In the caterpillar, molting and metamorphosis are under the control of the same hormone. What is the hormone? How are these separate developmental pathways controlled by the same hormone?
Answer:
The correct answer will be- 20-hydroxyecdysone
Explanation:
Ecdysone is a steroid hormone produced in the Arthropods which plays an important role in coordinating and regulating the different developmental transitions like larval moulting and the metamorphosis.
The other effector hormone also plays an important role in molting and metamorphosis called Juvenile hormone.
The presence of juvenile hormone prevents the ecdysone induce changes which regulate molting and gene expression during metamorphosis.
Therefore the presence of the juvenile hormone leads to the different stages of molting while the absence of juvenile hormone promotes the development of metamorphic stages like the development of pupa.
Thus, 20-hydroxyecdysone is the correct answer.
When examining the two complementary strands of nucleotides in one DNA molecule, it would be expected that the sum of ________ in one strand would be ________ the sum of the ________ in the other strand.
Answer: Purine; is equal to; pyrimidine
Explanation:
When a segment of DNA is examined then the nucleotide in one DNA molecule, it will be expected that the sum of purine in one strand will be equal to the sum of the pyrimidine in the other strand.
Adenine combines with Thymine and Cytosine combines with Guanine in 1:1 ratio.
More specifically the ratio of purine is to pyramidine is 1:1 in an organism. This pattern is seen in both the strands of the DNA.
Final answer:
When examining the two complementary strands of nucleotides in one DNA molecule, it would be expected that the sum of adenine (A) in one strand would be equal to the sum of the thymine (T) in the other strand, and vice versa.
Explanation:
Similarly, the sum of guanine (G) in one strand would be equal to the sum of the cytosine (C) in the complementary strand. This relationship is due to the rules of base pairing, where adenine always pairs with thymine, and guanine pairs with cytosine.
These rules are fundamental to the structure of DNA and its replication process, ensuring that each new molecule is an accurate copy of the original.
Zinc, an essential trace element for most organisms, is present in the active site of the enzyme carboxypeptidase. The zinc most likely functions as a(n)____________.
A) competitive inhibitor of the enzyme.
B) noncompetitive inhibitor of the enzyme.
C) allosteric activator of the enzyme.
D) cofactor necessary for enzyme activity.
E) coenzyme derived from a vitamin.
Answer:
D) cofactor necessary for enzyme activity.
Explanation:
A cofactor is an inorganic ion that is required as an additional chemical component by an enzyme. The ions such as Fe2+, Mg2+, or Zn2+ mostly serve as cofactors for one or more enzymes. The enzymes such as Cytochrome oxidase, catalase, and peroxidase require Fe2+ or Fe3+ ions as a cofactor. Some cofactors are very tightly bound to the active site of their enzymes and their removal renders them enzymes inactive. According to the given information, zinc is present in the active site of the enzyme carboxypeptidase which means that it is a cofactor for this enzyme.
Colin owns a dog food business, but wants to develop a canine fashion line as a way of expanding his portfolio and increasing sales and profits. One of the things customers love about Colin’s company is that all of the products are manufactured locally, creating jobs for young people. Though he does not have a lot of money and hopes that he won’t have to hire new staff, he thinks he might have to. Rather than developing a canine fashion line, which of the following alternatives can you suggest to Colin?a. Colin should use cannibalization strategy.b. Colin should use a market penetration strategy.c. Colin should outsource his company’s jobs to cut costs.d. Colin should consider selling his business to a larger pet food company.e. Colin should hire a marketing company to help him promote his products.
Answer:
I think it is B. Market Penetration strategy.
Explanation:
I'm very sorry if this is wrong but studying the two possible answer (A or B)
B seemed the most probable due to the definition of the Market Penetration Strategy. Which is essentially "To focus on current products and current markets in order to increase market share". (I hope I was able to help! :)
In the given scenario Colin should use a market penetration strategy. The correct option is b.
What is market penetration strategy?When a business uses existing products in existing markets, it is using a market penetration strategy in an effort to increase its market share.
This is how a business that currently has a product on the market can expand by boosting sales among existing customers.
Using market penetration techniques, a brand can introduce its current goods or services to a booming, in-demand market.
He fears he might need to hire more employees because he doesn't have much money and is hoping he won't.
The solution that can be advised to Colin instead of creating a dog fashion line so Colin should employ a market penetration strategy.
Thus, the correct option is b.
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Given the following cross between parents of the genotypes AABBCc × AabbCc, assume simple dominance for each trait and independent assortment. What proportion of the progeny will be expected to phenotypically resemble the first parent?
Answer:
3/4
Explanation:
The first parent with the "AABBCc" genotype would express the dominant trait for all the three loci since the mentioned loci exhibit simple dominance for each trait and independent assortment. Therefore, the desired progeny should also exhibit the dominant trait for all three loci.
Parents= AABBCc × AabbCc
AA x Aa= 1/2 AA (dominant): 1/2 Aa (dominant) = 100% dominant progeny
BB x bb = Bb (dominant) = 100% dominant progeny
Cc x Cc = 1/4 CC (dominant): 1/2 Cc (dominant): 1/4 cc (recessive) = 3/4 dominant : 1/4 recessive progeny
Therefore, the proportion of the progeny showing the dominant trait for all the three loci= 1 x 1 x 3/4 = 3/4
Final answer:
The proportion of progeny from the cross AABBCc × AabbCc that will phenotypically resemble the first parent is 3/4, considering the dominance of traits and independent assortment.
Explanation:
The question involves the cross between two parents with genotypes AABBCc and AabbCc, with simple dominance for each trait and independent assortment. To determine the proportion of progeny that would phenotypically resemble the first parent, we consider each gene independently due to the principle of independent assortment.
For gene A, the first parent is AA and the second parent is Aa. Since A is dominant, all offspring will have the dominant phenotype (A), just like the first parent.
For gene B, the first parent is BB and the second parent is bb. Since B is dominant, all offspring will be Bb, displaying the dominant phenotype as the first parent.
For gene C, the first parent is Cc and the second parent is Cc. This gene will produce offspring with a 3:1 phenotypic ratio in favor of the dominant C phenotype, according to Mendel's principle of dominance.
Because the offspring must display the dominant phenotype for all three traits to resemble the first parent phenotypically, we calculate the probability as a product of the probabilities for each gene, which is 1 (for A) × 1 (for B) × 3/4 (for C), giving a final proportion of 3/4.
During a stressful interval, ________. During a stressful interval, ________. the posterior pituitary gland secretes more growth hormones adrenocorticotropic hormone (ACTH) stimulates the adrenal cortex, and neurons of the sympathetic nervous system stimulate the adrenal medulla thyroid-stimulating hormone (TSH) stimulates the adrenal cortex and medulla to secrete acetylcholine the alpha cells of islets secrete insulin and simultaneously the beta cells of the islets secrete glucagon the calcium levels in the blood are increased due to actions of two antagonistic hormones, epinephrine and norepinephrine
Answer:
Adrenocorticotropic hormone (ACTH) stimulates the adrenal cortex, and neurons of the sympathetic nervous system stimulate the adrenal medulla.
Explanation:
Hormones are the chemical messengers of the body and maintain the body's homeostasis in different adverse conditions. The hormones work by the feedback regulation system.
To overcome from the stress condition the body releases the hormone known as epinephrine. The receptors sense the stressful condition and adrenal cortex is stimulated by adrenocorticotropic hormone that results in the release of adrenaline hormone. The hormone activates the sympathetic nervous system of the body to cope with the fight and flight conditions.
Thus, the correct answer is option (2).
What specialized cells do cnidarians use for protection
Answer:
Cnidarians have special cells called cnidocytes, or commonly known as stinging cells, which contain organelles called nematocysts, or stingers.
Cnidarians use specialized cells called cnidocytes, which contain nematocysts, for protection.
Explanation:Cnidarians use specialized cells called cnidocytes for protection. Cnidocytes are located around the mouth and tentacles and contain organelles called nematocysts. When triggered by a tactile stimulation, the nematocysts release a coiled thread and barb, along with toxins, which can immobilize or paralyze prey or predators. These cells are the defining feature of cnidarians and are responsible for their ability to capture prey and defend themselves.
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What role does RNA stability play in gene regulation? List three types of RNA structural elements that can affect the stability of an RNA molecule in eukaryotic cells. Explain how these elements act to influence RNA stability.
Answer:
RNA structural elements include double stranded regions of RNA formed by RNA folding upon itself to form a hairpin loop for example. These structures provide structural stability to RNA and allows for RNA to function as enzymes in some instances.
Another RNA structural element is the polyA tail formed at the end of mRNA sequence after mRNA maturation to prevent degradation of mature mRNA by RNase.
Another RNA structural element is the 5'cap formed at the 5' of mRNA sequence after mRNA maturation to prevent degradation of mature mRNA by RNase and also acts as a signalling target for transport of mRNA.
RNA stability affects gene regulation by controlling RNA degradation rates. Key elements include: 5' Cap: Protects RNA from degradation, Poly-A Tail: Enhances stability and translation, and RNA Secondary Structures: Can mask or expose degradation signals.
Three key RNA structural elements that affect stability in eukaryotic cells are:
5' Cap Structure: This modified guanine nucleotide at the RNA's 5' end protects it from degradation by exonucleases and facilitates translation initiation. A loss of the 5' cap leads to rapid degradation.3' Poly(A) Tail: A string of adenine nucleotides added to the 3' end stabilizes the RNA and enhances translation efficiency. Shortening of the poly(A) tail triggers degradation by exonucleases.Stem-Loop Structures: These secondary structures formed by base-pairing within the RNA molecule can influence stability. They can protect the RNA from degradation by shielding vulnerable regions or signal decay pathways when specific sequences are recognized by RNA-binding proteins or ribonucleases.These elements collectively regulate RNA half-life, impacting gene expression.
Mr. Rojas was looking through a microscope at some similar cells. He noticed that every cell had a different structure. What can he infer about these cells? * They are from different organisms. They have the same organelles. They have the same functions. They have different functions.
Answer:
They have different functions.
Explanation:
The cells having different structures may come from the same organism and having different structures may result from having different organelles. Due to the structure function correlation in cells, it is thus likely that the different structures arise from different functions, hence cells are specialised structurally to optimise performance in their functions.
Theodor W. Engelmann illuminated a filament of algae with light that passed through a prism, thus exposing different segments of algae to different wavelengths of light. He added aerobic bacteria and then noted in which areas the bacteria congregated. He noted that the largest groups were found in the areas illuminated by the red and blue light.If you ran the same experiment without passing light through a prism, what would you predict?A) There would be no difference in results.B) The bacteria would be relatively evenly distributed along the algal filaments.C) The number of bacteria present would decrease due to an increase in the carbon dioxide concentration.D) The number of bacteria present would increase due to an increase in the carbon dioxide concentration.E) The number of bacteria would decrease due to a decrease in the temperature of the water.
Answer:
The correct answer will be option-B
Explanation:
Theodor Engelmann performed the experiment to study the relationship between the production of oxygen in response to the wavelength of light. This was observed by the congregation of the bacteria in the red and blue wavelength.
In the given condition, if no light is passed through prism then the bacteria will not be accumulated in a particular area instead will be evenly distributed as the algae will produce the oxygen evenly and the concentration is not high in any particular area.
Thus, Option-B is the correct answer.
Final answer:
Without a prism to separate light into its components, aerobic bacteria would likely be relatively evenly distributed along the algal filaments since all areas would receive red and blue light uniformly, resulting in more uniform oxygen production across the algae.
Explanation:
The experiment conducted by Theodor W. Engelmann involved illuminating a filament of algae with light from a prism, which allowed different segments of the algae to be exposed to different wavelengths of light. The addition of aerobic bacteria to this setup led to the observation that the bacteria congregated in areas where the red and blue light illuminated the algae, indicating that these areas had the highest rates of photosynthesis and oxygen production.
If the same experiment were conducted without the prism, so that the algae were exposed to uniform white light containing all wavelengths, based on Engelmann's results, the prediction would be that the bacteria would be relatively evenly distributed along the algal filaments. This is because all areas would receive red and blue wavelengths equally, alongside other wavelengths, so the oxygen production should in theory be more uniform across the entire filament of algae without the prism separating the light into its component colors.
Therefore, the correct prediction for what would happen if the same experiment were run without passing light through a prism is: B) The bacteria would be relatively evenly distributed along the algal filaments.
Most of the cells examined during interval A have half as much DNA as those measured in interval C. Therefore, most of the cells at interval B must have been in:
a) G1 phase.
b) G0 phase.
c) M phase.
d) G2 phase.
e) S phase.
Answer:
e) S phase
Explanation:
The cell cycle is divided into various phases. These are in accordance with their order:
1.[tex]G_{1}[/tex] phase
In this phase, the cell grows in size and doubles its organelles. This phase is in fact a preparatory phase for cell division.
2.S phase
In this phase, the cell doubles its nuclear material (DNA). As cell divides into 2 daughter cells in mitosis, therefore it is necessary for cell to double its genetic material (DNA) which takes place in S phase. This genetic material is equally distributed among the daughter cells.
3.[tex]G_{2}[/tex] phase
In this phase, the cell further increases its store of proteins and organelles and prepares for the next phase i.e. M phase in which cell divides.
4.M phase
It consists of 2 processes i.e. mitosis and cytokinesis. In mitosis, the cell divides its genetic material (DNA) and in cytokinesis, the cell divides its cytoplasm resulting in 2 daughter cells.
[tex]G_{0}[/tex] phase
This is an inactive stage of cell cycle. In this stage cell does not divide further but remains quiescent. It performs its normal function in this phase. This phase is not a feature of all cells. Some cells like neurons (nerve cells) enter Go phase and remain there permanently whereas epithelial cells (cells lining body surfaces) do not enter Go phase and continuously divide. Cells enter Go phase after going through [tex]G_{1}[/tex] phase.
Back to the question,
If interval A contains half as much DNA as interval C then it implies that interval A is [tex]G_{1}[/tex] phase as in this phase of cell cycle, cell increases its size, and organelles but does not double its DNA content. It is further understood that interval C is [tex]G_{2}[/tex] phase as it contains double the DNA amount as in interval A. It can be understood that B interval is S phase, as in S phase cell doubles its DNA content and this phase occurs in between [tex]G_{1}[/tex] and [tex]G_{2}[/tex] phases.
The correct answer is (e) S phase.
The cell cycle consists of several phases: G1 (first gap), S (synthesis), G2 (second gap), and M (mitosis). During the S phase, DNA replication occurs, resulting in cells having twice as much DNA as they had in G1.
Given that cells in interval A have half as much DNA as those in interval C, we can infer the following:
- Cells in interval C must have completed DNA replication and are therefore in either G2 or M phase, where they have double the DNA content compared to G1.
- Cells in interval A, having half the DNA content of those in interval C, are likely in G1 phase, where they have not yet replicated their DNA.
Now, considering interval B, which is between intervals A and C, the cells in interval B must be in the process of moving from the G1 phase (with a single set of DNA) to the G2 or M phase (with a double set of DNA). The only phase in which DNA content is increasing from the G1 level to the G2 level is the S phase.
Therefore, most of the cells at interval B must have been in the S phase, where DNA synthesis is occurring, and the DNA content is increasing from the amount present in G1 to that present in G2/M. This is consistent with the observation that the DNA content in interval B is between that of interval A (G1 phase) and interval C (G2 or M phase).
To summarize:
- Interval A: G1 phase (single DNA content)
- Interval B: S phase (DNA content increasing)
- Interval C: G2 or M phase (double DNA content)
Hence, the correct answer is e. S phase.
Why is it essential that the regulatory mechanisms that activate glycogen synthesis also deactivate glycogen phosphorylase?
Final answer:
Regulatory mechanisms activate glycogen synthesis and deactivate glycogen phosphorylase simultaneously through reciprocal regulation to prevent a futile and energy-wasting cycle of simultaneous glycogen synthesis and breakdown. This coordinated process is crucial for energy homeostasis and is regulated allosterically and by hormones that control phosphorylation states of key enzymes.
Explanation:
It is essential that the regulatory mechanisms that activate glycogen synthesis also deactivate glycogen phosphorylase to prevent a futile cycle of synthesizing and breaking down glycogen at the same time, thus wasting energy. This is known as reciprocal regulation, a process ensuring efficiency and simplicity in metabolic control. Glycogen synthase, the enzyme responsible for glycogenesis, is active when dephosphorylated while glycogen phosphorylase, the enzyme for glycogenolysis, is active when phosphorylated. When insulin is present, it activates a phosphatase which simultaneously promotes the dephosphorylation and activation of glycogen synthase and the dephosphorylation and inhibition of glycogen phosphorylase. Conversely, in the presence of hormones like epinephrine or glucagon, a signaling cascade leads to the phosphorylation of enzymes, resulting in activated glycogen phosphorylase and inhibited glycogen synthase.
High-energy molecules like ATP and G6P allosterically inhibit glycogen phosphorylase, while low-energy molecules such as AMP activate it. This allosteric regulation aligns the activity of glycogen phosphorylase with the cell's energetic demands. Overall, the reciprocal regulation of glycogen metabolism by phosphorylation and dephosphorylation mechanisms is crucial for maintaining energy homeostasis within the cell.
In which of the following groups are most mRNAs monocistronic?
a.bacteria and archaea archaea
b.bacteria eukarya
c.All mRNAs are monocistronic.
Answer:
The answer is B
Explanation:
Bacteria eukarya
Codons are three-base sequences in mRNA that specify the addition of a single amino acid to the growing protein chain during translation. How do eukaryotic codons and prokaryotic codons compare?
Answer:
There is no difference between Eukaryotic and prokaryotic codons. All 61 codons and 20 amino acids are same. Start codons are also same.
The genetic code is essentially universal, with both eukaryotic and prokaryotic organisms using the same mRNA codons to specify the addition of amino acids or the termination of protein synthesis during translation. The start codon is typically AUG, and it also codes for the amino acid methionine. The uniformity of the genetic code across different life forms suggests that all life on Earth shares a common origin.
Explanation:Codons are three-nucleotide sequences in mRNA that specify the addition of a specific amino acid or the termination of protein synthesis during translation. There are 64 possible mRNA codons comprising of combinations of A, U, G, and C. Out of these, three are stop codons that indicate the termination of protein synthesis. Typically, AUG is the start codon for initiating translation, and it also encodes for the amino acid methionine.
The genetic code, which is the relationship between an mRNA codon and its corresponding amino acid, is essentially universal and alike for both eukaryotic and prokaryotic organisms, implying that all life on Earth shares a common origin. Such a universal genetic code strongly supports the theory of common descent for all life on Earth.
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During which processes does independent assortment of chromosomes occur?
Independent assortment occurs during anaphase I of meiosis I and to some extent during metaphase II and anaphase II, leading to the formation of gametes with diverse combinations of paternal and maternal chromosomes.
Explanation:Independent assortment of chromosomes occurs during meiosis, a critical process for producing genetically diverse gametes in eukaryotic organisms. Specifically, this event takes place during anaphase I of meiosis I, where bivalent chromosomes separate, and again, although indirectly, during metaphase II and anaphase II as sister chromatids segregate. During the phase of prophase I, chromosomes align randomly at the metaphase plate, setting the stage for the independent assortment. The physical basis for the law of independent assortment lies here, as different homologous pairs line up in random orientations, allowing any combination of paternal and maternal chromosomes to segregate into each gamete, thus increasing genetic variation.
Additionally, recombination or chromosomal cross-over during meiosis further increases genetic diversity by producing new combinations of alleles. At the end of meiosis II, four different gametic combinations are produced, each containing a haploid set of chromosomes. As the number of chromosomes increases, the number of possible assortment combinations increases dramatically, leading to a vast potential for genetic diversity.
If a person loses a large amount of water in a short period of time, he or she may die from dehydration. ADH can help reduce water loss through its interaction with its target cells in the
A) Anterior pituitary.
B) Posterior pituitary.
C) Adrenal gland.
D) Bladder.
E) Kidney.
Answer:
E because ADH is found is the hormones found in the kidney
When two frog species, Rana pipiens and Rana sylvatica, mate, the offspring die early in embryonic development. This is an example of ________.
a) hybrid sterility
b) reduced hybrid fertility
c) reduced hybrid viability
d) mechanical isolation
Answer:
The correct answer is c) reduced hybrid viability.
Explanation:
As the offspring dies early in embryonic development, it's possible to understand that the organism isn't viable, which is very common among hybrid organisms as well as their sterility or reduced fertility.
Which of the following definitions is INCORRECT? Which of the following definitions is INCORRECT? epidemic: a disease that is constantly present across the world endemic: a disease that is constantly present in a population pandemic: a disease that affects a large number of people in the world in a short time incidence: number of new cases of a disease
Answer:
epidemic: a disease that is constantly present across the world.
Explanation:
An epidemic is a sudden outbreak of a disease that spreads over a large area within a short period.
Endemic disease is a disease that is found regularly within an area at every point in time.
Pandemic disease is a disease that is prevalent over the whole world within a short period.
Incidence is the number of newly reported cases of a disease.
From the definition of each of the term. epidemic is the only one defined incorrectly of all the options.
The incorrect definition is that of an endemic disease, which is correctly defined as a disease that is always present at low incidence, not high incidence, in a population.
Explanation:The definition provided in the question that is INCORRECT is that of an endemic disease. The correct definition is that an endemic disease is one that is constantly present, usually at low incidence, in a population, not at high incidence as suggested in one of the erroneous definitions. In contrast, an epidemic is an outbreak of a disease that occurs in an unusually high number of individuals within a population at the same time. A pandemic is an epidemic that becomes widespread across multiple populations or continents, or even worldwide. Incidence refers to the number of new cases of a disease that appear in a population during a specific period of time. These terms are crucial for understanding the spread of diseases and for public health management.
Which of the following correctly lists in order the four stages of food processing in animals? View Available Hint(s) Which of the following correctly lists in order the four stages of food processing in animals? ingestion, digestion, absorption, elimination ingestion, digestion, elimination, absorption ingestion, absorption, elimination, digestion digestion, ingestion, absorption, elimination
Answer:
ingestion, digestion, absorption, elimination
Explanation:
Final answer:
The correct order of the four stages of food processing in animals is ingestion, digestion, absorption, and elimination, all vital for obtaining nutrition and managing waste.
Explanation:
The correct order of the four stages of food processing in animals is: ingestion, digestion, absorption, and elimination. During ingestion, food is taken into the body through the mouth. The process of digestion breaks down the food into smaller components that the body can use. Absorption is the stage where the body's cells take up the small molecules resulting from digestion. Finally, elimination removes indigestible substances from the body.
Each stage is crucial for obtaining nutrition and energy from food. Without ingestion, there is no food to process; without digestion, nutrients cannot be broken down; without absorption, digested nutrients cannot be utilized by the body; and without elimination, waste products would accumulate, leading to health issues.
Stamens are composed of:_______ Please choose the correct answer from the following choices, and then select the submit answer button.
A) stems and styles.
B) stigmas and ovaries.
C) anthers and filaments.
D) stigmas and filaments.
E) stems and anthers.
Answer:
C: anthers and filaments
Explanation:
The stamen is the male reproductive structure of flowers. It is made up anthers which houses the pollen grain and the filament which serves as a channel for transporting nutrient to nourish pollens.
The number of stamen a flower possesses varies from plant to plant. Some flowers have as few as one half while some have thousands of stamens. Collectively, stamens form the same flower are referred to as androecium. The stamens in some flowers join to form a tube, otherwise known as staminal tube.
The correct option is C.
A young female is unconscious after intentionally ingesting a large amount of aspirin. You will MOST likely find her respirations:
a. slow and deep.
b. deep and rapid.
c. slow and shallow.
d. rapid and shallow.
Answer:
Deep and rapid.
Explanation:
Respiration may be defined as the process of exchange of gases between the atmosphere, alveoli and body tissue. Two important process of respiration are inspiration and expiration.
Aspirin drug may also effect the rate of respiration in the individual. The breathing rate get increased after taking the excess dose of aspirin. The increase in the breathing rate results in the rapid and deep breathing of the individual.
Thus, the correct answer is option (b).
When a young female ingests a large amount of aspirin and becomes unconscious, the most likely finding in terms of respirations is rapid and shallow.
Explanation:When a young female is unconscious after intentionally ingesting a large amount of aspirin, the most likely finding in terms of respirations is rapid and shallow. Aspirin overdose can lead to respiratory alkalosis, causing the breathing rate to increase and become shallow due to the stimulation of the respiratory center in the brainstem. It is important to seek immediate medical help in such situations.
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Which of the pictured cell types influences neuronal function by ""mopping up"" leaked potassium ions and by recapturing and recycling released neurotransmitters? These cells also participate in information processing in the brain.
Answer:
Astrocytes
Explanation:
In the central nervous system, astrocytes, also collectively known as astroglia, are distinctive star-shaped glial cells.
It is not well known the ratio of astrocytes in the brain.
Research has found that the astrocyte proportion varies by area depending on the counting technique used and varies from 20% to 40% among all glia.
These fulfill many roles, namely biochemical reinforcement of tumor cells that build the blood-brain barrier, nutritional supply to the nervous tissue, extracellular ion balance maintenance, as well as a function in the brain and spinal cord repair and scarring cycle after severe injuries.
Pea flowers may be purple (P) or white (p). Pea seeds may be round (R) or wrinkled (r). What proportion of the offspring form a cross between purple-flowered round-seeded individuals (heterozygous for both traits) will have both white flowers and round seeds?
Answer:
3/16
Explanation:
The following is the punnet square which represents the probabilities of offsprings being produced for this cross:
PR Pr pR pr
PR PPRR PPRr PpRR PpRr
Pr PPRr PPrr PpRr Pprr
pR PpRR PpRr ppRR ppRr
pr PpRr Pprr ppRr pprr
This cross shows that the probability of the offsprings to have white flowers and rounded seeds would be 3/16.
There is a misconception that girls cannot inherit recessive X-linked disorders. Girls can in fact inherit these disorders, but at a lower rate than boys. The following choices give the genotypes for pairs of mothers and fathers. Which pair of parents could have a daughter with the recessive disorder a?
Choose one:a. XAXA × XAYb. XAXA × XaYc. XAXa × XAYd. XAXa × XaY
Answer:
d. [tex]X^{A}X^{a}[/tex] × [tex]X^{a}Y[/tex]
Explanation:
Since the disorder is a recessive one, the daughter will need to inherit two recessive alleles, each one from the two parents in order for the disorder to manifest.
A cross between [tex]X^{A}X^{a}[/tex] and [tex]X^{a}Y[/tex] will yield the following offspring;
[tex]X^{A}X^{a}[/tex] female, carrier of disorder
[tex]X^{a}X^{a}[/tex] female, affected with disorder
[tex]X^{a}Y[/tex] male, affected with disorder
[tex]X^{A}Y[/tex] male, unaffected
Hence, the only parental pair that can have a daughter with the recessive disorder is [tex]X^{A}X^{a}[/tex] × [tex]X^{a}Y[/tex]
The correct option is d.
Which structure does very little albumin escape from the blood during glomerular filtration?
a. Fenestrated endothelium
b. Basement membrane
c. Filtration pore
d. Juxtaglomerular apparatus
e. Macula densa
Answer:
b. Basement membrane
Explanation:
Together, the endothelial cells of the glomerular capillaries and the podocytes that completely surround the capillaries form a permeable barrier called filtration membrane or endothelial capsular membrane. This sandwich arrangement allows filtration of water and small solutes, but prevents filtration. of most plasma proteins, blood cells and platelets. The filtered substances move from the bloodstream through three barriers: a gIomerular endothelial cell, the basal lamina and a filtration slot formed by a podocyte.
1 The endothelial cells of the gIoméruIo are quite permeable because they have large perforations (pores) of 70 to 100 nm (0.07 to 0.1 ml) in diameter. These dimensions allow all solutes of the blood plasma to leave the gIomerular capillaries, but prevents blood cells and platelets from leaking. The mesangial cells, contractile cells that help regulate gIomerular filtration are located in the middle of the gomeromer capillaries and the groove formed between the afferent and efferent arterioles.
2 The basement membrane is a layer of acellular material between the endothelium and podocytes, which consists of fibruses in a gIucoprotein matrix; prevents filtration of large plasma proteins.
3 From each podocito thousands of extensions in the form of feet called pedicels that surround the glomerular capillaries extend. The spaces between the pedicels are called filtration cracks. A thin membrane, the crack membrane, extends through each filtration crack and allows the passage of molecules with a diameter smaller than 6 to 7 nm (0.006 to 0.OO7 mm), including water, glucose, vitamins, amino acids , very small plasma proteins, ammonia, urea and ions. Since the most abundant plasma protein, albumin, has a diameter of 7.1 mm, less than 1% of it crosses the membrane of the crack.
Dr. Aruna has diagnosed several clients with social anxiety disorder, and he is interested in finding a treatment plan that will be effective in the long term and have the fewest side effects. Dr. Aruna should try:
Answer:
Cognitive-behavioral therapy
Explanation:
Cognitive-behavioural therapy can be described as a very common type of psychotherapy which psychologists use to help patients undergoing anxiety disorders. It is a type of talk therapy. This kind of therapy trains your brain to become aware whenever you are thinking in a negative manner. It trains you to think in a positive manner instead. It helps a person to change the way they feel. Hence, cognitive-behavioural therapy is the best plan for patients with social- anxiety disorder.
Answer: Cognitive behavioral therapy
Explanation:
Cognitive behavioral therapy is a kind of talk therapy which is not only useful to treat mental problems, disorders and depression but it is also useful to reduce stress, overcome emotional unbalance. This is also helpful to overcome the social anxiety disorders.
This is associated with fewer side effects and the treatment is effective for long term or lifelong. This helps to cope with stress, grief and resolving the relationship conflicts.
According to the given situation, CBT is the correct treatment plan that can be used to treat the social anxiety disorder.
The sharp, inch-long thorns of the crown-of-thorns sea star are its spines. These spines, unlike those of most other sea stars, contain a potent toxin. If it were discovered that crown-of-thorns sea stars do not make this toxin themselves, then the most likely alternative would be that this toxin is _____.
A) derived from the nematocysts of its prey
B) absorbed from the surrounding seawater
C) an endotoxin of cellulose-digesting bacteria that inhabit the sea star's digestive glands
D) injected into individual thorns by mutualistic corals which live on the aboral surfaces of these sea stars
Answer:
The answer is A) derived from the nematocysts of its prey.
Explanation:
Sea stars mostly eat cnidarians (mollusks) which can contain a large number of nematocysts in their bodies. Therefore it would be possible that this toxin is absorbed from the prey.
1.List and describe the structural components of viruses.
2. Explain why viruses are obligate intracellular parasites.
3. Explain how a virus identifies its host cell.
4. Describe bacterial defenses against phages.
5. Distinguish between the lytic and lysogenic reproductive cycles, using phage as an example.
6. Describe the reproductive cycle of an HIV retrovirus.
7. What are several catagories of Virus and how do they differ?
8. List some characteristics that viruses share with living organisms and explain why viruses do not fit our usual definition of life.
9. Define and describe mobile genetic elements. Explain why plasmids, transposons, and viruses are all considered mobile genetic elements. Viruses, Viroids and Prions.
10. Explain how viral infections may cause disease.
11. Describe the three processes that lead to the emergence of new viral diseases.
12. Distinguish between the horizontal and vertical routes of viral transmission in plants.
13. Describe viroids and prions.
Answer:
A virus is a fragment of code embedded in a legitimate program. Virus are self-replicating and are designed to infect other programs. They can wreak havoc in a system by modifying or destroying files causing system crashes and program malfunctions. On reaching the target machine a virus dropper(usually trojan horse) inserts the virus into the system.
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Various types of virus :
File Virus : This type of virus infects the system by appending itself to the end of a file. It changes the start of a program so that the control jumps to its code. After the execution of its code, the control returns back to the main program. Its execution is not even noticed. It is also called Parasitic virus because it leaves no file intact but also leaves the host functional.
Boot sector Virus : It infects the boot sector of the system, executing every time system is booted and before operating system is loaded. It infects other bootable media like floppy disks. These are also known as memory virus as they do not infect file system.
boot sector virus flowchart
Macro Virus : Unlike most virus which are written in low-level language(like C or assembly language), these are written in high-level language like Visual Basic. These viruses are triggered when a program capable of executing a macro is run. For example, macro virus can be contained in spreadsheet files.
Source code Virus : It looks for source code and modifies it to include virus and to help spread it.
Polymorphic Virus : A virus signature is a pattern that can identify a virus(a series of bytes that make up virus code). So in order to avoid detection by antivirus a polymorphic virus changes each time it is installed. The functionality of virus remains same but its signature is changed.
Encrypted Virus : In order to avoid detection by antivirus, this type of virus exists in encrypted form. It carries a decryption algorithm along with it. So the virus first decrypts and then executes.
Stealth Virus : It is a very tricky virus as it changes the code that can be used to detect it. Hence, the detection of virus becomes very difficult. For example, it can change the read system call such that whenever user asks to read a code modified by virus, the original form of code is shown rather than infected code.
Tunneling Virus : This virus attempts to bypass detection by antivirus scanner by installing itself in the interrupt handler chain. Interception programs, which remain in the background of an operating system and catch viruses, become disabled during the course of a tunneling virus. Similar viruses install themselves in device drivers.
Multipartite Virus : This type of virus is able to infect multiple parts of a system including boot sector,memory and files. This makes it difficult to detect and contain.
Armored Virus : An armored virus is coded to make it difficult for antivirus to unravel and understand. It uses a variety of techniques to do so like fooling antivirus to believe that it lies somewhere else than its real location or using compression to complicate its code.
Answer: 1. nucleoid, capsid, envelope, enzymes.
A. Nucleoids represents the viral chromosomes, it is made up of single molecular which is either linear or coiling.
B. Capsid is the proteinacious covering around the virus that protects the nucleoids from damage and it consists of subunits called capsomere and campaigners.
C. Envelope is the loose membrane covering that occur in some viruses consisting of protein from lipids and carbohydrates(from host).
D. Enzymes are occasionally enzymes lysozymes is present in the region that comes in contact with host cell in bacteria pages.
2. Due to their nature of behaving like simple chemical compounds outside the living host.
3. Virus recognises it's host cell through the receptors they carry and virus can also trick the cell into taking them by bulk transport called endocytosis.
4. The development of CRISPR immune system to fight against phages.
5. Lytic involves reproduction of viruses by using host cell for virus manufacturing while lysogenic cycles involves integration of viral genome into host DNA and reproducing with it.
6. The life cycle of hic
a. Attachment and entry- this happens in two stages attachment and fusion.
B. Reverse Transcription- when hiv enters the cell it must be reverse transcribed into proviral DNA before integrating into the DNA of host.
C. Integration- after conversion into DNA, hiv integrate enzyme attaches itself to the proviral DNA strands it passes through the wall of cell nucleus.
D. Transcription and Translation- they become active when they encounter an infectious agent.
7. Bacterial viruses are classified under viruses that affect the ecoil.
-DNA phages of T series where the T phages of ecoil are larger lyric phages that contain single molecule of double stranded DNA
-Temprate pages: bacterial phage lambda
-Small DNA phage: the genome of some bacteria phage encoded 10-12 protein
-RNA phage- some e coil bacterial phage contains a genome composed of RNA instead of DNA.
B. ANIMAL viruses are classified by genome type and MNRA synthesis path way.
-DNA viruses (classes 1 and 2) commonly used in the study of DNA replication.
-RNA viruses(classes 3 and 4) have RNA genome and all animal cells belong to this class.
8. They are made of protein.
They contain genetic codes needed for reproduction.
-viruses do not fit our normal life because viral reproduction is not possible without a host cell and it doesn't consume energy to survive that is metabolic processes within the cell.
9. The DNA sequences which can move around the genome changing their number of copies or location often affecting the nearest genes.
- plasmids and trasposoons are mobile transferable genetic elements through bacterial conjugation which is a horizontal mechanism gene transfer while viruses can move due to their adaptive replication strategy.
10. They cause infectious diseases by entering a living and normal cell, thereby using the cell to reproduce other viruses of them selves and infecting the immune system of the living cell.
11. -introduction of viral pathogens into a new host species.
-establishment of pathogens in New host.
-dissemination of pathogens among large number of individual of new host species.
12. In a nut shell, transmission of viruses occur horizontally is the transmission of viruses among plants of same generation while vertical transmission is between full grown plants and the seed from it to ensure transmission from current generation to the next generation.
13. Virions is the infectious form of viruses as it exists outside the host cell designed for transmission of nucleic acid genome amongst hosts while a prion is neither bacterial nor fungal nor viral and contains no genetic materials.
Explanation:2. They have long and non living thing characteristics in nature.
4. This is an act of copying a snipet of phages they come in contact with into a memory to help build against subsequent attacks by viruses in future.
6. The life cycle of hiv consists of 4 stages which includes the process of entering , reverse integration where the RNA enzymes reverse transcribed into DNA , integration where it passes through the walls of the cell and finally the last stage where it becomes activated.
N.B
Most of the answers required explanations above in order to avoid repetition of answers.
Some nutrients and other types of particles can be taken into a cell when the cell’s plasma membrane encloses the particles and pinches off. True or False
Answer:
True
Explanation:
The particles takes up by the cell from extracellular side and allow their fusion with the plasma membrane. The process is called endocytosis. The large molecules that are polar in nature like hormones and proteins are absorbed by cell. The polar molecules are unable to cross plasma membrane that is hydrophobic in nature. Thus simple diffusion is not possible for polar molecules.
Thus, Some nutrients and other types of particles can be taken into a cell when the cell’s plasma membrane encloses the particles and pinches off is true.
Waste stream is a term describing Group of answer choices the steady production of all waste products that humans produce. the process of eliminating domestic and commercial waste. the production of solid waste, specifically. the intermittent production of particular wastes. the production of liquid waste, specifically.
Answer:
Waste stream is a term describing the process of eliminating domestic and commercial waste.
Explanation:
Waste stream refers to the journey of waste from its domestic or industrial origins to its final destiny. As waste moves through the line, it passes by different stages or areas that must be controlled by different agents. Waste stream analysis is an important tool to evaluate the management of materials or resources, allowing to identify contamination and polluting problems in a system. It might be the base of sustainability by solid waste use. Applying recycling strategies during different stages of waste stream might reduce the content of waste as it goes along the journey.
Waste stream is a term describing the steady production of all waste products that humans produce.
The term "waste stream" refers to the continuous and ongoing production of all types of waste materials generated by human activities.
It encompasses various forms of waste, including solid waste, liquid waste, and gaseous waste.
The waste stream concept recognizes that waste is generated at different stages and in various sectors of human activities, such as residential, commercial, industrial, and agricultural sectors.
Thus, analyzing the waste stream also helps identify areas where waste reduction efforts can be targeted and facilitates the development of policies and practices to promote a circular economy and reduce the environmental impact of waste.
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