as an organic compound that is utilized as the source of energy for use and storage at the cellular level.
It is the molecule that is present in the cells of all living entities to carry out specific metabolic activities. It apprehends chemical energy aquire from the breakdown of food and liberation it to fuel other cellular processes.
Therefore, ATP is an energy-containing molecule that is readily used around the cell to power its processes. Thus, the correct option is B.
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Importance of antimicrobial emulsion
Answer:
Citral is a typical essential oil used in the food, cosmetic, and drug industries and has shown antimicrobial activity against microorganisms. Citral is unstable and hydrophobic under normal storage conditions, so it can easily lose its bactericide activity. Nanoemulsion technology is an excellent way to hydrophilize, microencapsulate, and protect this compound. In our studies, we used a mixed surfactant to form citral-in-water nanoemulsions, and attempted to optimize the formula for preparing nanoemulsions. Citral-in-water nanoemulsions formed at So 0.4 to 0.6 and ultrasonic power of 18 W for 120 seconds resulted in a droplet size of < 100 nm for nanoemulsions. The observed antimicrobial activities were significantly affected by the formulation of the nanoemulsions. The observed relationship between the formulation and activity can lead to the rational design of nanoemulsion-based delivery systems for essential oils, based on the desired function of antimicrobials in the food, cosmetics, and agrochemical industries.
