What is the function of catalysts?

B. Club moss

Explanation:

This is because club moss is an seedless evergreen plants that have scale-like leaves.

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answer : club moss

explanation: Because they have vascular tissue, seedless vascular plants

are often larger than nonvascular plants. Vascular tissue is spe-

cialized to transport water to all of the cells in a plant.

Answer:

The answer is Internal energy

KE + PE = IE

Explanation:

The sum of potential and kinetic energy is refers to mechanical energy which is expressed by motion  

Answer:C for edge (internal energy)

Explanation:

Explanation:

Water can act as an acid and a base. As an acid, water donates H+, the hydrogen ion. As a base, water donates OH-, the hydroxide ion

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Answer:

The temperature is the key factor for prodigiosin production. It has been shown that , S. marcescens can produce this kind of pigment at about 25 °C, which however could not produce the pigments at elevated temperatures, especially till 37 °C

Explanation:

Answer:

D) ions are transported into and out of the cell by one process and not by the other

Answer:

Without the evaluation of sexual stages, fungi can be classified according to the type of colony they present, the speed of growth, the formation of pigments and the type of coloration.

Explanation:

Although the observation of sexual stages is extremely efficient for the classification of fungi in the laboratory. This type of analysis is not always possible to be carried out. In that case, scientists need to find other methods that allow for the classification of fungi. These methods are carried out with the help of a microscope, where scientists observe the morphology of the fungi and are able to classify them according to the type of colony they present, the speed of growth, the formation of pigments and the type of coloration.

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The answer is D, The U.S Public Health Service Policy. Its main purpose was providing protection for human subjects for research work which was conducted by federal agencies. You can read more about it on HHS.gov, as it is a very interesting regulation.

I hope this helps! :)

Issued in 1974, 45 CFR 46 raised to regulatory status US Public Health Service Policy. Option D

The restrictions known as the Common Rule for the protection of human subjects in research done by or supported by federal agencies are outlined in Title 45 of the Code of Federal restrictions, Part 46. These laws define moral principles and requirements for the protection of research subjects who are being used as human subjects.

The US Public Health Service (PHS) published the policy that would eventually become 45 CFR 46 in 1974. This policy established standards for the examination and approval of research involving human subjects as well as the legal foundation for the protection of those individuals. Informed consent, weighing risks and benefits, and the creation of Institutional Review Boards (IRBs) to regulate research techniques were all adopted.

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Answer:

h² = 0.6

Explanation:

Before answering the question, we need to know a few concepts.

Artificial selection is the selecting practice of a specific group of organisms in a population -that carry the traits of interest- to be the parents of the following generations.

Parental individuals carrying phenotypic values of interest are selected from the whole population. These parents interbreed, and a new generation is produced.

The selection differential, SD, is the difference between the mean value of the trait in the population (X₀) and the mean value of the parents, (Xs). So,  

SD = X₀ - Xs

Heritability in the strict sense, h², is the genetic component measure to which additive genetic variance contributes. The heritability might be used to determine how the population will respond to the selection done, R.  

h² = R/SD

The response to selection (R) refers to the metric value gained from the cross between the selected parents. R can be calculated by multiplying the heritability h², with the selection differential, SD.  

R = h²SD  

R also equals the difference between the new generation phenotypic value (X₁) and the original population phenotypic value (X₀),  

R = X₀ - X₁

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Now that we know these concepts and how to calculate them, we can solve the proposed problem.

Available data:  

According to what we sow previously, we need to find out the value of h².

We know that h² = R/SD, so we need to get R and SD first.  

R = X₁ - X₀

R = 27 - 30

R = -3

SD = Xs -  X₀  

SD = 25 - 30

SD = -5

Knowing this, we can calculate h²

h² = R/SD

h² = -3/-5  

h² = 0.6

Answer:

Explanation:

The cerebrum is divided into two hemispheres or halves and is the biggest portion of the brain. The cerebrum is in charge of voluntary movement, speech, intellect, memory, emotion, and sensory processing, among other things.

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Answer:

I posted some pictures, I hope it helps!

Answer:

Carbohydrates may be defined as energy-rich foods such as sugars and starches. if consumed in excess or not properly metabolized in the body, carbohydrates may lead to obesity and which may also lead a person to severe number of diseases.

Chitins are components of the exoskeletons of foods such as shrimps, crabs and snails, and insects.

Chitin's use as a component in healthy foods is based on its health benefits.

For example it promotes weight loss , prevents obesity, relieving constipation and preventing inflammation associated with refined carbohydrates, cookies and candies

Answer:

28 × 10⁶ colonies/ml

Explanation:

Let C be the concentration of bacterial in the initial stock. When it is diluted by a factor of 10⁶, the new concentration is C' = C/10⁶.

When 2 ml of this concentration is spread on a Petri dish of solidified media, 56 colonies were produced. The number of colonies, n after spreading the 2 ml of C' is C' × 2 ml = 2C' = 2C/10⁶.

So, n = 2C/10⁶.

Since the number of colonies after spreading on a Petri dish of solidified media is 56, n = 56 colonies.

So, 2C/10⁶ = 56

Making C subject of the formula, we have

C = 56 × 10⁶/2

C = 28 × 10⁶ colonies/ml

So, the initial concentration of bacteria is 28 × 10⁶ colonies/ml

Answer:

1. Functions:

a. Umbilical artery >> carries deoxygenated blood from the fetus to the placenta

b. Umbilical vein >> transports oxygenated blood from the placenta to the fetus

c. Ductus venosus >> allows oxygenated blood from the placenta to bypass the liver

d. Ductus arterious >> allows most of the blood from the right ventricle to bypass the fetus's non-functioning lungs

e. Foramen ovale >> oxygenated blood from the umbilical vein to bypass the pulmonary circulation

2. After the bird:

1. Umbilical artery >> medial umbilical ligament

2. Umbilical vein >> round ligament of the liver

3. Ductus venosus  >> ligamentum venosum

4. Ductus arteriosus >> ligamentum arteriosum

5. Foramen ovale >> fossa ovalis

Explanation:

The umbilical artery is a paired artery localized in the abdominal and pelvic regions, which carries deoxygenated blood from the fetus to the placenta through the umbilical cord. The medial umbilical ligament is the obliterated part of the umbilical artery that arises from the internal iliac arteries. In utero, the umbilical arteries carry waste products back to the placenta, whereas the umbilical vein carries oxygenated blood from the placenta to the fetus. The round ligament of the liver (also known as ligamentum teres hepatis) is a remnant of the umbilical vein that  exists in the embryonic stage, it connects the left lobe of the liver to the umbilicus. The ductus venosus is a slender shunt that allows oxygenated blood from the placenta to bypass the liver, it connects the intra-hepatic portion of the umbilical vein to the inferior vena cava. The ligamentum venosum is an extrahepatic, slender, and fibrous remnant of the fetal ductus venosus that travels between the left portal vein and the inferior vena cava. The ductus arteriosus is a fetal artery that connects the aorta to the pulmonary artery. The ligamentum arteriosum is a nonfunctional vestige of the ductus arteriosus, it is attached to the superior surface of the pulmonary trunk. The foramen ovale is an oval-shaped, small, opening in the wall (septum) between the two upper chambers of the heart. The fossa ovalis is a vestige stricture of the foramen ovale of the embryonic heart, which forms a depression in the right atrium of the heart.

Answer:

A- Slippery Slope

Explanation:

A slippery slope fallacy is when someone claims, without proper evidence, that an action will lead to an often catastrophic consequence via a series of events.

Here, the statement claims that if semi-automatics are banned, so will handguns and rifles, however, doesn't provide any evidence that this is the case.

Answer:

d) Send a detailed email

Explanation:

Send a detailed email is the best way to deliver an update to her because in the email he can send all the information in detail form which can satisfy his owner. He can't call or message because it takes too much time to provide information so email is the best way to provide information. Booking a one-hour meeting is not worth it and the reason for this is that there is no big presentation which takes one hour of description one email is enough for it.

Answer:

C. They are relatively rigid structures.

Explanation:

The plasma membrane is a semipermeable lipid bilayer mainly composed of amphipathic phospholipids, which acts as a barrier to separate the cytoplasm from the extracellular environment. This barrier is considered a fluid combination of proteins, lipids and carbohydrates. The fluid mosaic model states that the plasma membrane is a mosaic of lipids (especially phospholipids and cholesterol), as well as proteins and carbohydrates (glycolipids and glycoproteins) that are constantly moving. This fluidity alters the rotation and diffusion of proteins and other components within the plasma membrane, thereby affecting their functions. Membrane fluidity has been experimentally demonstrated by a variety of techniques. (e.g., X-ray diffraction, labeling techniques, calorimetry, etc).

Answer:

The correct answer is - B.often bind their ligands via multiple weak interactions.

Explanation:

Lectins are specific types of proteins that identify and bind to specific carbohydrates present on the cell surfaces. They have an essential role in interactions and communication between various cells for identification and recognition.

Binding sites of lectins on the surface of one cell bind to the Carbohydrates on the surface of another cell. A lectin usually has two or more binding sites for carbohydrate units.

Answer:

Heart failure — sometimes known as congestive heart failure — occurs when the heart muscle doesn't pump blood as well as it should. When this happens, blood often backs up and fluid can build up in the lungs, causing shortness of breath.

Explanation:

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Answer:

D. 1 in a trillion bases

Explanation:

A mutagen agent can change the genetic information of organisms increasing mutations over the natural level. Mutagens cause changes in the bases, and pairing bases, that compose DNI strands.

A mistake in the process of DNI copy during cell division might cause genetic changes in daughter cells. Defects DNI replication might be inherited if it occurs in germinal cells. But it can also cause many significant epigenetic changes.

Many of these changes can be detected on time by enzymes such as DNI polymerase. This enzyme can correct these mistakes or at least some of them, moving from 3´to 5´direction, and eliminating the mistakes.

The highly effective replication system, together with the action of enzymes, makes it rare to occur a mistake in DNI replication. Generally speaking, the mistaken rates in DNI replication are very low, meaning that only one in a trillion times occurs a mistaken DNI copy.

Answer: Bacteria gain resistance to drugs because of mutations (permanent and random changes to their DNA) which means they have changed DNA coding, giving them the ability to resist the drug fighting them off. As a result, they survive and reproduce. Over time, more and more bacteria are generated as the DNA code for resistance is passed on over generations. This results in bacteria having the ability to resist drugs. This is particularly prevalent with antibiotics.

Bacteria develop drug resistance through genetic mutations, acquisition of resistance genes, production of inactivating enzymes, efflux pumps, and biofilm formation.

Bacteria can develop resistance to drugs through several mechanisms. One common way is through genetic mutations or acquisition of resistance genes. Mutations can occur in the bacterial DNA, leading to changes in the target site of the drug, rendering it ineffective. Resistance genes can be transferred between bacteria through horizontal gene transfer, allowing the recipient bacteria to acquire resistance traits.

Another mechanism is the production of enzymes that can inactivate the drug. Bacteria can produce enzymes, such as beta-lactamases, that break down antibiotics like penicillin, preventing them from functioning properly. Bacteria can develop efflux pumps that actively pump out drugs from their cells, reducing their concentration and effectiveness. This mechanism helps bacteria evade the lethal effects of antibiotics. Biofilm formation provides a protective environment for bacteria, making them less susceptible to drugs and immune system attacks.

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The correct question is:

How do bacteria develop resistance to drugs?

Answer:

The correct answer is -

1. c, and d.

2. a, b, and e.

Explanation:

Anaerobic respiration is the short and quick way of producing energy, however, it produces less amount of energy than aerobic respiration and produces lactic acid as a byproduct. It takes place in the cytoplasm of the cell and working or acting muscles depend on such respiration.

Aerobic respiration is the main and important respiration process that takes place in mitochondria and it takes time to produce ATPs that are much more than anaerobic respiration. Resting muscles get energy by this type of respiration.

Answer:

It indicates that core histone genes were present in the last common ancestor of yeasts, plants, and animals

Explanation:

Histones are highly basic proteins that can strongly interact with DNA, which is packaged into nucleosomes, the basic structural and functional unit of chromatin. Each nucleosome is composed of approximately 147 base pairs of DNA wrapped around a core of eight histone proteins (two copies of four types of histones H3, H4, H2A, H2B). These core histones are evolutionarily conserved across eukaryotic kingdoms in terms of sequence and structure. Therefore, DNA-packaging into nucleosomes is considered a constraint for the evolution of core histones. Moreover, the presence of conserved core histones in eukaryotic kingdoms (e.g., yeast, plant, and animal kingdoms) is strong evidence that histone-mediated DNA packaging was presumably present in the last common ancestor of eukaryotic genomes.

Answer:

The potassium equilibrium potential would increase, meaning that more K+ would be leaving the cell.

Explanation:

Let us assume that the only ion transported through the cell membrane is K+. We need to use the Nernst equation to know the destiny of the ion.

Nernst equation:

E = 58 millivolts/z. [Log₁₀ (C-out/C- in)

Where,    

• E = Equilibrium potential

• 58 millivolts/z = Constant

• z = Ion charge + positive or negative symbol

• C-out = Ion concentration out of the cell

• C-In = Ion concentration inside the cell

By convenience, in the Nerts equation, the bigger concentration value corresponds to the numerator and the smaller concentration value to the denominator.

Now let us see the provided values,

• z = Ion charge + positive or negative symbol ⇒ +1 ⇒ K+

• C-out = Ion concentration out of the cell ⇒ 5 mEq/L

• C-In = Ion concentration inside the cell ⇒ 150 mEq/L

E = 58 millivolts/z. [Log₁₀ (Ion in/Ion out)

E = 58 millivolts/+1. [Log₁₀ (150 mEq/L / 5 mEq/L)

E = 58 millivolts (Log₁₀ 30)

E = 58 millivolts (1.477)

E = 85.67 millivolts

85.7 mV is the absolute value of equilibrium potential.

E = 58 millivolts/z. [Log₁₀ (Ion in/Ion out)

E = 58 millivolts/+1. [Log₁₀ (150 mEq/L / 3.5 mEq/L)

E = 58 millivolts (Log₁₀ 42.85)

E = 58 millivolts (1.63)

E = 94.65 millivolts

94.7 mV is the absolute value of equilibrium potential.

If the extracellular concentration of potassium is changed from 5.0 to 3.5 mEq/L, there will be an increase in the membrane potential from 85.7 to 94.7 mV. The increase in the equilibrium potential will result in more potassium diffusing out of the cell, turning the cell interior less positive than before.  

The potassium equilibrium ability might increase, which means that greater K+ might be leaving the cell.

Let us expect that the handiest ion transported via the cell membrane is K+. We want to apply the Nernst equation to recognize the future of the ion.

Nernst equation:

[tex]E = 58 millivolts/z. [Log₁0 (C-out/ -in)[/tex]

Where,

For convenience, withinside the Nerts equation, the larger awareness fee corresponds to the numerator and the smaller awareness fee to the denominator. Now allow us to see the supplied values,

[tex]z=lon charge + effective or terrible ⇒+1 ⇒ K+\\C - out = lon awareness out of the cell ⇒5 mEq/L\\C-ln= lon awareness withinside the cell ⇒150 mEq/LE = 58 millivolts/+ 1.[Log 10 ( 50mEq / L / 5mEq/L)\\E = 58millivolts (Log30)\\E = 58 millivolts (1.477)[/tex][tex]E = 85.67 millivolts\\85.7 mV is absolutely the fee of equilibrium capability.\\E = 58 millivolts/z. [Log10 (lon in/lon out)\\E =58 millivolts/+1. [Log 10 (a 100and 50 mEq/L / 3.five mEq/L)\\E =58 millivolts (Log10 42.85)\\E = 58 millivolts (1.63)\\E = 94.65 millivolts94.7 mV is absolutely the fee of equilibrium capability.[/tex]

If the extracellular attention of potassium is modified from 5.0 to 3.5mEq/L, there can be a growth withinside the membrane capability from 85.7 to 94.7 mV. The growth withinside the equilibrium capability will bring about extra potassium diffusing out of the cell, turning the cell indoors much less high quality than before.

Thus it is clear from this that the potassium equilibrium potential is affected.

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Answer:

Plants can use ammonia as a nitrogen source. After ammonium fixation, the ammonia and ammonium that is formed will be transferred further, during the nitrification process. Aerobic bacteria use oxygen to convert these compounds.

the answer for this would be E

The photic zone has the most nutrients closer to land, an area with sufficient light for photosynthesis, has an abundance of photosynthetic organisms, and is very shallow, Thus, the correct option for this question is E, i.e. all of the following.

The photic zone present in the surface layer of the aquatic ecosystem significantly receives sunlight for photosynthesis. It is the topmost layer that receives sunlight. Hence, this zone is also known as the Sunlight zone.

According to the question, due to the sufficient availability of sunlight, this area performs a good rate of photosynthesis and occupies an abundance of photosynthetic organisms. Phytoplanktons are the characteristic members of this region. As this region has sufficient availability of light, the concentration of nutrients is highly rich.

Therefore,  according to the photic zone, the correct option for this question is E, i.e. all of the following.

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Answer: The xylem distributes water and dissolved minerals upward through the plant, from the roots to the leaves. The phloem carries food downward from the leaves to the roots. Xylem cells constitute the major part of a mature woody stem or root.

Answer:

"scientific investigations."