Between an open air space and inside laboratory, which one has
more fungi and bacteria and why?

Answers

Answer 1

Both an open-air space and an inside laboratory can have varying amounts of fungi and bacteria present. However, an inside laboratory is typically more controlled and sterile, meaning that there are likely fewer fungi and bacteria present.

Comparison between indoor and outdoor environments

It's important to note that the specific microbial communities present in each environment can vary widely depending on a number of factors, and it's difficult to make a general comparison between indoor and outdoor environments without more specific information about the conditions present in each setting.

We can assume that an inside laboratory is typically more controlled and sterile, meaning that there are likely fewer fungi and bacteria present. In contrast, open-air space is exposed to a wider variety of environmental factors, such as wind, rain, and soil, which can introduce and spread different types of fungi and bacteria. Therefore, it is generally assumed that an open-air space has more fungi and bacteria present than an inside laboratory.

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Related Questions

The blood is composed of two portions - a ____ portion made up of the various groups of blood cells and a _____ portion in which the cells are suspended.

Answers

The blood is composed of two portions - a cellular portion made up of the various groups of blood cells and a liquid portion in which the cells are suspended.

The cellular portion is made up of red blood cells, white blood cells, and platelets. The liquid portion, also known as plasma, is made up of water, electrolytes, proteins, and other substances. Together, these two portions make up the blood and allow it to carry out its vital functions within the body.

Blood is a bodily fluid in the circulatory system of humans and other vertebrates that distributes vital chemicals such as nutrition and oxygen to the cells, and moves metabolic waste products away from those same cells.

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Two heterozygous red flowers (white flowers are recessive) are crossed. What are the gentoypes and phenotypes of the offspring?
75% homozygous dominant; 25% heterozygous
50% homozygous dominant; 50% heterozygous
25% homozygous dominant; 25% homozygous recessive; 50% heterozygous

Answers

The correct answer is 25% homozygous dominant; 25% homozygous recessive; 50% heterozygous.

We can use a Punnett square to determine the genotypes and phenotypes of the offspring from this cross. Here is the Punnett square for the cross between two heterozygous red flowers:

|  | R | r |
|---|---|---|
| R | RR | Rr |
| r | Rr | rr |

From this Punnett square, we can see that there are four possible genotypes for the offspring: RR, Rr, Rr, and rr. This means that the genotypes of the offspring are 25% homozygous dominant (RR), 25% homozygous recessive (rr), and 50% heterozygous (Rr).

The phenotypes of the offspring are determined by their genotypes. The homozygous dominant (RR) and heterozygous (Rr) offspring will have red flowers, while the homozygous recessive (rr) offspring will have white flowers. This means that the phenotypes of the offspring are 75% red flowers and 25% white flowers.

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1. Outline and describe the two (2) checkpoints that determine
the fate of B cells during development in
the bone marrow.
2. Draw and describe the process by which self-reactive immature
B cells are p

Answers

1. The two checkpoints that determine the fate of B cells during development in the bone marrow are the pre-B cell receptor checkpoint and the B cell receptor checkpoint.


2. The process by which self-reactive immature B cells are prevented from maturing and potentially causing autoimmune disease is called negative selection.

1. The pre-B cell receptor checkpoint ensures that the B cell has successfully rearranged its heavy chain gene segments to produce a functional pre-B cell receptor. If this checkpoint is passed, the B cell can proceed to the next stage of development.

The B cell receptor checkpoint ensures that the B cell has successfully rearranged its light chain gene segments to produce a functional B cell receptor. If this checkpoint is passed, the B cell can proceed to the next stage of development and eventually leave the bone marrow to mature in the spleen.

2. This process involves the testing of the B cell receptor for self-reactivity.

If the B cell receptor binds to self-antigens with high affinity, the B cell undergoes one of three fates: receptor editing, anergy, or apoptosis. Receptor editing involves the rearrangement of the light chain gene segments to produce a new B cell receptor that is not self-reactive.

Anergy involves the inactivation of the B cell so that it cannot respond to antigen stimulation. Apoptosis involves the programmed cell death of the B cell. Through these mechanisms, negative selection ensures that self-reactive B cells do not mature and cause autoimmune disease.

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TOPIC : CELL PHYSIOLOGY
1. Describe the docking system of the secretory vesicle to the target organelle or plasma membrane.
2. What are the different ways of how substances pass through the cell membrane?

Answers

1. The docking system of the secretory vesicle to the target organelle or plasma membrane is a process known as exocytosis which involves the fusion of the secretory vesicle with the target organelle or plasma membrane, resulting in the release of the vesicle's contents into the extracellular space or into the target organelle.

2. The different ways of how substances pass through the cell membrane are simple diffusion, facilitated diffusion, active transport, endocytosis, and exocytosis.

1. Exocytosis is mediated by a set of proteins known as SNAREs (soluble N-ethylmaleimide-sensitive factor attachment protein receptors) that are found on both the secretory vesicle and the target organelle or plasma membrane. The SNAREs on the secretory vesicle (v-SNAREs) interact with the SNAREs on the target organelle or plasma membrane (t-SNAREs) to form a complex that pulls the two membranes together, leading to their fusion and the release of the vesicle's contents.

2. There are several different ways that substances can pass through the cell membrane, including:

- Simple diffusion: Small, non-polar molecules can pass directly through the lipid bilayer of the cell membrane without the assistance of any proteins.

- Facilitated diffusion: Larger or polar molecules can pass through the cell membrane with the assistance of transport proteins, such as channels or carriers.

- Active transport: Substances can be transported against their concentration gradient (from an area of low concentration to an area of high concentration) with the assistance of transport proteins and the expenditure of energy in the form of ATP.

- Endocytosis: Substances can be taken into the cell by the formation of vesicles from the cell membrane.

- Exocytosis: Substances can be released from the cell by the fusion of vesicles with the cell membrane.

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How does the sympathetic nervous system affect HR? What neurotransmitter is involved? What nerves are involved?

Answers

The sympathetic nervous system increases heart rate (HR) by releasing the hormone adrenaline, which stimulates the heart to beat faster. The neurotransmitter involved in this process is norepinephrine. The nerves involved in this process are the sympathetic nerves, which originate in the spinal cord and travel to the heart.

The sympathetic nervous system is one of the two main divisions of the autonomic nervous system (ANS), with the other being the parasympathetic nervous system (PNS). The ANS controls involuntary functions such as heart rate, breathing, and digestion.

When the body is under stress or in danger, the sympathetic nervous system is activated. This triggers a cascade of physiological responses, including an increase in heart rate, which prepares the body for fight or flight. The neurotransmitter involved in this process is norepinephrine, which is released from sympathetic nerve endings and acts on the heart to increase its rate of contraction.

The sympathetic nerves that regulate heart rate originate in the spinal cord and travel to the heart via the sympathetic chain. These nerves release norepinephrine, which acts on beta-adrenergic receptors in the heart to increase the rate and force of contractions.

In summary, the sympathetic nervous system increases heart rate through the release of norepinephrine from sympathetic nerve endings, which acts on beta-adrenergic receptors in the heart. The sympathetic nerves that regulate heart rate originate in the spinal cord and travel to the heart via the sympathetic chain.

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A number of tests are used to identify a bacterial pathogen taken from human patients. Research and describe a biochemical test that is used to help determine of the genus and species of a bacteria. What bacterial types are usually identified with the use of the test you chose?

Answers

A number of tests are used to identify a bacterial pathogen taken from human patients. The biochemical test that is used to help determine of the genus and species of a bacteria is the API 20E test.

Bacterial types are usually identified with the use of the test the bacteria from Enterobacteriaceae family

A biochemical test that is commonly used to help determine the genus and species of bacteria is the API 20E test. This test is a series of 20 biochemical tests that are used to identify members of the Enterobacteriaceae family, which includes common bacterial pathogens such as Escherichia coli, Salmonella, and Shigella.

The API 20E test is performed by inoculating a bacterial sample onto a series of 20 small test tubes that each contain a different substrate. The bacteria are then incubated for 24-48 hours, and the results are observed. Each test tube contains a different substrate that the bacteria may or may not be able to metabolize, and the pattern of positive and negative results can be used to identify the specific genus and species of the bacteria.

In conclusion, the API 20E test is a biochemical test that is commonly used to help determine genus and species of bacteria. It is particularly useful for identifying members of the Enterobacteriaceae family, which includes common bacterial pathogens such as Escherichia coli, Salmonella, and Shigella.

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(part 1): analyzing your gene of interest, you able able to determine the lengths of the introns, axons, and untranslated regions. assume the poly A tail is exactly 100 bases long 5' UTR: 138 bases, Exon 1: 372 bases, Intron 1: 287 bases, Exon 2: 214 bases, Intron 2: 392 nucleotides, Exon 3: 371 bases, Intron 3: 251 nucleotides, exon 4: 325 bases, Intron 4: 198 bases, exon 5: 297 bases, 3'UTR 108 bases
1: what is the length of the primary transcript?
2: when examining the protein products of this gene, you find 2 distinct protein products. one is 335 amino acids long, the other is 372 amino acids long. what process causes these 2 different proteins and what are the lengths of each mature mRNAs? List EACH component of EACH mRNA, which exon, introns, and modifications will be found in each?

Answers

Q1: The length of the primary transcript is 2853 bases.Q2: The lengths of the two mature mRNAs are 1628 bases and 1925 bases, respectively.

The length of the primary transcript is the sum of the lengths of all the components of the gene, including the 5' UTR, exons, introns, and 3' UTR. Therefore, the length of the primary transcript is:

5' UTR (138 bases) + Exon 1 (372 bases) + Intron 1 (287 bases) + Exon 2 (214 bases) + Intron 2 (392 bases) + Exon 3 (371 bases) + Intron 3 (251 bases) + Exon 4 (325 bases) + Intron 4 (198 bases) + Exon 5 (297 bases) + 3' UTR (108 bases) = 2853 bases

The process that causes the two different protein products is alternative splicing, which is the process of removing introns and joining exons in different combinations to produce different mature mRNAs from the same primary transcript.
The length of each mature mRNA is the sum of the lengths of the components that are included in each mRNA, plus the length of the poly A tail (100 bases). The components of each mRNA are the 5' UTR, the exons that are included, and the 3' UTR.

For the mRNA that produces the protein product that is 335 amino acids long, the length of the mRNA is:
5' UTR (138 bases) + Exon 1 (372 bases) + Exon 2 (214 bases) + Exon 3 (371 bases) + Exon 4 (325 bases) + 3' UTR (108 bases) + Poly A tail (100 bases) = 1628 basesFor the mRNA that produces the protein product that is 372 amino acids long, the length of the mRNA is:
5' UTR (138 bases) + Exon 1 (372 bases) + Exon 2 (214 bases) + Exon 3 (371 bases) + Exon 4 (325 bases) + Exon 5 (297 bases) + 3' UTR (108 bases) + Poly A tail (100 bases) = 1925 bases

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Plants get water from the soil through their________ and it gets
up into the plant through tissue called_________ .Carbon dioxide
gets into the leaves through________ and sunlight is absorbed by
the g

Answers

Plants get water from the soil through their roots and it gets up into the plant through tissue called xylem. Carbon dioxide gets into the leaves through stomata and sunlight is absorbed by the chlorophyll in the leaves. These are all essential processes for the plant to carry out photosynthesis, which is the process of converting sunlight into energy in the form of glucose. The water, carbon dioxide, and sunlight are all used in the chemical reaction that produces glucose and oxygen, which the plant uses for energy and growth.

Plants get water from the soil through their roots and it gets up into the plant through a tissue called the xylem. Carbon dioxide gets into the leaves through stomata and sunlight is absorbed by the chlorophyll in the leaves.

In most lаnd plаnts, wаter enters the roots аnd is trаnsported up to the leаves through speciаlized cells known аs xylem. Plаnts hаve а wаxy cuticle on their leаves to prevent desiccаtion or drying out.

Cаrbon dioxide аnd oxygen cаnnot pаss through the cuticle, but move in аnd out of leаves through openings cаlled stomаtа. Guаrd cells control the opening аnd closing of stomаtа. When stomаtа аre open to аllow gаses to cross the leаf surfаce, the plаnt loses wаter vаpor to the аtmosphere.

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Calculate the constant for an absorbance of 12.7, a concentration of 3.4, and a path length of 7.6. Round to 1 decimal place; if the answer is pure decimal (say, 0.1), type the leading zero in your answer (that is, 0.1 instead of .1).

Answers

Answer: Lower

Explanation:

The Beer-Lambert Law relates the absorbance of a solution to its concentration and path length:

A = εcl

Where:

A is the absorbance

ε is the molar absorptivity (a constant for a particular substance and wavelength)

c is the concentration in mol/L

l is the path length in cm

We can rearrange this equation to solve for ε:

ε = A / (cl)

Plugging in the given values, we get:

ε = 12.7 / (3.4 x 7.6)

ε ≈ 0.496

Rounding to 1 decimal place, the constant is approximately 0.5.

The constant for an absorbance of 12.7, a concentration of 3.4, and a path length of 7.6 is 0.5

To calculate the constant for an absorbance of 12.7, a concentration of 3.4, and a path length of 7.6, we can use the Beer-Lambert Law equation:

                            A = εlc

Where A is the absorbance, ε is the molar absorptivity or constant, l is the path length, and c is the concentration. Rearranging the equation to solve for ε, we get:

ε = A/(lc)

Plugging in the given values:
ε = 12.7 / (3.4 * 7.6)
ε = 0.49
Rounding to 1 decimal place, the constant is 0.5.

Therefore, the answer is 0.5.

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Why is IgM more efficient at activating complement than IgG?

Answers

IgM is more efficient at activating complement than IgG due to its pentameric structure, which allows for the simultaneous binding of multiple antigens.

This leads to a clustering effect that enhances complement activation by bringing complement proteins into close proximity. Additionally, IgM has a higher number of complement-binding sites than IgG, which also contributes to its increased efficiency.

IgG, on the other hand, has a flexible Y-shaped structure that allows it to bind to a wider range of antigens, but it requires higher antigen density for efficient complement activation. Overall, the structural differences between IgM and IgG contribute to their differential abilities to activate the complement system.

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*class is forensics laboratory*
write a laboratory policy for how any case comtaining paper
currency will be handled? the policy should avoid putting undue
suspicion on innocent suspects.

Answers

In a forensic laboratory, it is important to have a clear policy on how to handle any case containing paper currency in order to avoid putting undue suspicion on innocent suspects.

The following laboratory policy can be used to ensure that all cases involving paper currency are handled fairly and accurately:

All paper currency should be handled with gloves to avoid contaminating any potential evidence.The paper currency should be photographed and documented before any further examination is conducted.Any potential evidence on the paper currency, such as fingerprints or DNA, should be collected and analyzed in accordance with standard forensic laboratory procedures.All evidence collected from the paper currency should be securely stored and properly labeled to avoid any potential mix-ups or contamination.The results of any analysis conducted on the paper currency should be thoroughly documented and reported to the appropriate authorities.

By following this laboratory policy, we can ensure that any case involving paper currency is handled in a fair and accurate manner, without putting undue suspicion on innocent suspects.

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The table to the right shows the average temperature each season.Which would be a challenging place to live if you were a farmer? Why?

Answers

Answer:

Region C

Explanation:

A challenging place to live as a farmer would be an area that experiences extreme temperatures or unpredictable weather patterns that can damage crops. Additionally, places with frequent droughts or floods can also be difficult for farmers to sustain their crops. Other factors that can make farming challenging include poor soil quality, limited access to water resources, and high levels of pests and diseases. Ultimately, the challenges faced by farmers can vary depending on the location and local conditions, so it is essential to consider specific factors when evaluating the difficulty of farming in a particular area.

Select the correct answer. The product of two numbers is 21. If the first number is -3, which equation represents this situation and what is the second number? A. The equation that represents this situation is x − 3 = 21. The second number is 24. B. The equation that represents this situation is 3x = 21. The second number is 7. C. The equation that represents this situation is -3x = 21. The second number is -7. D. The equation that represents this situation is -3 + x = 21. The second number is 18.

Answers

In this case, -3x = 21 is the appropriate equation to use. What second number, according to the equation, best describes this circumstance

The correct answer is C

What kind of equation might you use?

An equation is an algebraic statement that proves two formulas are equal in algebra, and this is how it is most commonly used. For instance, the equation 3x + 5 = 14 contains two expressions, 3x + 5 and 14, which are separated by the 'equal' sign.

What in mathematics is meant by an equation?

When two expressions are joined by an equal sign, a mathematical statement is called an equation. An equation is something like 3x - 5 = 16.

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refers to the fluid portion that is separated by centrifugation from the red blood cells, white blood cells and platelets. This has fibrinogen and could be collected using an anticoagulant tube. is called?

Answers

The term that refers to the fluid portion that is separated by centrifugation from the red blood cells, white blood cells, and platelets, has fibrinogen and could be collected using an anticoagulant tube is called plasma.

Plasma contains fibrinogen, which is a protein that helps with blood clotting. Plasma could be collected using an anticoagulant tube. When blood is collected in an anticoagulant tube, the anticoagulant prevents the blood from clotting, allowing the plasma to be separated from the other components of the blood. This plasma can then be used for a variety of medical purposes, including transfusions and the production of medications.

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Would it be possible for the following types of transporters to
create a concentration gradient across a plasma membrane? Explain
your reasoning!!!
a. ATPase-Pump
b. Symporter
c. Antiporter
d. Unipor

Answers

The following types of transporters to create a concentration gradient across a plasma membrane ATPase-Pump, symporter, antiporter, and unipor. Yes it possible to create a concentration gradient.

ATPase-Pump uses ATP energy to move ions across the membrane against their concentration gradient, which creates a concentration gradient across the plasma membrane. ATPase-Pump is also known as the Sodium-Potassium pump. It is a vital pump for maintaining the concentration of sodium ions outside the cell and potassium ions inside the cell. It helps in creating an electrochemical gradient in which the concentration of sodium ions outside the cell is higher than inside and the concentration of potassium ions inside the cell is higher than outside.

Symporter, this transporter moves two different molecules simultaneously across the membrane in the same direction. This creates a concentration gradient across the membrane as the concentration of these molecules differs inside and outside the cell, example - glucose-Na+ symporter in the intestinal cells. Antiporter, this transporter moves two different molecules simultaneously across the membrane in the opposite direction. This creates a concentration gradient across the membrane as the concentration of these molecules differs inside and outside the cell. Example - Na+/H+ antiporter found in the kidney cells.

Unipor, this transporter moves only one type of molecule across the membrane. It works according to the concentration gradient of that molecule. Example - Aquaporins that move water molecules across the membrane in response to the concentration gradient. Hence, ATPase-Pump, symporter, antiporter, and unipor all can create a concentration gradient across the plasma membrane.

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Phosphofructokinase, a controlling enzyme of glycolysis, is inhibited by _____ and activated by _____.
A. ATP;ADP
B. ADP;ATP
C. ADP;NADH
D. NADH;ATP
E. NADH;ADP

Answers

Phosphofructokinase, a controlling enzyme of glycolysis, is inhibited by ATP and activated by ADP. The correct option is A.

Phosphofructokinase, a key regulatory enzyme of glycolysis, is inhibited by ATP and citrate, and activated by ADP and AMP.

ATP and citrate are both indicators of adequate energy reserves in the cell, while ADP and AMP are indicators of low energy reserves. When ATP and citrate levels are high, it signals that the cell has enough energy and there is no need to produce more through glycolysis.

In response, phosphofructokinase is inhibited, slowing down glycolysis. Conversely, when ADP and AMP levels are high, it signals that the cell is low on energy and needs to produce more. In response, phosphofructokinase is activated, increasing the rate of glycolysis to produce more ATP. Thus, the correct option is A. ATP;ADP

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Discuss the etiology, signs and symptoms, diagnostic tests, and
treatment of stroke.

Answers

Stroke is a medical condition caused by the interruption of the blood supply to the brain due to a blocked or ruptured artery.

Signs and symptoms of stroke include sudden numbness or weakness of the face, arms or legs, confusion or difficulty understanding, difficulty speaking or slurred speech, difficulty seeing out of one or both eyes, difficulty walking, dizziness, loss of balance or coordination, and a severe headache with no known cause.

Diagnostic tests for stroke include MRI, CT scan, carotid ultrasound, echocardiogram, and arterial blood gases.

Treatment of stroke usually depends on the type of stroke, severity of the condition, and the underlying cause. Some treatments include medications to break up clots, surgery to open up blocked arteries, rehabilitation to help with recovery, physical therapy, and lifestyle changes to reduce the risk of stroke in the future.

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For Hobbes, the Categorical Imperative is an alternative, secular foundation for morality, based on reason. True False Question 9 2 pts For Bentham, the best guide for one's actions is doing whatever

Answers

The given statement, "For Hobbes, the Categorical Imperative is an alternative, secular foundation for morality, based on reason," is false (F) because the Categorical Imperative is actually a concept developed by Immanuel Kant, not Thomas Hobbes.

The given statement, "For Bentham, the best guide for one's actions is doing whatever," is true (T) because Bentham believe doing whatever maximizes overall happiness or pleasure and minimizes overall pain or suffering.

The Explanation to Each Answer

Thomas Hobbes was a philosopher who believed that morality is based on self-interest and the desire for self-preservation. He argued that individuals enter into a social contract to create a peaceful and stable society, and that the government has the authority to enforce this contract. This view is known as the "social contract theory." However, Hobbes did not use the concept of the Categorical Imperative, which is actually a concept developed by Immanuel Kant. The Categorical Imperative is an alternative foundation for morality that is based on reason and emphasizes the importance of treating others as ends in themselves, rather than as means to an end.

Bentham was a philosopher who believed that the best guide for one's actions is maximizing overall happiness or pleasure and minimizing overall pain or suffering. He developed the principle of utilitarianism which holds that an action is moral if it promotes the greatest happiness for the greatest number of people. Bentham believed that by applying reason to moral decision-making, individuals and society could achieve the greatest good.

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Write a short essay to discuss which type of 2D-NMR can yield this type of result and describe the difference between the two structures while specifically invoking the number of restraints and the resolution (in angstroms).

Answers

Nuclear Magnetic Resonance spectroscopy (NMR) is a powerful tool for elucidating the structure of organic compounds in solution. Two-dimensional NMR techniques such as COSY (correlated spectroscopy) and NOESY (nuclear Overhauser effect spectroscopy) provide additional information beyond one-dimensional spectra, allowing for the identification of proton-proton correlations and distances between nuclei.

In the case of two structures with differing chemical environments, NOESY is the preferred 2D-NMR technique to yield the necessary results. NOESY is capable of detecting correlations between nuclei that are in spatial proximity to each other, allowing for the determination of molecular structure based on the distance and orientation of these nuclei. This information is then used to create a 3D model of the molecule.

The difference between the two structures can be elucidated through the number of restraints and the resolution. The number of restraints in NOESY spectra refers to the number of measured distances between protons, which correspond to the number of NOE (nuclear Overhauser effect) restraints that can be used to build the 3D structure. The greater the number of NOE restraints, the greater the accuracy of the structure. The resolution, measured in angstroms, refers to the precision of the measured distances between protons.

In the case of the two structures, the one with the higher number of NOE restraints and higher resolution will have a more accurate and precise structure. The increased number of NOE restraints allows for a more detailed and accurate depiction of the molecular structure, while the higher resolution allows for more precise measurements of distances between protons. This results in a more defined and accurate representation of the molecule's overall structure.

In summary, NOESY is the preferred 2D-NMR technique to yield results for structures with differing chemical environments. The number of restraints and resolution are key factors in determining the accuracy and precision of the resulting molecular structure. The structure with the higher number of NOE restraints and higher resolution will be more accurate and precise than the other.

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Describe blood flow through the mammalian heart, beginning with return from the systemic circuit, out and back from the pulmonary circuit, then back out to the systemic circuit. Include major vessels, heart chambers, and valves involved. Indicate locations in which the blood is oxygen-rich vs. oxygen-poor.

Answers

Blood flow through the mammalian heart begins with the return of oxygen-poor blood from the systemic circuit through the superior and inferior vena cavae into the right atrium. From there, the blood flows through the tricuspid valve into the right ventricle.

The right ventricle then pumps the blood through the pulmonary valve into the pulmonary artery, which carries the blood to the lungs for oxygenation.

Once the blood is oxygen-rich, it returns to the heart through the pulmonary veins and enters the left atrium. From there, the blood flows through the mitral valve into the left ventricle.

The left ventricle then pumps the oxygen-rich blood through the aortic valve into the aorta, which carries the blood out to the systemic circuit to deliver oxygen to the body's tissues.

In summary, the major vessels involved in blood flow through the mammalian heart are the vena cavae, pulmonary artery, pulmonary veins, and aorta. The heart chambers involved are the right atrium, right ventricle, left atrium, and left ventricle.

The valves involved are the tricuspid valve, pulmonary valve, mitral valve, and aortic valve. The locations in which the blood is oxygen-poor are the vena cavae, right atrium, right ventricle, and pulmonary artery. The locations in which the blood is oxygen-rich are the pulmonary veins, left atrium, left ventricle, and aorta.

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1.maltose
2 fructose
3 icing sugar
4 cornstarch
5 whipping cream
6 gelatin
7 milk
8 vagetable oul
9 mystery solution
10 water
Hot plate, 500 mL beaker, 7 test tubes, Test Tube Holder, Test Tube Rack, Distilled Water, Biuret’s Solution, Iodine, and Benedict’s Solution, Marker, Solutions,
Masking Tape, spot plate, graduated cylinders, droppers, dropper bottles.
Method
Part A- Testing for Mono and Disaccharides
1. Turn an electric plate on high and place a 500 mL beaker half full of water, to make a hot water bath (about 80 degrees Celsius). 2. Measure 3 mL of water and of each of the provided solutions (Not #5,#6 or #7) using a graduated cylinder. Place in clean test tubes and label each tube.
3. Add 15-20 drops of Benedict’s Solution to each test tube (this is about 1mL). 4. Place the test tubes in the hot water bath and note your observation. Use a test tube holder to move the tubes in and out of the bath. Observe for 6 min and record your any colour changes in a chart.
Colour of Benedict’s Reagent
Approximate Sugar Concentration (%)
blue
0
Light green
0.5-1.5
Green to yellow
1.0-2.0
orange
1.5-2.0
Red to red brown
>2.0
Part B –Testing for Starch
Place a drop of distilled water and a drop of iodine in a well on the spot plate.
Fill the wells of the spot plate with a drop of each testing solution (Not #5, #6 or #7). Place one drop of iodine in each solution noting the colour before and after the iodine is added. Iodine turns a blue/purple/black when mixed with a starch.
Part C – Testing for Protein
Measure 2 mL of water into a clean labelled test tube. Repeat this for your solutions (Not for #1 #2,, #3)
Add 2 mL of Biuret reagent to each test tube and tap the test tube to mix the contents. Record any colour changes. Biuret reagent reacts with the peptide bonds that join amino acids together, producing colour changes from blue (indicating no protein) to pink (+), violet (++) and purple (+++). The + sign indicates the relative amounts of the peptide bonds present.
Part D – Testing for Fat
Using a graduated cylinder, measure 3 mL each of #10, #5, #6, #8, #9 into clean labelled test tubes. Clean the graduated cylinder after each pour.
Add 6 drops of Sudan IV indicator to each test tube. Stopper the the test tubes and shake vigorously for 2 mins. Lipids turn Sudan IV from a pink to a red colour. Polar compounds will not cause the the Sudan IV to change colour.
Record the colour of your mixtures on the chart.
Many experiments have controls. What was used as a control? Why is it ideal to have a control? 2. What macromolecule(s) was/were present in the unknown solution? How do you know?

Answers

The control in this experiment was the distilled water. It is ideal to have a control because it provides a baseline for comparison and helps to eliminate any possible external factors that may influence the results. By comparing the results of the control with the results of the other solutions, we can determine if the changes observed in the other solutions are due to the presence of the macromolecules being tested for.

Based on the results of the experiment, the macromolecule(s) present in the unknown solution can be determined by observing the colour changes that occurred when the different reagents were added. If the unknown solution turned a different colour than the control when Benedict's Solution was added, it indicates the presence of mono or disaccharides. If the unknown solution turned a different colour than the control when iodine was added, it indicates the presence of starch. If the unknown solution turned a different colour than the control when Biuret reagent was added, it indicates the presence of protein. If the unknown solution turned a different colour than the control when Sudan IV indicator was added, it indicates the presence of fat. The specific colour changes that occurred can be compared to the colour charts provided in the experiment to determine the approximate concentration of the macromolecule(s) present in the unknown solution.

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Describe the three types of survivorship curves and explain what a
survivorship curves says about a population.

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A survivorship curve is used to illustrate the death rate of a population at various ages or stages in life. The three types of survivorship curves are Type I, Type II, and Type III, which are all described below.

A graph of the number of surviving individuals in a cohort over time is called a survivorship curve.

Types of survivorship curves:

Type I: This is when the death rate is high among older individuals (low death rate when young), indicating that there are few offspring, but that those offspring have a high survival rate. Humans and elephants are examples of this.

Type II: This is when the death rate is consistent across all age groups. Lizards and hydra are examples of this.

Type III: This is when the death rate is high among younger individuals (low death rate when older), indicating that many offspring are produced, but only a few survive. Oysters and insects are examples of this.

What does a survivorship curve say about a population?

Survivorship curves may be used to forecast population growth and predict life expectancy. Survival rates for specific species can be compared using survivorship curves. When a population is booming, it might indicate that birth rates are high and mortality rates are low, while a population that is dwindling might indicate that survival rates are low and death rates are high.

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Assume that the molecular clock ticks at a rate of 5 X 10-9 bp substitutions per bp per year. On a volcanic island you find two species of Drosophila, descended from one species that colonized the island some time after it first rose out of the ocean. You sequence the Adh genes of the two species and find they show 5 synonymous substitutions in 1 kbp.
a. 5x 10-3 b. 2 x 10-2 c. 1.5 x 10-3 d. 10 x 10-3 d. 5.5 x 10-2
What is the frequency of synonymous substitutions between the two species?

Answers

The frequency of synonymous substitutions between the two species is 5 X 10^-3 substitutions per kbp, or answer choice a. 5x 10^-3.

Determine The synonymous substitutions

The frequency of synonymous substitutions between the two species can be calculated by using the molecular clock rate and the number of synonymous substitutions found in the Adh genes.

First, we need to convert the molecular clock rate to substitutions per kbp per year:

5 X 10^-9 bp substitutions per bp per year X 1000 bp per kbp = 5 X 10^-6 substitutions per kbp per year

Next, we can use the number of synonymous substitutions found in the Adh genes (5) and the molecular clock rate (5 X 10^-6 substitutions per kbp per year) to calculate the time since the two species diverged:

5 synonymous substitutions / 5 X 10^-6 substitutions per kbp per year = 1 X 10^6 years

Finally, we can use the time since the two species diverged (1 X 10^6 years) and the molecular clock rate (5 X 10^-6 substitutions per kbp per year) to calculate the frequency of synonymous substitutions between the two species:

1 X 10^6 years X 5 X 10^-6 substitutions per kbp per year = 5 X 10^-3 substitutions per kbp

Therefore, the frequency of synonymous substitutions between the two species is 5 X 10^-3 substitutions per kbp, or answer choice a. 5x 10^-3.

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students examine images of certain species of bat commonly found in texas using the bat dichotomous key they conclude that the bat species is a mexican free tailed bat due to its large round ears based on the dichotomous key in addition to the shape of the bats ears which other set of features should the student look for to confirm the identity of the bat

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Tail length, fur color, and wing shape and size are other features of Mexican free-tailed bats.

What are the features of Mexican free-tailed bats?

Tail length: Mexican free-tailed bats have relatively long tails, which are longer than their body length.

Fur color: These bats have fur that is dark brown or gray-brown on the back and lighter on the belly.

Wing shape and size: Mexican free-tailed bats have long, narrow wings that are pointed at the tip. The wingspan can be up to 12 inches.

By examining these features in addition to the shape of the bat's ears, the students can confirm the identity of the bat species as a Mexican free-tailed bat.

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You have heard someone use the term Diapedesis. What are they referring to?Select one:a. movement of WBC's into tissue from blood vesselsb. Vasoactive mediators causing blood vessel dilationc. Edemad.Production of Interferons

Answers

The term Diapedesis refers to the movement of WBC's (white blood cells) into tissue from blood vessels. So the correct answer is option a. movement of WBC's into tissue from blood vessels.

Diapedesis is an important process that occurs during the inflammatory response, which is the body's response to injury or infection. It allows white blood cells to leave the blood vessels and move into the affected tissue where they can help to fight infection or repair damage. Diapedesis is facilitated by the interaction of adhesion molecules on the surface of white blood cells and the endothelial cells that line the blood vessels.

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1. Briefly describe the principle for protein quantification
methods: Isobaric Tag for Relative and Absolute Quantitation
(iTRAQ)/Tandem Mass Tag (TMT).

Answers

The principle for protein quantification methods like Isobaric Tag for Relative and Absolute Quantitation (iTRAQ)/Tandem Mass Tag (TMT) is based on the labelling of proteins with tags that have a unique mass signature. This enables the identification and quantification of the tagged proteins by mass spectrometry.

What is protein quantification?

Protein quantification is a method of measuring the concentration of a protein in a sample. It is critical in biochemistry and molecular biology studies for determining the effects of drugs, identifying biomarkers, and understanding diseases. The principle for protein quantification methods like iTRAQ/TMT involves the use of isobaric tags that are chemically identical but have different mass signatures. The tags are attached to the amino acid side chain of peptides, resulting in isobaric peptide sets. Following trypsin digestion, these isobaric peptide sets are mixed and analyzed by mass spectrometry.

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What would be the outcome of Ames test if too much histidine
were added to the top agar?

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The outcome of the Ames test if too much histidine were added to the top agar would be that there would be a higher number of revertant colonies appearing on the plate. This would make it difficult to determine if the test substance is mutagenic or not, as the excess histidine could be masking the effects of the test substance.

The Ames test is a bacterial reverse mutation assay that is used to determine if a test substance is mutagenic or not. It is performed by treating bacterial cells with the test substance and then plating them onto an agar medium that lacks histidine. The bacterial cells used in the test are histidine-dependent mutants that cannot grow without histidine in the medium. However, if the test substance is mutagenic, it can cause mutations in the bacterial cells that allow them to revert back to being able to grow without histidine. These revertant colonies are then counted to determine if the test substance is mutagenic or not.

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Explain the benefits and shortcomings of using prostatic specific antigen (PSA) levels to diagnose benign prostatic
hyperplasia (BPH)? Could an elevated PSA level be seen with any other medical conditions

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Prostatic specific antigen (PSA) levels can be used to diagnose benign prostatic hyperplasia (BPH) as it is an important indicator of the level of inflammation in the prostate.

The benefits of using PSA to diagnose BPH are that it can detect subtle increases in the levels of the PSA protein before BPH symptoms appear. It is also a noninvasive test.
The shortcomings of using PSA levels to diagnose BPH include a potential for false positives or false negatives due to PSA levels varying in other medical conditions.

An elevated PSA level can be seen with other medical conditions such as prostatitis, urinary tract infection, prostate cancer, and recent ejaculation. Therefore, PSA levels alone should not be used as a definitive diagnosis for BPH and should be used in conjunction with other tests and evaluations.

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You are a graduate student in behavioral pharmacology, and your lab is conducting a drug discrimination study, an operant procedure in which rats are trained to identify drugs with
stimulus properties similar to those of a training drug. The primary goal of the present study is to test several experimental compounds for their similarity to clozapine, an important treatment for schizophrenia. The compounds to be tested have been sent to your advisor as part of a contract awarded from a drug company. The generalization testing portion of the study is nearing completion, with only one dose-response curve left to obtain. During routine feeding, you notice that 8 of the 10 animals in the study have developed tumor-like growths at the site of injection on the stomach. Additionally, these animals have begun losing weight. Finally, you note that the animals do not exhibit any behaviors suggesting that they are experiencing any discomfort. Concerned, you mention the growths and weight loss to your advisor, who instructs you to continue with generalization testing. He is concerned that having to train a new set of animals in order to test one drug would waste large amounts of research time and resources and may cause problems in interpreting the results. He further states that the animals will be euthanized as soon as the testing phase of the study is completed in less than a month and that the animals will be fine until then. Is your advisor's suggested course of action legally and ethically appropriate? If not, what should be done in this case? What are your obligations in this situation?

Answers

It is not legally or ethically appropriate for your advisor to continue with generalization testing without further investigation and medical care for the animals.

According to the Guide for the Care and Use of Laboratory Animals (8th edition) by the National Research Council (NRC), “it is the responsibility of the investigator to ensure the health and well-being of the animals used in his/her research”. You should immediately inform your advisor and discuss further action, such as consulting a veterinarian and consulting with Institutional Animal Care and Use Committee (IACUC). If the growths and weight loss are not signs of normal aging and instead may be indicative of a medical condition, the animals should be evaluated and treated as soon as possible.

Additionally, since the animal's safety and welfare is at risk, the generalization testing should be halted until the animals are medically cleared. Your obligations in this situation are to prioritize the animals' welfare, which includes providing medical care if necessary and halting the generalization testing if needed.

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Consider the amino acids phenylalanine and tyrosine. a. Draw their complete structures b. In the context of a protein, which of these has a side chain that can be (and IS) readily modified by phosphorylation (i.e. addition of a phosphate group)? c. Sketch a hypothetical titration curve for tyrosine, being sure to include approximate pkas of all ionizable groups and a reasonable guess for the pl. Clearly label axes.

Answers

The complete structures of phenylalanine and tyrosine are shown in the figures below. Tyrosine has a side chain that can be (and is) readily modified by phosphorylation. This is because it has a hydroxyl (-OH) group on its side chain that can be modified by the addition of a phosphate group.

Tyrosine titration curve

The hypothetical titration curve for tyrosine is shown in the figure below:  
 
The approximate pKas of all ionizable groups are:
- COOH group: 2.2
- NH3+ group: 9.2
- OH group: 10.5

The approximate pI (isoelectric point) of tyrosine is 5.7.

The axes are clearly labeled, with pH on the y-axis and the volume of titrant added on the x-axis

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