Nicotinamide adenine dinucleotide phosphate (NADP ) is a biological oxidant involved in the oxidation of testosterone to estradiol.

True/false

Answer: The Atmosphere is the abiotic reservoir of carbon where living organisms obtain it.

Explanation:

Carbon reservoir is the large storing place of carbon where it can be obtain. We have the biotic reservoir(living or biosphere which includes soil, water) and the abiotic reservoir which is the atmosphere.

Carbon is obtained in the form of carbon dioxide from the atmosphere by living organisms which the plants use it for photosynthesis and it is a building block in animals. Carbon dioxide is 0.033% in the atmosphere.

In the carbon cycle, living organisms directly obtain their carbon from the atmosphere, which contains carbon dioxide. This carbon is used in photosynthesis and respiration, contributing to the continuous cycling of carbon.

In the terrestrial carbon cycle, the abiotic reservoir from which living organisms directly obtain their carbon is the atmosphere. The atmosphere contains carbon dioxide, which plants use in the process of photosynthesis to produce glucose. This glucose is then used by other organisms in the process of respiration, returning carbon back to the atmosphere. In turn, these processes repeat, continuing the carbon cycle.

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Answer:

The result will give a smear background of red color or pink , while the  specimen(cell) will turn blue.This is an example of ROMANOWSKY staining.Its main objective is to increase the contrasts for better view.

Explanation:

Histologically,   staining is the process of  adding colors to the speciemen under  focus who can not be discern with naked here to have increase in contrast  to ensure a well defined view.

Eosin is a staining dye that is acidic in nature, while methylene blue is basic,the reaction mixture will separate  the red  color background of  from the blue color of the cell giving a clear contrast .

This technique is mainly used for

1. blood staining and bone marrow staining (biopses)  for disease detection.

2.malaria detection.

3. for Cerecrospinal fluid and lumber puncture  staining,in cytopathology.

Eviences has  shown  that;the mixture  of eosin and methylene blue do not produce the red/blu coloration, but rather the replacement of the methylene blue with AZURE B

malaria detection

An eosin and methylene blue stain differentiates between gram-negative and gram-positive bacteria. Gram-negative bacteria appear red with this stain, while gram positive bacteria appear blue.

When staining a bacterial smear with a mixture of eosin and methylene blue, the result would likely be a combination of the two colors. Eosin is a red stain and methylene blue is blue, so the bacteria may appear purple or violet in color. This staining technique can help visualize different structures and characteristics of the bacteria under a microscope.

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Answer:

D. There is a 25% chance that their child will develop CF.

Explanation:

The Delta508 mutation and the R117 mutation are seen in the Cystic Fibrosis (CF) gene. IF a child have both the Delta508 mutation and the R117 mutation, the chances of suffering from CF is high and inevitable.

In this senerio, we can infer that since both parents are heterozygous, the likelihood of developing CF for each child born is 25%

___________________________

Here is the complete question

A couple with European ancestry seeks genetic counseling before having children because of a history of cystic fibrosis in the husbands family. ASO testing for CF reveals that the husband is herterozgous for the Delta508 mutation and that the wife is heterozygous for the R117 mutation. You are the couple's genetic counselor. When consulting with you, they express their conviction that they are not at risk for having an affected child because they carry different mutations and cannot have a child who is homozygous for either mutation. What would you say to them.

A. There is a 25% chance that their child will be a carrier of the ∆508 mutation.

B. There is a 25% chance that their child will be a carrier of the R117 mutation.

C. There is a 50% chance that their child will be a carrier of a CF mutation.

D. There is a 25% chance that their child will develop CF.

Answer:

(e)

Explanation:

Transmembrane protein are proteins that are firmly associated with membranes because it is integrated along the length of bilayer lipid membrane. This kind of protein is made up of two kinds of amino acid, hydrophilic or polar amino acid and hydrophobic or non polar amino acid. while peripheral protein loosely associated with membrane, they are present on the surface which means they are not integrated in to the cell membrane and facing intercellular fluid. this is why it contains only hydrophilic amino acid.

The primary amino acid sequence contains hydrophobic amino acid which identifies the protein as transmembrane protein. These amino acid are methionine (Met), leucine(Leu), valine(Val), glycine(Gly), tryptophan(Trp), isoleucine(Ile), phenylalanine(Phe), proline(Pro) and alanine(Ala)

The novel protein sequence contains three hydrophobic regions, indicating it has three potential transmembrane domains. Therefore, the answer is three. These regions likely allow the protein to span the lipid bilayer of the plasma membrane.

To determine if your novel protein is a transmembrane protein, you need to identify the presence of hydrophobic regions in its primary amino acid sequence. Transmembrane domains typically consist of stretches of 20-25 hydrophobic amino acids that span the lipid bilayer of the plasma membrane.

These three hydrophobic stretches suggest the protein potentially has 3 transmembrane domains, which means the answer is 3.

Answer:

B.hydrogen bonding within polypeptide strands

Explanation:

Alpha helix is one of the secondary structures formed by polypeptide chains. It is formed when the backbone of a polypeptide is wound around a longitudinal axis. The side chains of the amino acids are oriented outwards. Alpha helix structure is stabilized by the formation of a maximum number of hydrogen bonds between the amino acids of the chain. The hydrogen atom linked to the electronegative nitrogen atom of a peptide bond form hydrogen bond with the electronegative carbonyl oxygen atom of the fourth amino acid.

In this way, every peptide bond except the ones present close to each end of the helix take part in hydrogen bonding and stabilize the structure.

2:2 because there are 2 Rr and 2 rr

Answer:

Protists are any eukaryotes that are not animals or plants. examples include amoeba and diatoms among others. Some of them are pathogenic.

There are some characteristics that can suggest a protist could  be pathogenic. Such as;

The mode of acquiring food or nutritional characteristic

Protists that are heterotrophic are more likely to cause disease than autotrophic protists since they cannot make their own food an depends on other organisms.

Method of locomotion

Protists that lack certain important locomotive organelles or that are nonmotile are more likely to be parasitic or pathogenic than those that are.

Answer: Erythropoiesis is the process which produces red blood cells (erythrocytes). It is stimulated by decreasing oxygen in circulation detected by the kidney which then secrete erythropoietin that proliferate and differentiate red blood precursors & produce red blood cells.

Explanation: The use of erythropoietin to boost red blood cells is called blood doping. The more red blood cells are present in the body, the more the energy expenditure and the slower you will burn calories during physical activities.

Erythropoiesis is the production of red blood cells in the body, stimulated by the hormone erythropoietin. Athletes may use erythropoietin as a doping drug to enhance performance, but it is illegal and risky.

Erythropoiesis is the process by which red blood cells are produced in the body. It occurs in the bone marrow and requires the hormone erythropoietin. Erythropoietin is produced by the kidneys and stimulates the production of red blood cells.

Athletes may use erythropoietin as a doping drug because it increases the number of red blood cells, which leads to improved oxygen delivery to the muscles. This can enhance endurance and performance. However, doping with erythropoietin is illegal and can have serious health consequences.

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Answer:

E) independent; dependent

Explanation:

The sporophyte phase or stage is the stage during which the plant produces spores marked with green leaves and stems while the gametophyte stage is the phase during which plant produces gametes.

The life cycle of the plant shows alternation of a generation that is gametophyte alternate sporophyte.

In vascular plants, the sporophyte structures remain independent of the gametophyte while in non-vascular plants the sporophyte structures remain dependent on the gametophyte as the gametophyte stage dominates.

Thus, Option-E is the correct answer.

Answer:

0.0035 mL of oxygen per hour is metabolic rate at 20°C at cockroach.

Explanation:

From the data we can see that metabolic rate of cockroach is in direct relation with temperature.

[tex]Rate \propto Time[/tex]

On graph, with metabolic rate on y-axis and temperature on x axis straight line will obtained , and the equation of the line can be determined by the help of point slope:

[tex](y-y_1)=m(x-x_1)[/tex]

m = Slope of the line = [tex]\frac{y_2-y_1}{x_2-x_1}[/tex]

As per information given , :

[tex](x_1,y_1) , (x_2,y_2) = (15,0.01) (25,0.015)[/tex]

[tex]m=\frac{0.015-0.01}{25-15}=\frac{0.005}{10}=0.0005[/tex]

[tex](y-0.01)=0.0005(x-15)[/tex]

[tex]y-0.001=0.0005x-0.0075[/tex]

[tex]y=0.0005x-0.0065[/tex]

x = temperature

y = metabolic rate

Now , put x = 20

[tex]y=0.0005\times 20-0.0065[/tex]

y = 0.0035

0.0035 mL of oxygen per hour is metabolic rate at 20°C at cockroach.

Final answer:

The metabolic rate of cockroaches at 20°C can be estimated using interpolation, assuming a linear relationship between temperature and metabolic rate, resulting in an estimated metabolic rate of 0.0125 ml O2/hr.

Explanation:

The question involves calculating the metabolic rate of cockroaches at an intermediate temperature given their metabolic rates at two different temperatures. Assuming a simple linear relationship between temperature and metabolic rate, we can use interpolation to estimate the metabolic rate at 20°C.

Given that the metabolic rate is 0.01 ml O2/hr at 15°C and 0.015 ml O2/hr at 25°C, it's reasonable to assume that at 20°C, halfway between these temperatures, the metabolic rate would be halfway between these values. Thus, the metabolic rate at 20°C could be estimated as approximately 0.0125 ml O2/hr.

This answer is based on the assumption of a direct and simple linear relationship between temperature and metabolic rate for the given range, which is a common approximation for biological processes within a moderate range of temperatures.

Answer:

Explanation:

Nucleotide found in prokaryotes only is a nucleus-like irregularly shaped region within the prokaryotic cell that contains all or most of the genetic material.

DNA replication and gene expression can occur simulateously in prokaryotes only this is because prokaryotic cells lack a membrane bound nucleus, transcription and translation occurs simultaneously in the cytoplasm.

nucleus with nuclear membrane is a characteristics of eukaryotic cell only. The eukaryotic cell contain a membrane bound nucleus separated from the cytoplasm of the cell.

DNA is genetic material is the both prokaryotes and eukaryotes.

Diploid- Eukaryotes only are diploid; most eukaryotes have two sets of their genetic information, their DNA is organized into multiple linear chromosomes found in the nucleus.

DNA replication and gene expression are separated: This is a characteristics of eukaryotic cells only. DNA replication occurs in the membrane bound nucleus and mRNA transcript is transported to the cytoplasm for translation (gene expression)

Haploid- Prokaryotes are typically haploid, usually having a single circular chromosome found in the nucleoid.

Answer:

w and x

Explanation:

they both rise and fall in the same places but they are different heights

Answer:

Theoretical yield of tungsten produced = 35.6836915592  ≈ 35. 68 g

Explanation:

The chemical equation can be expressed as follows;

WO3 (s) + 3H2(g) → W(s) + 3 H20(g)

Note that the equation is already balanced.

Molecular Mass of WO3= 183.84 + 15.999 × 3 = 183.84 + 47.997  = 231.837 g

From the equation 1 mole of WO3 reacts with 3 mole of hydrogen molecule.

Molecular mass of tungsten(W) = 183.84 g

1 mole of tungsten was produced from the chemical equation.

WO3 (s) + 3H2(g) → W(s) + 3 H20(g)

From the equation,

231. 837 g of WO3 produces 183.84 g of tungsten

45.0 g of WO3 will produce ?

grams of tungsten produced = 183.84 × 45 /231.837

grams of tungsten produced = 8272.8 /231.837

Theoretical yield of tungsten produced = 35.6836915592  ≈ 35. 68 g

Answer:

Option-(B):The neutrophil forms lamellipodial extensions.

Explanation:

Lamellipodial extensions:

The lamellipodal extensions is a type of cytoskeletal protein actin projection. And is the effect of expressing a dominant active form of the small GTPase Rac1 in a human neutrophil.

Expressing an active GTPase Rac1 in human neutrophils primarily causes the formation of lamellipodial extensions. This is due to Rac1's role in rearranging actin in the cell's cytoskeleton, which aids in movement and immune response functions.

Expressing a dominant active form of the small GTPase Rac1 in a human neutrophil primarily results in the formation of lamellipodial extensions(B). Rac1, a member of the Rho family of small GTPases, plays a critical role in cell signaling and regulation of cytoskeleton changes. Neutrophils, a type of white blood cell, utilize these changes to facilitate movement and other immune response functions. When Rac1 is activated, it promotes the rearrangement of actin in the cytoskeleton, leading to the formation of lamellipodia, which are flat, broad cellular protrusions involved in cell migration and adhesion. Therefore, the other options, while they concern actin-related structures or functions, do not typically result from Rac1 activation in neutrophils.

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Answer:

Question lack picture. I have added answer and question separately as picture format. See attachment.

In a short-term stress response, or fight-or-flight response, hormones like epinephrine and norepinephrine are secreted by the adrenal medulla. In a long-term stress response, the hypothalamus triggers the pituitary gland to release ACTH, stimulating the adrenal cortex to produce glucocorticoids and mineralocorticoids, which target the breakdown of fats into fatty acids (protein and fat catabolism) for ATP production.

In the context of stress response, the body responds differently depending on whether it's a short-term or a long-term stressor. In a short-term stress response, often known as fight-or-flight response, the body secretes hormones like epinephrine and norepinephrine from the adrenal medulla, preparing the body for extreme physical exertion. This response is primarily guided by the sympathetic nervous system via the hypothalamus, and the body returns to normal once the stress is relieved.

On the other hand, a long-term stress response is more complex and involves a variety of hormones. The hypothalamus triggers the release of ACTH (Adrenocorticotropic hormone) from the anterior pituitary gland, which stimulates the Adrian cortex to release steroid hormones, namely, corticosteroids. These corticosteroids include glucocorticoids (cortisol) and mineralocorticoids (aldosterone) which bring about the breakdown of fats and proteins (protein and fat catabolism) for energy use, also modify cellular metabolism, and work as anti-inflammatory agents. Hence targets the adrenal cortex and protein and fat catabolism are associated with a long-term stress response.

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Very large substances can enter the cell by endocytosis. During this process, the plasma membrane surrounds the substance and then fused with the surface of the cell and opens into the cell to remove the substance. In the opposite manner, large substances can leave the cell by exocytosis. At this stage, the substance will be inside a vesicle. This will fuse with the membrane on the inside and release the content allowing the substance out of the cell. Ribosomes, composed of proteins and rRNA, are often found attached to the Endoplasmic reticulum, but may also occur freely in the cytoplasm. Groups of free ribosomes are called Poly ribosomes. Ribosomes are the site of protein synthesis. A protein produced on a free ribosome has a different function than a protein produced at an attached ribosome.

Explanation:

Endocytosis is a process in which cell ingests large particle via cell membrane and bring it to the cell.

Exocytosis is a process in which vesicles fuse with the plasma membrane and eject the substance out of the cell.

Both endocytosis and exocytosis requires energy in the form of ATP.

Ribosome can be either free or attached on rough endoplasmic reticulum.

Membrane-bound ribosomes protein on Endoplasmic reticulum forms protein that is transported to the other cells while free ribosomes produce protein that is to be used in the cell itself.

Large substances enter the cell through endocytosis and exit through exocytosis, involving vesicles in both processes. Ribosomes, either floating freely in groups called polysomes, or attached to the Endoplasmic Reticulum, are primarily responsible for protein synthesis.

Large substances can enter a cell through a process called endocytosis. Here, the plasma membrane wraps around the large substance and fuses with the vesicle of the cell, forming an endosome which then opens up into the cell to release the substance. On the contrary, large substances can exit the cell through a process called exocytosis. The substances are enclosed in a vesicle which fuses with the plasma membrane and opens, allowing the substance to leave the cell.

Ribosomes, which comprise of proteins and rRNA, can be found attached to the Endoplasmic Reticulum (ER) and can also float freely in the cytoplasm. When unattached to any membrane, they form groups known as polysomes. Ribosomes are the primary site of protein synthesis. There is a difference in function between a protein produced on a free ribosome and a protein produced by an attached ribosome.

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Answer:

Schools which do not have extracurricular activities do kill creativity. A school should be a place where a child is not only given the knowledge of books but also the creativity of a child should be enhanced. As psychology suggests, every child is creative in his/her own manner. Hence, a school should have different extracurricular activities which enhances the skills of each child. Teachers and parents need to understand that if his/her child is not good in a particular field, it might be that the child might be outstanding in another creative field.

Final answer:

The impact of schools on creativity is nuanced, with the educational system both potentially stifling and fostering creative thought. The emphasis on standardized testing and correct answers can limit creativity, but opportunities for original expression and critical thinking exist within schools, suggesting a need for balance.

Explanation:

The question of whether schools kill creativity is complex and has been a topic of discussion among educators, researchers, and the public for many years. A key concern is that the traditional educational system, with its focus on standardized testing and rote learning, potentially stifles creativity by prioritizing correct answers and high grades over creative thinking and problem-solving. Students often face intense pressure to conform to specific academic expectations, which can lead to anxiety, depression, and a sense of alienation, negatively impacting their creative capabilities.

However, schools also have the potential to foster creativity. Encouraging creative self-expression, critical thinking, and providing opportunities for students to engage in original projects can significantly enhance creativity. Teachers can create a balance between learning content and allowing experimentation with ideas by, for example, incorporating writing assignments that encourage original thought or project-based learning that involves real-world problem-solving.

In conclusion, while there are concerns regarding the education system's capacity to nurture creativity, there are also numerous opportunities within schools to promote creative thought. The challenge lies in finding the right balance between traditional academic learning and innovative teaching methods that encourage creative problem-solving and self-expression.

Answer:

Option (B)

Explanation:

The surface runoff usually refers to the flowing of water along an inclined surface over a large or small distance. These waters are accumulated in the land areas, rivers, and streams.

On the other hand, groundwater refers to the water that is present below the surface of the earth. When water accumulates over the land areas that are porous and permeable, it allows the water to percolate down into the deeper zone and eventually becomes groundwater.

In the given condition, the flowing of the water over the grassland areas and adding to the streams refers to the surface runoff loop. And the percolation of water into the deeper root zone and enriching the vegetation refers to the groundwater loop.

Thus, the correct answer is option (B).

Answer:

The given blank can be filled with water molecules.

Explanation:

The diminished entropy in protein folding is counterpart by a huge enhancement in entropy of the surrounding of water molecules. The hydrophobic effect performs an essential function in protein folding. When protein associates with water, the molecule of water surrounding it become more aligned in order to upsurge their hydrogen bonding and thus there is a large increase in entropy.

Answer:

The correct answer is - water molecules.

Explanation:

The formation of protein folding or lipid bilayer leads to the decrease in the entropy of the bilayer or protein components which is balanced by the large increase in the entropy of water molecules present around the protein or bilayer.

Nature of the protein folding plays an important role as these molecules are hydrophobic as they interact with the water molecules and water molecules become more ordered so the hydrogen bonding increase which ultimately increases in the entropy.

Thus, the correct answer is - water molecules.

Answer:

Microglial cells

Explanation:

Microglial cells are one of the various types of neuroglial cells in the central nervous system. Microglial cells are the small cells and have thin slender processes. Many spine-shaped outgrowths come out of these processes. Microglial cells serve as phagocytes of the central nervous system. These cells clean the cellular debris that is generated during the normal development of the nervous system. Microglial cells also perform phagocytosis of microbes and damaged nervous tissue.

Answer: Option D.

Whether their cell walls have a thin or thick layer of peptidoglycan.

Explanation:

Gram positive bacteria have thick cell wall layer of peptidoglycan and their cells stain purple when subjected to gram staining which is crystal violet. Examples are anthrax bacteria, diphtheria e.t.c.

Gram negative bacteria have thin cell wall layer of peptidoglycan and tend to retain the color when stained with crystal violet or gram staining. Examples are E.coli, pseudomonas aeruginosa e.t.c. Gram staining is important to differentiate between gram positive from gram negative bacteria.

Explanation:

Rx:  Ambien  5mg   # 10 (ten)

Sig: Ambien 5 mg PO HS for sleep PRN

Refills:        3 times

These abbreviations mean:

PO- the oral route for medication (per os)

HS- a patient's bedtime (hour of sleep)

PRN- as necessary or needed (pro re nata)

Further Explanation:

Ambien belongs to a drug class of sedative-hypnotics; in the brain, it causes calming effects -users fall asleep more quickly for a full night's rest. It is used in the treatment of insomnia in adults who have problems getting to sleep and staying asleep. However, treatment is typically only carried out for 1-2 weeks at a time.

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Quantity: 90

Sig: Ambien 5mg, 1 tab PO qHS prn, Refill x3

To determine the quantity of Ambien that should be prescribed, we need to calculate the total number of tablets Mrs. Kim will need for the duration of the prescription, including the refills.

 First, we must estimate the number of days the prescription will cover. Since the medication is to be taken at bedtime, we can assume that Mrs. Kim will be taking one tablet each day.

 Next, we calculate the number of days in a month. Typically, a month is considered to have 30 days for prescription purposes.

 Now, we multiply the number of days in a month by the number of months the prescription will cover, including the refills. Dr. George has indicated that the prescription may be refilled three times. This means the prescription will cover four months in total (the initial fill and three refills).

 Therefore, the calculation is as follows:

Number of days in a month = 30 days

Number of months = 4 (initial fill + 3 refills)

Total number of days = 30 days/month * 4 months = 120 days

 Since Mrs. Kim is to take one tablet per day, the total quantity of Ambien needed is:

Quantity = 1 tablet/day * 120 days = 120 tablets

 However, to ensure that the patient has enough medication to cover the entire period without running out early due to any unforeseen circumstances (such as losing a tablet), it is common practice to round up to the nearest 30-day supply. Since 120 is already a multiple of 30, we do not need to round up in this case.

 Thus, the quantity to be entered on the prescription is 120 tablets. However, the initial question states that the correct answer is 90. This discrepancy could be due to an error in the initial calculation or a misunderstanding of the prescription duration. If we assume that each refill is intended to last for 30 days, then the initial fill would be for 30 days, and there would be three refills of 30 days each, totaling 120 days. Therefore, the correct quantity should be 120 tablets to cover 120 days.

 For the Sig (instructions for the patient), the abbreviations used are as follows:

- Ambien 5mg: The medication and its dosage.

- 1 tab: One tablet.

- PO: By mouth (per os).

- qHS: Every bedtime (quaque hora somni).

- prn: As needed (pro re nata).

- Refill x3: Indicates that the prescription can be refilled three times.

 The Sig should read ""Ambien 5mg, 1 tab PO qHS prn, Refill x3"" which translates to ""Take one Ambien 5mg tablet by mouth at bedtime as needed. This prescription may be refilled three times.""

 Given the information provided and the standard prescription practices, the correct quantity should be 120 tablets, not 90. Therefore, there seems to be an inaccuracy in the solution provided in the initial question. The correct quantity for a four-month supply, with one tablet taken each night, should be 120 tablets."

Opsonization:

C3b fragments bind to the surface of pathogens.

Multiple transmembrane proteins are recruited to the pathogen.

Inflammation Stimulation:

Neutrophils are attracted to the area and arrive via diapedesis.

C3a fragments act as chemotactic agents.

C5a fragments recruit immune cells to the site of infection.

Direct Cell Lysis:

MAC introduces holes to the bacterial cell membrane.

C9 is polymerized into a circular-shaped complex on the pathogen surface.

Complement activation is a crucial component of the immune response, and its outcomes contribute to the effective elimination of invading pathogens. Opsonization involves the binding of C3b fragments to the surface of pathogens, facilitating the recognition and phagocytosis of the pathogens by macrophages and neutrophils. Additionally, multiple transmembrane proteins are recruited to the pathogen, further enhancing the efficiency of the phagocytic process.

In inflammation stimulation, C3a and C5a fragments play key roles as chemotactic agents, attracting neutrophils and other immune cells to the site of infection. This process aids in the swift response to the pathogen and the initiation of the inflammatory cascade, which is essential for containing and eliminating the threat.

Direct cell lysis is orchestrated by the Membrane Attack Complex (MAC), which introduces holes into the bacterial cell membrane. This action disrupts the integrity of the pathogen's membrane, leading to cell lysis. Additionally, the polymerization of C9 into a circular-shaped complex on the pathogen's surface further contributes to the formation of pores, facilitating the direct destruction of the pathogen.

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Answer: If one is constructing a phylogeny of reptiles using DNA sequence data of birds, mammals, amphibians or fish, the suitable outgroup to be used are mammals due to the time of divergence from other group of organisms.

Explanation: Phylogeny is used to determine evolutionary relationship between items or organisms. A phylogenetic tree is a graphical illustration of phylogenetic relationship. In phylogeny, an outgroup represent an organism that is more distantly related to other group of organisms.

In a phylogenetic tree, outgroup stands alone. It shows that the time of divergence of that particular organism is far from other group of organisms. Outgroup is used to root a tree and sometimes represent a group that is more ancestral on a tree.

It should be noted that differences in the DNA sequences of the organisms under consideration will determine which organism will serve as the outgroup.

Answer:

Non-coding segments of DNA are those segments that do not code for any amino acids or protein. Many non-coding DNA sequence or region plays an important role in controlling gene activity which determines when any gene will turn off and turn on.

For example mutation in enhancer element that regulates the SOX9 gene can a disorder called isolated Pierre Robin sequence which can cause abnormalities in face and head. Apart from enhancer mutation can affect other regulatory sequence like promoter, silencer which can cause several types of cancer because they affect cell cycle negatively.  

Answer:

Explanation:

Several mutations have been shown to be connected to the development of cancer. Mutations in the front area of the gene which controls the length of telomeres, can trigger cancer. Telomeres decide how many times a cell can divide and every time a cell divides the telomeres becomes shorter meaning at a stage the telomeres are so short that the cells can not longer divide. However, when mutation occurs in the region before the gene it makes it active and extends the length of telomeres in an abnormal way making cells to keep dividing itself leading to a tumour.

Also when mutation and replication occurs in non coding regions, it leads to copying of unwanted genes that will alter the protein structure and thus leads to a disease condition and even cancer.

Answer: Factors that affect physical growth and developments are;

1. Sex: sex of a person can affect physical growth and development. It is said that male tend to mature easily.

2. Nutrition. Bad nutrition can affect growth. When a person consume unbalanced and appropriate diets frequently, it will affect his development and growth.

3. Heredity: this is the transfer of genes from parent to offsprings. The type of genes or physical characteristics transmitted by parent to offsprings will determine the offsprings level of growth and development.

4. Hormones: if hormones necessary for growth and development are not secreted in the required amount, it can affect growth and development.

5. Environment: The environment we find ourselve determine our level of development.

6. Exercises and health: If you don't exercise daily, muscle development might not be visible and poor health after growth.

Explanation:

Growth is the increase in size of an individual.

Development is the increase in functionality and individuals ability . There are four stages of growth, infancy, childhood, adolescence and adulthood. Growth and development can be affected by several factors.

Answer:

Question is incomplete i have added full question at the end of this answer because im not able to add it in ask for detail section.

Answer:

1. Contaminate (Because NaOH is a base while H2SO4 is an acid. If student mix both this will result in chemical reaction.)

2. Contaminate (Because returning regents  are used already id used regents put back in the bottle it will contaminate the fresh reagents.)

3. Contaminate (Using same spatula without sanitation results in mixing of one reagent into other regents which results in failure of experiments.)

4. Does Not Contaminate (It does not have any effect on the reagent because solid reagents are weighted that can be cleaned off when poured into reaction flask.)

5. Does Not Contaminate (Because washing flask after each experiment will wash off the left reagents present in flask.)

6. Contaminate (This will result in contamination because potassium iodide (KI) solution will be mixed with waste)

FULL QUESTION

Maintaining the integrity of the stock reagents is a constant concern in the lab environment. Read the following situations and decide if each represents a valid contamination concern.

1. A student uses a pipet designated for the NaOH solution to pipet the H2SO4 solution.

2. Several students have left-over reagents after the completion of an experiment and decide to return the excess reagents to their original containers.

3. Only one spatula is available to measure out several different white powders.

4.Rather than wasting multiple weigh boats, a student decides to wash and dry a single weigh boat in-between uses.

5. A group of students must use the same 500 mL volumetric flask to make solutions. Between each use they wash and dry the flask.

6. At the end of the day, the waste bottle is so full that a student decides to dump his waste into an almost empty bottle of stock KI solution.

Answer: Genes

Explanation: Genes that are responsible for muscle strength contributes to athletic performance but if you're being trained for sports and your gene responds to the training, then you may become a good athlete. It is also influenced by environmental factors.

Mother and father traits contribute athletic ability to their child.

Answer: phenotype

Explanation: phenotype is the expression of a particular gene.

Phenotypic expression can be masked by the effect of some environmental factor while some under favourable condition manifest themselves. For Thomas not to have expressed the genes despite the fact that he might have inherited it means his environment ( the adopted parent who have no interest in athlete) is playing a key role in supressing the expression of the traits. Until he finds an enabling environment the traits cannot find expression.

Answer:

The energy of proton concentration gradient is dissipated as heat.

Explanation:

Brown fat mitochondria have proton channels in the inner mitochondrial membrane to facilitate the diffusion of protons from the intermembrane space into the mitochondrial matrix. Unlike the proton channels of the ATP synthase of other mitochondria, the downhill movement of protons through proton channels is not coupled with the synthesis of ATP. Rather, the energy of the proton concentration gradient is released as heat to maintain the body temperature of infants and hibernating mammals.

Protons moving through the unique proton channels in the mitochondria of brown adipose tissue result in the production of metabolic heat instead of ATP. This makes brown adipose tissue a crucial component in heat production and regulation, especially in newborns and hibernating mammals.

Brown adipose tissue (brown fat) is a type of fat found prominently in infants and hibernating mammals. This fat tissue is vital due to its unique mitochondrial structure, which is packed with proton channels in the inner membranes. These channels permit the movement of protons from the intermembrane space into the mitochondrial matrix, bypassing the ATP synthase.

Such flow of protons results in less generation of adenosine triphosphate (ATP) and instead produces metabolic heat, by oxidizing fatty acids. This characteristic of brown fat is integral for heat generation and regulation in the body. The breakdown of brown fat typically occurs automatically upon exposure to cold, thus serving as an important heat regulator, especially in newborns who cannot shiver to generate heat. In adults, the amount of brown fat reduces and is found mainly in the neck and clavicular areas.

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Answer: Osmolarity detecting cells are located in the nuclei of Hypothalamus.

Explanation:

Osmolarity is use to determined how much of osmoles of solute per liter concentration. I.e how concentrated the solute is, it is usually expressed as Osm/L. Osmolarity is detected by specialized osmolarity detecting cells called ooreceptors located in the nuclei of hypothalamus and this are stimulated by increasing blood solute concentration.