Psychological and physiological dependence on a drug is called tolerance.
a. True
b. False

Answer:

A. True

Explanation:

Each homologous chromosome pair consists of two homologous chromosomes; one maternal and the other paternal. Prophase-I of meiosis-I includes synapsis (pairing) of homologous chromosomes.

The paired homologous chromosomes are aligned at the equator of the cell during metaphase-I.

During anaphase I, the homologous chromosomes separate from each other due to the dissolution of the synaptonemal complex and move towards opposite poles.

The segregation of homologous chromosomes of a homologous pair is random and creates different combinations of alleles in the daughter cells.

Answer:  

Companion cell is a type of specialized parenchyma cell, which is located in the phloem of the flowering plants.

Each of the companion cell is usually associated with the sieve element. The main function of the companion cell is uncertain but it regulates the activity of sieve tube.

It plays a major role in the loading and unloading of the sugar molecules into the sieve element. As the sieve tubes do not have nucleus and ribosomes so they need companion cell to help in the transportation of sugar molecules.

Answer:

The correct option is: b. Hypothalamus

Explanation:

Hypothalamus is a small structure present below the thalamus of the brain. It is part of the central nervous system, which contains several nuclei having diverse functions.

It is responsible for producing and secreting neurohormones that stimulates or inhibits the secretion of the pituitary hormones.

The hypothalamus is the primary location of sensory receptors and the thermostat of the body. It is responsible for monitoring the internal body temperature and serum osmolarity.

Answer:

The functional unit of a muscle cell is known as a "sarcomere" which covers the distance between the two Z-lines.

The sarcomere consists of A-zone which contains both thick and thin filaments, H-zone which contains only thick filaments and I band which contains only thin filaments.

During muscle contraction, thick and thin myofibrils slide over each other due to which the distance between two z-line get reduced. The H and I zone gets shortened but A zone does not shorten but come close to other A- bands and disappears.

Answer:

It is a learned behavior.

Explanation:

Typically, tool use implies a complex behavior that requires training and observing someone else who already has that particular ability. Thus, this ability shall be learned by one generation from the previous one, which shows and teaches this behavior to the next one.

Answer:

d. (1) acetyl-CoA and oxaloacetate; (2) acetyl-CoA carboxylase

Explanation:

ATP- Citrate lyase (ACLY) is an enzyme present in the cytosol. ATP- Citrate lyase (ACLY) is involved in breaking down the citrate, once it has arrived in the cytosol, into acetyl-CoA and oxaloacetate.

Acetyl-CoA carboxylase is produced as a result of increased levels of citrate in the cytosol. This enzyme plays an important role in the regulation of the synthesis of fatty acids as well as degradation of the fatty acids.  

Answer:

c. connective tissues

Explanation:

Adipose tissue is a connective tissue which is composed of cells called adipocytes. Since it a loose connective tissue located beneath the skin and around internal organs.

The main function of this tissue is to store fat, although it also cushions and insulates the body. It provides protective covering around internal organs. Since it stores fat it also functions as a reserve of nutrients.

Adipose tissue also acts as an endocrine organ. It has the ability to make a number of biologically active compounds that regulate metabolic homeostasis.

Answer:

The protein Spo11 performs an essential function in starting the process of recombination. If yeast exhibits a mutation on spo11- allele, it turns non-functional and it cannot generate the protein spo11 anymore. As a consequence, it cannot go through any recombination process, and the spores generated from it would be inviable because of aneuploidy. Aneuploidy refers to a condition in which the cell comprises an unusual number of chromosomes.

The spo11 yeast strain mutant would likely display impaired meiotic recombination due to the mutation causing the Spo11 protein to be nonfunctional. This could lead to difficulties in sexual reproduction and potential growth defects, similar to those observed in arginine mutants. The phenotype of this yeast may also change in response to environmental signals.

The yeast strain with the Spo11 allele mutated to nonfunctionality may display altered meiosis since the Spo11 protein generally functions to trigger double stranded breaks in DNA, initiating a critical part of meiosis. Loss of this protein may lead to impaired meiotic recombination, potentially affecting the yeast strain's ability to sexually reproduce. Furthermore, considering Beadle and Tatum's experiment involving mutants incapable of producing certain amino acids, the spo11 mutant could have observable growth defects on certain media, similar to the growth defects observed in arginine mutants.

Depending upon the environmental trigger for this allele's expression, the genotype of this strain might not change, but its phenotype could differ significantly depending on the conditions, much like how the phenotype of S. mutans alters based on sugar presence.

However, it's essential to note these effects could vary based on the dominance of the mutant allele and its interaction with other proteins in the cell.

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The correct answer is A. True

Explanation:

The external genitalia or vulva includes different organs and structures such as the inner lips, the mons pubis, the urinary meatus, among others that are external rather than internal as it occurs in the case of the uterus, the cervix or the ovaries. All of these structures are linked mainly to reproduction or the urinary system in the case of the urinary meatus. There, it is true the female external genitalia is collectively called the vulva as all the organs and structures in this zone are commonly known by this name.

The term  refers to the collective external female genitalia, including the labia majora and minora, , and the openings of the urethra and. Therefore, the answer to the question is true.

The term is indeed used in biology and health to refer to the collective external female genitalia. This includes anatomical structures such as the labia majora (outer lips), labia minora (inner lips), and the openings of the urethra . So in answer to the question, the female external genitalia are collectively called the , we can confidently affirm that this is true.

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Answer:

The molecule on the right; the middle carbon is asymmetric

Explanation:

Without additional information on the substituents attached to the carbon atoms, it's not possible to confirm the presence of an asymmetric carbon in the molecule described, which has a carbonyl group on the second carbon of a five-carbon chain. The provided question seems to be about classifying hydrocarbons rather than determining chirality.

To determine which molecule has an asymmetric carbon, we need to examine the structure provided. An asymmetric carbon atom, often referred to as a chiral carbon, is one that has four different groups attached to it. However, based on the information provided, which only details a molecule with a carbonyl group on the second carbon of a five-carbon chain, there's no mention or indication of an asymmetric carbon. Generally, asymmetric carbons are found in molecules with diverse substituents, which is not explicitly stated in the given description.

For a carbon to be asymmetric, it usually cannot be part of a carbonyl group because the carbonyl carbon is double-bonded to an oxygen, leaving only two other groups that can be attached, which is insufficient for chirality (asymmetry). Therefore, we cannot confirm the presence of an asymmetric carbon without additional information on the substituents attached to the carbon atoms.

Lastly, the provided question seems to be about classifying hydrocarbons as aliphatic or aromatic, and as alkanes, alkenes, or alkynes if they are aliphatic. This is unrelated to the presence of asymmetric carbons, which is more relevant to discussions of stereochemistry in organic compounds.

Answer:

The correct answer is option c. Haploid spores.

Explanation:

In the plants, spores are normally unicellular and haploid and produced by the process of meiosis in the sporophytic body of the plant.

These haploid spores undergo the mitotic division and develop a new individual called gametophyte that forms gametes eventually.

Thus, the correct answer is option C. Haploid spores.

In plants, meiosis produces haploid gametes and haploid spores, both of which carry a single complete set of chromosomes.

In plants, haploid gametes and haploid spores are produced by the process of meiosis. Meiosis is a type of cell division that results in four daughter cells each with half the number of chromosomes of the parent cell, as in the production of gametes and plant spores. Haploid cells contain one complete set of chromosomes, whereas diploid cells carry two complete sets of chromosomes. Hence, haploid gametes and haploid spores are the ones produced by meiosis in plants.

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Answer:

The correct answer is option c. "formation of protocells that use DNA to direct the polymerization of amino acids".

Explanation:

The polymerization of amino acids is a complex biological function that involves multiple proteins and cofactors working together at different locations in the cell. Even though there are some scientific advances in the production of synthetic protocells (cell like structures made from synthetic particles), the formation of protocells able to synthesize amino acids from DNA directly has not been accomplished.

Answer:

A dice generally has 6 sides, numbered from 1 - 6. Answer all the following:

(a) There is only one side number 6 out of the 6 sides. The probability would be 1/6.

(b) There are 3 sides that have an even number (2, 4, 6). The probability would be 3/6. Simplify. Divide 3 to both the numerator and denominator:

(3/6)/(3/3) = 1/2

1/2 would be your simplified answer.

(c)There are 2 numbers divisible by 3 (3, 6). The probability would be 2/6. Simplify. Divide 2 from both the numerator and denominator:

(2/6)/(2/2) = 1/3

1/3 would be your simplified answer.

(d) You have 2 dices. You are trying to roll for 2 6's. There are one 6 in each dice. Remember, one dice has the probability of 1/6 to get a 6. The two dices are independent variables, meaning that one dice would not affect the other. Multiply the two fractions together:

(1/6)(1/6) = (1 * 1)/(6 * 6) = 1/36

1/36 would be your answer.

(e) You have 2 dices. One has to roll a odd number, while the other a even number.

Remember, there are 6 sides:

1, 3, 5 are the odd sides (3 odd sides in all)

2, 4, 6 are the even sides (3 even sides in all)

You have a 1/2 chance of rolling an odd number on one dice, and 1/2 a chance to roll an even number on the other. Multiply the two numbers together:

(1/2)(1/2) = (1 * 1)/(2 * 2) = 1/4

1/4 is your answer.

(f) You are trying to roll matching numbers. Take only ONE pair of matching numbers. (for example, 1). In this case, it is the same as (d). Multiply the probability of getting 1 (out of 6 sides) with two dices worth of sides:

(1/6)(1/6) = (1 * 1)/(6 * 6) = 1/36

1/36 would be your answer.

(g) You are rolling for numbers over 4 for both of the dices (5, 6) Remember, there are 6 side in all. The fraction for numbers over 4 on a dice is 2/6, or 1/3. You have 2 dices, so multiply two fractions:

(1/3)(1/3) = (1 * 1)/(3 * 3) = 1/9

1/9 would be your answer.

~

Final answer:

The probabilities of specific outcomes when rolling dice vary, from rolling a single number like a 6 at 1/6 chance, to more complex scenarios like rolling matching numbers on two dice at 1/6. These examples help illustrate basic principles of probability through the lens of a simple, everyday random experiment.

Explanation:

Understanding probabilities with dice throws covers various basic probability concepts. Here's a breakdown of each part of the question:

a) The probability of rolling a 6 on a single die is 1/6, as there is one favorable outcome out of six possible outcomes.

b) An even number (2, 4, or 6) has a 1/2 probability of being rolled because there are three favorable outcomes out of six.

c) The chance of rolling a number divisible by 3 (either a 3 or a 6) is also 1/3, as there are two favorable outcomes out of six.

d) For rolling two 6s with a pair of dice, the probability is 1/36, since each die has a 1/6 chance of landing on 6, and the outcomes are independent.

e) Rolling an even number on one die and an odd number on the other has a probability of 1/2, considering the independent outcomes and that half the numbers on a die are even, and half are odd.

f) The chance of rolling matching numbers on both dice is 1/6, as there are six possible matching outcomes (1-1, 2-2, and so on) out of 36 total outcomes.

g) Rolling two numbers both over 4 (either 5 or 6) has a probability of 1/9, since there are four favorable outcomes (5-5, 5-6, 6-5, 6-6) out of 36 possible outcomes.

These calculations are fundamental to understanding how probabilities work in simple random experiments like dice rolling.

Answer:

None of the above.

Explanation:

t RNA (transfer RNA) contains the anticodon that are complimentary to the codon of the mRNA molecule. These codons codes for the particular amino acid by the process of translation.

The tRNA contains the anti-codon UGA. The UGA anti-codon is a stop codon and does not code for any amino acid. The UGA codon acts as the signal for the termination of the translation process.

Thus, the correct answer is option (d).

Answer:

The most prominent types of membrane transport are 1) Passive 2) Active 3) Endocytosis/exocytosis.

Explanation:

1) Passive Transport-

It can be further classified into-

a. Diffusion- In diffusion small molecules or lipid soluble molecules pass through the membrane of phospholipid from their high concentration to the areas of their lower concentration.

b. Osmosis- In this process water molecules move through a membrane which is selectively permeable, from their higher concentration to their lower concentration.

c. Facilitated diffusion- In this process molecules take help of the protein channels to move across the membrane.

2) Active Transport- In this process the cell uses its energy(ATP) to move the substance across against their concentration gradient.

3) Endocytosis and Exocytosis( For movement of large particles)-

The cells take up the substance from outside by a process called Endocytosis. Endocytosis can be further classified into-

Phagocyytosis- if the substance taken up is solid

Pinocytosis- If the substance taken up is fluid

Exocytosis- the cell excretes by exocytosis

Answer:

The Correct Answer is True.

Explanation:

The United States has passed a law in 1973 to protect its biodiversity

has three major parts:

Final answer:

Biodiversity hotspots are identified by the presence of high numbers of endemic species and significant human-induced disturbance; just counting endemic species is not the sole criteria.

Explanation:

Biodiversity hotspots are specific geographical areas that not only house a high number of species but are also characterized by a large number of endemic species. The statement that scientists always identify biodiversity hotspots by counting the number of endemic species is false. While the number of endemic species is a critical factor, the original criteria for a hotspot also included the condition that 70 percent of the area is disturbed by human activity. Thus, a biodiversity hotspot is identified not merely by the count of endemic species, but also by the level of threat posed to these species due to human disturbance.

By definition, in a biodiversity hotspot, you are most likely to find B. A large number of endemic species. This term refers to species that are native to a specific geographical area and are not found naturally elsewhere.

Merkel discs and tactile corpuscles are found in the hairless skin layers like in lips and finger tips among the body parts.

Explanation:

Merkel cells was first discovered by Friedrich Merkel in 1875 and hence the cells were named after him. They are found in the epidermal layer of the cell.

They are found in the layer of the skin which helps in transmitting tactile signals. There are various types of tactile mechanoreceptors that work together during the process of touch. The Markel cells are encapsulated when touched.

Merkel discs and tactile corpuscles, types of mechanoreceptors found in the skin, are most commonly found in areas with high tactile sensitivity like the fingertips, lips, and genital areas. These areas generally require higher density of these receptors for tasks that demand detailed tactile feedback.

Merkel discs and tactile corpuscles are types of mechanoreceptors found in the skin. These play a crucial role in the sense of touch and help us detect pressure, vibration, and texture.

The areas of the body where you would expect to find the most Merkel discs and tactile corpuscles are those that have a high degree of tactile sensitivity, such as the fingertips, the lips, and the genital areas. These areas require a greater density of Merkel discs and tactile corpuscles to provide the precision required for tasks that demand detailed tactile feedback.

The fingertips, hands, and face (especially the lips) are primary locations for these sensory receptors because our hands and face are often used to explore our environment and require fine tactile discrimination. The genital areas also have a high concentration of these receptors due to their role in sexual response.

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Answer:

This disorder is caused by a recessive allele, its inheritance is sex-linked.

Explanation:

The disorder affects boys born to unaffected parents, this means that at least one of them has to be a carrier of the allele which causes the disease. So, if the allele was dominant, it would express in parents and sons. But, in this case, only is expressed in sons, so it can´t be dominant. Moreover, this condition has a sex-linked inheritance because it is always seen in boys and never in girls. This happens due to boys only have an X chromosome, so if they inherit the recessive allele of the disease, they will express it. On the other hand, girls have two X chromosomes, so if they inherit one copy of the recessive allele, they will be carriers and they won't be affected.  

This disorder is never seen in females because they need to have two copies of the recessive allele. However, to have double copy, they should inherit one copy from their mothers and one copy of their fathers, but boys with the allele are affected and they die in early teens without having progeny.  Therefore, a girl can't have a "carrier-father", so they will never have two copies to express the disorder.

Answer:

D) Pseudomonas

Explanation:

Pseudomonas is the group of gram negative bacteria which are considered opportunistic pathogens. This means that they infect in presence of an already existing ailment like cystic fibrosis. They usually target immuno compromised people.

Normally they are present on soil, water and skin. They are a major cause of hospital acquired infections like sepsis syndrome and ventilator-associated pneumonia. They thrive in moist conditions hence can be found on medical equipment causing cross infections in hospitals. They are resistant to many antibiotics hence the treatment is difficult.

Answer:a. dimerization and phosphorylation.

Explanation:

Phosphorylation as well as dimerization are responsible for the activation of the receptor tyrosine kinase.

The dimer forms when two monomers combine together. This process is known as dimerization.

This process activates the monomers and basically add up the ATP molecule to the tyrosine. Proteins are added up to the phosphorylated tyrosine which causes the structural change in the protein. This leads to the cellular response.

The activation of receptor tyrosine kinases is primarily characterized by dimerization and phosphorylation. This process occurs when a ligand binds to the receptor, causing two receptors to pair up and cross-phosphorylate each other on specific tyrosine residues.

The activation of receptor tyrosine kinases is primarily characterized by dimerization and phosphorylation. Once a ligand binds to the receptor, two receptors pair up or 'dimerize'. The cytoplasmic sides of the receptors come close together and cross-phosphorylate each other on specific tyrosine residues. This acts as a 'flag' to other proteins in the cell, signalling them to interact with the phosphorylated tyrosines of the receptor.

Option B does not apply to receptor tyrosine kinases directly, as IP3 binding is usually seen in other signalling pathways such as G-protein coupled receptors. Options C and D describe other signal cascades that can potentially be triggered by the activation of the receptor but are not part of the direct characterization of receptor tyrosine kinase activation.

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Answer:

H+

Explanation:

Hemoglobin is the major protein of red blood cells. It has many exposed amino groups and carboxylic groups at its surface. These NH3 and COOH groups serve as weak acids and bases respectively and allow hemoglobin to serve as a buffer to maintain the pH of the RBC cytoplasm.

As the exposed amino groups of hemoglobin protein bind to the H+ ions, the free H+ concentration of the cytoplasm of RBC is reduced leading to a buffer action to maintain the pH.

Answer:

Answer is D

Explanation:  Experiments in food industry have demonstrated that Clostridium botulinum spores are highly resistant to radiation, so then Clostridium botilinum in canned food,must be controlled by high temperatures (around 120 Celsius degrees)or acid pH (lower than 4,7) in canned food.

Final answer:

The incorrect statement is that cans of sauerkraut are usually irradiated. Instead, the salt and acidic pH inhibit the growth of C. botulinum, and pressure canning is used to destroy the spores.

Explanation:

The statement regarding the control of Clostridium botulinum in canned sauerkraut that is not correct is:

c. cans of sauerkraut are usually irradiated to destroy any spores or bacteria present in the can

The correct methods for ensuring the safety of canned sauerkraut involve the salt in the sauerkraut inhibiting the growth of C. botulinum, and the acidic pH preventing the botulinum spores from germinating. Furthermore, while pressure canning is a recommended practice as it reaches high temperatures sufficient to destroy C. botulinum spores, irradiation is not a standard practice for sauerkraut. The natural fermentation process and adequate canning procedures both play significant roles in making canned sauerkraut safe from botulism.

Answer:

Apoptosis does not involve:

c. lysis of the cell

Explanation:

Apoptosis is a programmed cell death that occurs under normal physiological conditions and in a controlled manner. Normally seen in cell turnover, embryogenesis, also involved in processes of immune, nervous and endocrine systems.

The main morphological and biochemical changes seen during the apoptosis are the fragmentation of DNA by endonucleases, nuclear, chromatin and cytoplasmatic condensation, apoptotic bodies formation (membrane bound-vesicles form of cell parts) and the phagocytosis (digestion) of those bodies by the scavenger cells.

Apoptosis is regulated by cell- signaling pathways, the caspases, a family of cysteine proteases, are the ones involved in the process.

In the process there is no lysis of the cell as this could lead to a inflammatory response (just happens in necrosis) which would affect contiguous cells, and will involve immune cells. In apoptosis there is just a membrane blebbing, but it does not loss its integrity.

Apoptosis is the programmed cell death process involving DNA fragmentation, cell-signaling pathways, and digestion of cellular contents by scavenger cells. However, it does not involve the lysis of the cell which is associated with necrosis.

Apoptosis, a form of programmed cell death, entails several processes. These include: a. fragmentation of the DNA, which triggers the cell's self-destruct mechanism; b. cell-signaling pathways, which relay the message to commit self-destruction; and d. digestion of cellular contents by scavenger cells, which consume and clean away the debris left behind by the dying cell. However, process c. lysis of the cell is not part of apoptosis. In lysis, the cell burst and spills its contents into the surrounding environment, which is a characteristic of another type of cell-death called necrosis.

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Answer:

Dissolve 111 grams of CaCl2 in 1 Liter of water and it would give 1 M solution.

Explanation:

This is simple, if you want to make 1M solution of any substance just take one mole of that substance and dissolve it in one decimeter cube or one liter of water.

And one mole of a substance equal to the molecular weight of that substance taken in grams.

So you have to dissolve 111 grams of CaCl2 in one liter water and it would make 1 M solution of CaCl2.

To prepare a 1 M solution of a substance, you need to dissolve the molecular weight of the substance (expressed in grams) in enough water to make 1 Liter of solution. Therefore, to make a 1 M solution of CaCl2, dissolve 111 grams of it in less than 1 Liter of water and then add more water until the volume is 1 Liter.

To make 1 Liter of a 1 M (Molar) solution of CaCl2(Calcium Chloride), which has a molecular weight of approximately 111g/mol, you would need to dissolve 111 grams of CaCl2 in enough water to make 1 Liter of solution.

This is because the molarity of a solution is defined as the number of moles of solute per liter of solution. In this case, 1 mole of CaCl2 is 111g, so to have 1 M solution, you need 1 mole (or 111g) of CaCl2 per 1 Liter of solution. Therefore, you should add 111g of CaCl2 to less than 1 Liter of water, then add water until the volume reaches 1 Liter.

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Answer:

Gene mutation:

Gene mutation may be defined as change or alteration in the nucleotide sequence of the gene.  DNA replication errors and mutagens are responsible for the gene mutation. A single gene may affected in this mutation. For example: Sickle cell anemia is caused by the replacement of GAG to GUG results in the formation of valine instead of glutamic acid. The shape of RBCs changed from the biconcave to the sickle cell shape.

Chromosome mutation:

Chromosome mutation may be defined as the change or the alteration of chromosome number or structure. Chromosome mutation occurs due to errors during crossing over. The segment of the chromosome may get changed. Large number of genes may get affected. For example: Down syndrome occurs due to the presence of extra chromosome at chromosome 21.

Answer:

(B) Measure the number of tumors formed on plants, which are exposed to Agrobacterium for different lengths of time.

Explanation:

The problem question should be something like:

how long a plant must be exposed to these bacteria to become infected?

so the experimental design needs to include a time variable. The other answers do not include this variable.

Answer:

A) Cytokine are the different type of the proteins, peptides and glycoproteins released by immune system like interleukin, interferon, chemokine and growth factors. These cells affect other cells by various means. These are cell signalling molecule regulates hematopoiesis and immunity. Example: interlukins.

B) Interleukins are a group of the cytokines that mediate cell communication and help in cell differentiation, cell growth and regulating motility. As they are cytokines and play role in cell signalling these proteins are essential in immune response. Interleukins 1 is an example of interleukins.

C) Chemokine are also protein molecules that are cytokines, acts as chemoattractants as they attracts other cells to the site of the infection or injury so they can destroy or kill pathogens or microbes. Example : CCR1

D) Chemoattractants are the cytokines that lead the migration of the immunity cells to the site of infection or injury. Chemokine is an example of chemoattractant.

Cytokines are signaling molecules like TNF-alpha that influence immune responses. Interleukins, such as IL-2, are a type of cytokine that regulate the immune system. Chemokines, including IL-8, direct immune cell migration, and chemoattractants like C5a attract immune cells to infection sites.

The student has asked for definitions and examples of cytokines, interleukins, chemokines, and chemoattractants:

Answer:

True

Explanation:

The thylakoid membrane of chloroplast has electron carriers embedded in it. During light-dependent reactions of photosynthesis, the electrons move from water molecules to the PSII and then through electron carriers to the PSI and finally to NADP+.

Movement of electrons through the intermediate carrier is accompanied by the pumping of protons from stroma to the thylakoid lumen. This creates an electrochemical gradient along the thylakoid membrane.

The protons are moved back from the thylakoid lumen into the stroma down the concentration gradient through proton channels known as "CFo".

As the protons move down their concentration gradient, the energy is used to phosphorylate the ADP into ATP.

Answer:

they release substances

Explanation:

also known as inflammatory mediators

Answer:

Vasconstriction

Explanation:

Answer:

The correct answer is option d.

Explanation:

FSH or follicle-stimulating hormone refers to a gonadotropin, which is produced and secreted by the anterior pituitary gland's gonadotropic cells. It monitors the growth, development, reproductive procedures, and pubertal maturation of the body.  

In males, FSH instigates Sertoli cells to produce ABPs or androgen-binding proteins, regulated by negative feedback mechanism of inhibin on the anterior pituitary. Mainly, the stimulation of Sertoli cells by FSH maintains spermatogenesis and instigates secretion of inhibin B.  

FSH receptors are located in the sertoli cells in males.

The correct answer is D. FSH receptors are located in the sertoli cells.

In males, FSH (follicle-stimulating hormone) is secreted by the pituitary gland and plays a crucial role in the production of sperm. FSH receptors are primarily located in the sertoli cells, which are found in the seminiferous tubules of the testes. These cells support and nourish developing sperm cells, and FSH binding to its receptors on the sertoli cells stimulates spermatogenesis.

Therefore, option D is the correct answer.

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