Answer:
Explanation:
An office can be referred to a building or rooms or a room which is arranged and set up with furnitures and things needed for commercial use or professional work to achieve a goal.
An office is also refered to position of authority occupied by someone to perform some tasks.
Answer:
here's your answer
Explanation:
The office helps in controlling the activities of different people and department of an organisation. Through controlling it ensures that the various activities of business are performed with much accuracy. Memory Center: Office protects important information of past in a safe manner.
which is more vulnerable to disturbances, a simple food web with only a few species or a more complex one
Answer:
few species
Explanation:
in a complex one im not sure as to how the question measures complexity but a complex one may have more options and more things to adapt to
Select all of the characteristic monosaccharide features that are absent in this molecule: H O H CO о -C-H H-C -O CH H-C- -O H H-C-H H
Answer:
Carboxyl functional group.
Explanation:
Carboxyl functional group is the feature or characteristic which us absent in the monosaccharide. The carboxyl group is an organic, functional group comprising of a carbon atom which is double-bonded to an oxygen atom and have singly bond with a hydroxyl group. In other words we can say that it is a carbonyl group (C=O) that has a hydroxyl group (O-H) which is attached to the carbon atom.
Explain how the results from testing Barley High Lysine (BHL) are related to the primary, secondary and tertiary structure of protein?
Answer:
Enhanced derivatives of barley chymotrypsin inhibitor-2 are known as Barley high lysine (BHL) proteins. BHL has tryptophan, threonine, and isoleucine in a folded structure similar as the tertiary structure of protein.
Derivatives of BHL are digestible in gastric juice and other fluid. Denaturation of these proteins shows proteins in gastric fluid minus pepsin which is looks as the secondary structure of protein. It has four soluble protein groups –
a) Albumins
b) globulin fraction in embryo and scutellary proteins
c) Prolamins
d) Glutelin
In 1985 a biologist counted 750 pine trees in a 250 hectare forest. Using similar counting techniques, the biologist counted 1,250 pine trees in 1990 and 1,500 pines in 1995.
What was the average change of the size of the population from 1985 to 1995?
What was the density of pine trees each year that they were counted?
What was the average change of density from 1985 to 1995?
Answer:
Suppose that we have a given function f(x)
The average rate of change of the function between two values x₁ and x₂ is given by:
[tex]r = \frac{f(x_2) - f(x_1)}{x_2 - x_1}[/tex]
a) We want to find the average (rate) of change on the size of population from 1985 to 1995.
We have that:
f(1985) = 750
f(1995) = 1500
Then we have:
[tex]r = \frac{1500 - 750}{1995 - 1985} = 750/10 = 75[/tex]
This means that the population of trees increases, in average, at a rate of 75 trees per year.
b) What is the density of trees each year that they were counted?
This will be equal to the quotient between the number of trees and the area.
1985: number of trees = 750 pines
area = 250 ha
Then the density is:
D(1985) = (750 pines)/(250 ha) = 3 pines/ha
So 1985, there were 3 pines per hectare.
1990: number of trees = 1250 pines
area = 250 ha
Then the density is:
D(1990) = (1250 pines)/(250 ha) = 5 pines/ha
1995: number of trees = 1500 pines
area = 250 ha
The density is:
D(1995) = (1500 pines)/(250 ha) = 6 pines/ha
3) now we want to get the average change between 1985 and 1995 in the density, this will be:
[tex]r = \frac{D(1995) - D(1885)}{1995 - 1985} = \frac{6 pines/ha - 3pines/ha}{10} = 0.3 pines/ha[/tex]
So, on average, each year the number of pines per hectare increases by 0.3
What must happen before meiosis can begin ?
3. A bacterial isolate from a urine specimen was grown in culture, Gram stained, and then tested for its ability to ferment sugars and hydrolyze various subtrates. What approach to bacterial identification is this an example of
Answer:
Phenotypic approach for bacterial identification
Explanation:
Bacterial identification can be done by conventional methods, which are based on phenotypical characteristics. These methods are much affordable and reasonable.
Phenotypical identification is based on bacteria´s observable characteristics, such as their morphology, development, and biochemical/metabolic properties.
It is important to consider that these methods do not provide absolute certainty. They can only indicate the genera or species to which the bacteria under study may belong.
Some primary evidence is usually used for fast bacteria identification:
Gram staining, morphology, growth at different media or different incubation atmospheres, glucose fermentation, spores production, motion, aerobiosis/anaerobiosis, among others.Knowing that the bacteria in the exposed example was isolated and grown in culture, then Gram-stained and tested for biochemical reaction, we can assume that the approach for its identification is phenotypic.
urine is made of nutrients and water
Answer:
what's the question over here?!
Aspirin may still irritate the stomach as a side effect. What functional group in the molecule is responsible for this effect
Answer:
The functional group that irritates the stomach is known as the salicylic acid molecule.
Explanation:
The energy source for active transport is ________ , while the force driving facilitated diffusion is ________.
Answer:
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Explanation:
Ariella is a 12-year-old whose BMI-for-age is above the 95th percentile. What conclusion could be drawn? a. She is at risk for undernutrition b. Her body fat percentage is in an unhealthy range. c. She is going through a growth spurt. d. She is at risk for future obesity.
Answer:
She is at risk for future obesity
Explanation:
The correct conclusion could be drawn is she is at risk for future obesity. BMI is a body mass index that is derived from the mass and height of the person.
How BMI is calculated?The formula is BMI = kg/m2
Where kg is a person’s weight in kilograms and m² their height in meters squared.
BMI categoriesUnderweight = <18.5Normal weight = 18.5–24.9Overweight = 25–29.9Obesity = BMI of 30 or greaterThus, it is concluded that Ariella is at a risk for future obesity according to her BMI percentile.
Learn more about BMI here:
https://brainly.com/question/2064513
#SPJ2
Which of the following events contributes to the termination of a signal generated by the binding of a ligand to a receptor tyrosine kinase? Only one answer is correct
1. Phosphatases hydrolyze key phosphorylated residues.
2. Kinases hydrolyze key phosphorylated residues.
3. Kinases phosphorylate key residues.
4. The receptor tyrosine kinase dimerizes.
5. GTPase removes a phosphoryl group from GTP.
6. Phosphatases phosphorylate key residues.
Answer:
1. Phosphatases hydrolyze key phosphorylated residues.
Explanation:
Receptor Tyrosine Kinases (RTKs) are high-affinity transmembrane protein receptors that bind to a wide variety of ligands (e.g., growth factors, cytokines, hormones). These receptors (RTKs) have a transmembrane domain and therefore these proteins act as membrane receptors, as well as exhibit catalytic activity. The intracellular C terminal region of RTKs contains catalytic domains responsible for both autophosphorylation and tyrosine phosphorylation of their protein substrates. Moreover, Protein Tyrosine Phosphatase (PTPs) are critical enzymes that remove phosphate groups from tyrosine residues in different substrates (including RTKs), thereby regulating key signaling pathways such as cell proliferation, cell differentiation, and cell-cell adhesion.
Outermost layer of virion fullfills which of the following functions maturation biosynthesis release attachment
4: Åcil fast bacteria are
a. Neisseria
b. Staphylococci
e !vlycobacteria
d. All of the above
。
Answer:
c. Mycobacteria
Explanation:
The mycobacteria is an acid fast bacteria. So, option (c) is the correct answer.
Los mamíferos marinos, que a diferencia de los peces tienen sangre caliente, han perdido el pelaje como adaptación a la vida en el agua. Para compensar la falta de la cubierta de pelo desarrollaron gruesas capas de grasa debajo de la piel que ayudan a evitar la pérdida de calor corporal. ¿Con qué función de los lípidos se relaciona esta adaptación? Justificar.
Answer:
Please Translate it in english, so i can easyly answer it.
Explanation:
thank you.
Mutations that result in the death of the individual are known as
Answer:
Mutations that cause the death of an organism are called lethals — and it doesn't get more negative than that.
what is defferenciated cell and and undeferenciated cell.with example
Answer:
Differentiated cells refer to the specialized cells in multicellular organisms modified to carry out a particular function, such as transporting a certain substance or executing a specific task, while undifferentiated cells refer to the stem cells capable of giving rise to indefinitely more cells of the same type,.
example of differentiated cell=zygote to embryo then a comlex foetusexample of undifferentiated cell= embryonic stem cell.hope it helpsstay safe healthy and happy.The membrane potential that occurs when neurotransmitters bind to their receptors is called _______.
Answer:
action potential
Explanation:
Transmission of a signal within a neuron (from dendrite to axon terminal) is carried by a brief reversal of the resting membrane potential called an action potential. When neurotransmitter molecules bind to receptors located on a neuron's dendrites, ion channels open.
Answer:
I guess action potential is the correct one
_____________ is a post-translational modification that occurs in the endoplasmic reticulum or Golgi apparatus to make proteins functional.
Answer:
Lipidation is a post-translational modification that occurs in the endoplasmic reticulum or Golgi apparatus to make proteins functional.
What are physical and chemical properties
According to a famous article by Max Kleiber, the scaling of the metabolic rate or energy consumption for mammals Pmetab (measured in kcal/day) with the body mass Mb (measured in kilograms) is Pmetab=70M0.75b.
a. What is the specific metabolic rate, i.e. the metabolic rate per kilogram of body mass for the following animals?a mouse whose mass is 30g. Give your answer to three significant figures.
b. What is the specific metabolic rate, i.e. the metabolic rate per kilogram of body mass for the following animals?a cow whose mass is 300kg. Give your answer to three significant figures.
Answer:
Explanation:
Given that:
The scaling of the specific metabolic rate is:
[tex]=\dfrac{P_{metab}}{m}[/tex]
where;
[tex]P_{metab} = 70M_b^{0.75}[/tex]
[tex]=\dfrac{ 70M_b^{0.75}}{m}[/tex]
(a) For a mouse whose mass = 30 g = 0.030 kg
The specific metabolic rate [tex]=\dfrac{ 70*(0.030)^{0.75}}{(0.030)}[/tex]
= 168.197 kg
≅ 168 kg to 3 significant fig.
(a) For a cow whose mass = 300 kg
The specific metabolic rate [tex]=\dfrac{ 70*(300)^{0.75}}{(300)}[/tex]
= 16.8197 kg
≅ 16.8 kg to 3 significant fig.
what is the difference between cell elongation and plasmolysis?
What is the phase that not all cells enter, but it is a phase where cells are not actively dividing
Answer:
The phase that not all cells enter, but is a phase where cells are not actively dividing is called the G0 or non-dividing phase.
Answer:
G₀ phase
Explanation:
Some types of cells divided slowly or not at all. These cells may exit the G₁ phase and enter a resting phase called G₀ phase. In G₀ phase, a cell is not actively preparing to divide. It's simply doing its job. For some cells, G₀ is a permanent state, while others may restart cell division given the right signals.
Even after taking into account all of the variants discovered by this study, there are still likely to be other variants that you have not discovered. The affected and unaffected cohorts are roughly the same size and are drawn from a similar population with similar ancestry. All of the following modifications to the study are likely to allow identification of new variants associated with the disease, EXCEPT ____________.
Question Completion with Options:
a. increasing the size of your cohorts.
b. increasing the diversity of your cohorts.
c. increasing the size of your affected cohort.
d. increasing the diversity of your affected cohort.
Answer:
All of the following modifications to the study are likely to allow identification of new variants associated with the disease, EXCEPT
a. increasing the size of your cohorts.
Explanation:
Option A is chosen because increasing the size of the cohorts will be equally divided between the affected and unaffected cohorts since the study discovered that these two classes are roughly the same size. Therefore, new variants of the disease may not likely be identified and isolated unless the other modification options are followed instead of Option A.
what best describes technology's use in science?
Technology, since the 19th century, has very much helped the development of Scientific theories. It has helped Scientists use telescopes to look into germs or zoom into the stars, it has also helped Chemists determine the weight of, say, a mashed up apple with soda.
In most unsaturated fatty acids found in animal fats, vegetable oils, and biological membranes, the cis isomer predominates. true false
Answer:
The correct answer is - true.
Explanation:
Most unsaturated fats are oils and normally are biological membranes or plant-based such as vegetables. These fats mostly have cis isomer in them. Unsaturated fats help to lower blood cholesterol levels.
However, few oils show saturated fats but these are few in number, the majority are made up of unsaturated fats that have double bonds.
what is the function of bile
Answer:
Bile breaks down fats into fatty acid during digestion.
Answer:
Bile helps with digestion. It breaks down fat into fatty acids. Which can be taken into rhe body by the digestive tract. Bile contains:mostly cholesterol
Absolute zero temperature is not the zero energy temperature.explain
The absolute zero is the temperature at which molecular motion ceases i.e translational kinetic of molecules become zero, but the molecules possess the potential energy, according to the Kinetic theory of gases. Thus, the absolute zero is not the zero energy temperature.
The most common Class II variant is the F508del variant, which is missing the 508th amino acid in the protein but has an otherwise normal sequence. Loss of this amino acid leads to misfolding of the protein, and the misfolded protein is not properly transported to the membrane. Based on this information, how many nucleotides are deleted in the CFTR allele encoding this variant?
Answer:
The correct answer is - 3 (two codons of 507th that are U, C, and the first nucleotide of 508th codon U).
Explanation:
Each amino acid is coded by a three-nucleotide base long codon. So in the case of the F508del variant, there is a deletion of three nucleotides that is encoded for phenylalanine amino acid.
In this isoleucine is encoded by 507th codon that is AUG and 508th codon is UUU codes for Isoleucine.
In this case, two codons of 507th are U, C, and the first nucleotide of 508th codon U. So overall three nucleotides are deleted in the F508del variant..
we should conserve environment give reason
Answer:
for healthy living and long life
Explain the processes that occur in the nucleus.
Answer:
A nuclear reactor is driven by the splitting of atoms, a process called fission, where a particle (a 'neutron') is fired at an atom, which then fissions into two smaller atoms and some additional neutrons. ... The fissioning of atoms in the chain reaction also releases a large amount of energy as heat.
