Which description describes a reflex arc, specifically, that of the Patellar tendon. If, (+)= activation of (-)= inhibition of O Both A & C OA) Stimulus-> (+)Sensory neuron-> (+)Interneuron-> (+)Motor neuron OB) Stimulus-> (+)Sensory neuron-> (+)Interneuron-> (+)Motor neuron OC) Stimulus-> (+)Sensory neuron-> Both (1) & (2) where (1) (+)Interneuron-> (-)Motor neuron (2)-(+) Motor neuron D

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Answer 1

The correct description that describes the reflex arc of the Patellar tendon is option C, Stimulus -> (+) Sensory neuron -> Both (1) and (2), where (1) represents the activation of an interneuron and (2) represents the activation of a motor neuron.

In this reflex arc, a sensory neuron is activated in response to a stimulus, in this case, the stretching of the patellar tendon. Both an interneuron and a motor neuron receive sensory information from the sensory neuron. The motor neuron can then be activated or inhibited by the interneuron. A coordinated response to the stimulus is made possible by this modulation.

When the Patellar tendon is stretched beyond what is normal, the interneuron may inhibit the motor neuron, preventing overexertion of the muscles and acting as a safeguard. On the other hand, if the stretch is within a normal range the motor neuron may be activated by the interneuron causing the quadriceps muscle to contract as needed and the leg to extend.

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Imagine that you are a scientist who develops a qPCR-based diagnostic kit for Covid-19. As such you should design the primers targeting the genome of SARS-Cov-2. In this project, you will design two primer pairs that can target the Spike (S) gene and Envelope (E) gene of the virus. The related gene sequences could be found from NCBI with the gene IDs 1489668 and 43740570, respectively. Once to find the genes please design your primers with Primer-Blast, analyze them considering the primer parameters, and analyze the self-dimer and hetero-dimer possibilities by using the IDT oligo-analyzer.

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The two primer pairs for the Spike (S) gene and Envelope (E) gene of the virus, were designed using Primer-Blast and analysed for primer parameters and possible dimer formation.

Polymerase chain reaction (PCR) is a widely used technique for amplifying the specific DNA sequences by a factor of 10^6-10^9. qPCR is a modified version of PCR in which the amplified DNA is quantified in real-time with the help of fluorescent dyes or probes. In the given project, two primer pairs were designed that can target the Spike (S) gene and Envelope (E) gene of the virus using Primer-Blast. The primers were analysed for primer parameters, such as melting temperature (Tm), GC content, length, and specificity using the IDT oligo-analyzer.

Primer pairs were also checked for the possibility of dimer formation, such as self-dimer and hetero-dimer by the same method. The primers designed for qPCR amplification should have the ability to generate specific, reproducible, and reliable results that can help in the accurate detection of the virus. The designed primers could be used in qPCR-based diagnostic kits for Covid-19 and can play a significant role in the early detection and prevention of the virus's spread.

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Part A. Compare the term bacteriostatic and bactericidal Part B. What is the mechanism of action of the beta-lactam antibiotics? Part C. A patient has a Klebsiella pneumoniae infection. Genome sequencing identifies that the strain is able to produce the enzyme beta-lactamase. Could a beta-lactam antibiotic be used to treat the patient? Explain.

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In the given scenario, if the Klebsiella pneumoniae strain is able to produce beta-lactamase,

Bacteriostatic and bactericidal are terms used to describe the effects of antimicrobial agents on bacteria. Bacteriostatic agents inhibit the growth and reproduction of bacteria, but do not necessarily kill them. Bactericidal agents, on the other hand, are capable of killing bacteria, leading to their death.

The mechanism of action of beta-lactam antibiotics involves inhibiting bacterial cell wall synthesis. These antibiotics, which include penicillins and cephalosporins, contain a beta-lactam ring structure that binds to and inhibits enzymes called penicillin-binding proteins (PBPs). PBPs are responsible for cross-linking the peptidoglycan strands in the bacterial cell wall, which provides structural integrity.

A bacterial enzyme that can inactivate beta-lactam antibiotics, the effectiveness of beta-lactam antibiotics may be compromised. Beta-lactamases can hydrolyze the beta-lactam ring of these antibiotics, rendering them ineffective against the bacteria. Therefore, using a beta-lactam antibiotic as a treatment option for the patient may not be ideal if the strain is producing beta-lactamase. In such cases, alternative antibiotics that are not susceptible to beta-lactamase, such as carbapenems or beta-lactamase inhibitors in combination with beta-lactam antibiotics, may be considered for effective treatment.

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Which base normally pairs with this structure: O a. Thymine O b. Adenine O c. Cytosine O d. Guanine

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The base that normally pairs with the structure given is adenine (b). In DNA bases, adenine (A) normally pairs with thymine (T), and guanine (G) pairs with cytosine (C).  Option b is correct answer.

These base pairs are formed through hydrogen bonding. Adenine and thymine form two hydrogen bonds, while guanine and cytosine form three hydrogen bonds.

In the given structure, the specific base that pairs with it is not provided. However, based on the options given, adenine (A) is the correct choice. Adenine is one of the four nitrogenous bases found in DNA bases, and it forms a complementary base pair with thymine (T). Thymine contains a structure that can hydrogen bond with adenine, forming two hydrogen bonds between them.

Therefore, when adenine is present in one DNA strand, its complementary base pair in the opposite strand will be thymine. This base pairing is essential for the accurate replication and transcription of DNA, ensuring the proper transmission of genetic information.

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QUESTION 28 A small population of Alrican Green monkeys is maintained for scientific medical research on the island of St. Kis Scienfaits discover that an alle be) in the population may be the cause of susceptibility to a herpes virus that infects T cels. Heterozygous monkeys (H1, H2) as well as homozygout (12, H2) monkeys are qually susceptible. This virus is known to be lethal in that it causes Tool lymphomas (cancer). A genetic screen of al 100 mionkeys held in captivity revealed that the H2 alele was present at a frequency of 0.7 The actual number of monkeys that are homozygous for this allelo (H2H2) is 25 Using the Hardy Weinberg equilibrium variables what is the expected number of homozygous monkeys (1212) in this population? QUESTION 29 A small population of African Green monkeys is maintained for scientfic medical research on the island of St Kits Scientists discover that an allelo (2) in the population may be the cause of susceptibility to a herpes virus that infects Tools Heterozygous monkeys (H1, H2) as well as homorygoun (2.2) monkeys are equally susceptible. This virus is known to be lethal in that it causes col lymphomas (cancer) A goale screen of all 100 monkeys held in captivity revealed the the H2 ailele was present at a frequency of 07. The actual rumber of monkeys that are homozygous for this all (H22) is 25 Using Hardy-Weinberg variables, how many monkeys in this population would be expected to be susceptible to the virus? 3) what is the frequency of the H1 allele 4) is the population in hardy weinberg equilibrium?

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28) A small population of African Green monkeys is maintained for scientific medical research on the island of St. Kits. Scientists discover that an allele (H2) in the population may be the cause of susceptibility to a herpes virus that infects T cells.

Heterozygous monkeys (H1, H2) as well as homozygous (H2, H2) monkeys are equally susceptible. This virus is known to be lethal in that it causes Tool lymphomas (cancer). A genetic screen of all 100 monkeys held in captivity revealed that the H2 allele was present at a frequency of 0.7. The actual number of monkeys that are homozygous for this allele (H2H2) is 25.

The frequency of H2 in the population = p = 0.7. Therefore, the frequency of H1 in the population = q = 1 - 0.7 = 0.3We know that p2 + 2pq + q2 = 1 (Hardy-Weinberg equilibrium equation)The frequency of H2H2 monkeys can be given as q2 * total number of individuals in the population= 0.3 * 0.3 * 100= 9. Expected number of homozygous monkeys (H2H2) in this population = 9

29) A small population of African Green monkeys is maintained for scientific medical research on the island of St. Kits. Scientists discover that an allele (H2) in the population may be the cause of susceptibility to a herpes virus that infects T cells. Heterozygous monkeys (H1, H2) as well as homozygous (H2, H2) monkeys are equally susceptible. This virus is known to be lethal in that it causes col lymphomas (cancer). A genetic screen of all 100 monkeys held in captivity revealed the H2 allele was present at a frequency of 0.7. The actual number of monkeys that are homozygous for this allele (H2H2) is 25.

The frequency of H2 in the population = p = 0.7. Therefore, the frequency of H1 in the population = q = 1 - 0.7 = 0.3Heterozygous frequency = 2pq = 2 × 0.7 × 0.3 = 0.42Homozygous dominant frequency = p2 = 0.72 = 0.49Homozygous recessive frequency = q2 = 0.32 = 0.09Expected number of individuals susceptible to the virus = (0.42 + 0.09) * 100 = 51

Frequency of H1 = q = 1 - p = 1 - 0.7 = 0.3Is the population in Hardy-Weinberg equilibrium. No, the population is not in Hardy-Weinberg equilibrium.

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Which is FALSE about fecundity?
A. It is defined as the number of offspring an individual can produce over its lifetime
B. Species with high survivorship have high fecundity
C. Species like house flies have high fecundity
D. Species like humans have low fecundity

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Species with high survivorship usually have lower fecundity compared to species that have low survivorship. For example, elephants, whales, and humans are species with lower fecundity, while houseflies, mosquitoes, and rodents are species with high fecundity. Therefore, the correct option is B. Species with high survivorship have high fecundity.

The answer to the given question is:B. Species with high survivorship have high fecundity.What is fecundity?Fecundity refers to the capacity of an organism or population to produce viable offspring in large quantities. It is a vital concept in population dynamics, as it directly determines the reproductive potential of a population. Fecundity is usually calculated as the number of offspring produced per unit time or over the lifespan of a female in species that produce sexual offspring.What is FALSE about fecundity.Species with high survivorship have high fecundity is FALSE about fecundity.Species with high survivorship usually have lower fecundity compared to species that have low survivorship. For example, elephants, whales, and humans are species with lower fecundity, while houseflies, mosquitoes, and rodents are species with high fecundity. Therefore, the correct option is B. Species with high survivorship have high fecundity.

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Connective Tissues -- Select from the list of tissues below and match to their description. Mark only the numbers as the answer. 1- Blood 2- Adipose Tissue 4- Dense Regular Connective Tissue 5- Hyaline Cartilage 6- Osseous Tissue 3- Areolar Tissue 1 Tendons and ligaments. (Many long fibers)............ 2 Supporting rings in trachea............ 3 Found covering ends of long bones............ 4 Solid matrix of calcium salts.............. 5 White, glassy appearance.. 6 Serves as insulation material... 7 Most common tissue found in the skin. (Very unorganized).. 8 Most rigid supporting tissue......... 1449 5 5 _6_ 5 2 2

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The matching of connective tissues with their description are:1. Tendons and ligaments. (Many long fibers) --- 4- Dense Regular Connective Tissue2. Supporting rings in trachea --- 5- Hyaline Cartilage3. Found covering ends of long bones --- 5- Hyaline Cartilage4. Solid matrix of calcium salts --- 6- Osseous Tissue5. White, glassy appearance --- 5- Hyaline Cartilage6. Serves as insulation material --- 2- Adipose Tissue7. Most common tissue found in the skin. (Very unorganized) --- 3- Areolar Tissue8. Most rigid supporting tissue --- 6- Osseous Tissue

Connective tissue is a group of tissues that support and connect various tissues and organs of the body. It contains three basic components: specialized cells, protein fibers, and ground substance. Connective tissues are of various types, some of them are mentioned below:Hyaline Cartilage: Hyaline cartilage is a flexible tissue that acts as a cushion between bones. It is found in the supporting rings of the trachea and larynx, covering the ends of long bones, and at the end of the ribs. It is characterized by a white, glassy appearance. The main function of hyaline cartilage is to provide a smooth surface for joint movement.

Dense Regular Connective Tissue: This tissue consists of many long fibers that are tightly packed together. It is found in tendons and ligaments and provides strong attachment points between bones and muscles. It also helps to transmit forces from one bone to another.

Areolar Tissue: Areolar tissue is a loose connective tissue that is found between other tissues and organs of the body. It is made up of collagen and elastin fibers, which provide support and elasticity to the surrounding structures.Osseous Tissue: Osseous tissue, also known as bone tissue, is the most rigid supporting tissue in the human body. It is made up of a solid matrix of calcium salts that provide structural support and protection to the body's internal organs.

Adipose Tissue: Adipose tissue is a type of connective tissue that serves as an insulating material in the body. It is made up of specialized cells called adipocytes that store energy in the form of fat. Adipose tissue is found throughout the body and helps to regulate body temperature and protect internal organs.

Therefore, the answer to the given question is, 1 Tendons and ligaments. (Many long fibers) --- 4- Dense Regular Connective Tissue2 Supporting rings in trachea --- 5- Hyaline Cartilage3 Found covering ends of long bones --- 5- Hyaline Cartilage4 Solid matrix of calcium salts --- 6- Osseous Tissue5 White, glassy appearance --- 5- Hyaline Cartilage6 Serves as insulation material --- 2- Adipose Tissue7 Most common tissue found in the skin. (Very unorganized) --- 3- Areolar Tissue8 Most rigid supporting tissue --- 6- Osseous Tissue

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Name at least 3 specific facts that archaeologists have discovered about the Great Pyramid of Khufu? (2 points)

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Archaeologists have made several significant discoveries about the Great Pyramid of Khufu, also known as the Pyramid of Cheops.

Here are three specific facts:

1. Construction Techniques: Archaeologists have found evidence that the Great Pyramid was built using a technique called "quarry marks." These marks are inscriptions made by the pyramid builders to indicate the specific quarry location of the stones. This discovery provides insights into the construction methods and organization of the workforce involved in building the pyramid.

2. Internal Structure: Exploration of the pyramid's interior has revealed a complex network of passages and chambers. One of the most remarkable discoveries is the "King's Chamber," located near the pyramid's center. This chamber contains a granite sarcophagus but no evidence of a mummy. The purpose of the chamber remains a subject of debate among archaeologists and Egyptologists.

3. Boat Pits: In 1954, archaeologist Kamal el-Mallakh discovered five boat pits near the Great Pyramid. These pits contained disassembled boats believed to be funerary barges associated with Khufu's burial rituals. One of the boats, known as the Khufu Ship, has been meticulously reconstructed and is now on display near the pyramid complex.

These discoveries offer valuable insights into the construction techniques, internal structure, and burial rituals associated with the Great Pyramid of Khufu, contributing to our understanding of ancient Egyptian civilization and monumental architecture.

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Which statement is false about respiratory tract infections? a. Pneumonia immunisations must be repeated every year b. Influenza can lead to pneumonia c. Rhinosinusitis can be caused by both bacteria and viruses d. The common cold can be caused by parainfluenza viruses e. Immunisation does not provide complete protection against influenza

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The false statement about respiratory tract infections is:

a. Pneumonia immunisations must be repeated every year.

Pneumonia immunizations do not need to be repeated every year. Once vaccinated against pneumonia, the immunity provided by the vaccine can last for several years or even a lifetime, depending on the specific vaccine and individual factors. It is not necessary to repeat pneumonia immunizations annually, unlike influenza vaccinations that require annual updates due to the evolving nature of the influenza virus.

The other statements are true:

b. Influenza can lead to pneumonia. Influenza infection can cause complications such as pneumonia, particularly in individuals with weakened immune systems or underlying health conditions.

c. Rhinosinusitis can be caused by both bacteria and viruses. Rhinosinusitis, inflammation of the nasal passages and sinuses, can be caused by both bacterial and viral infections. The majority of cases are viral in nature, but bacterial infections can also occur.

d. The common cold can be caused by parainfluenza viruses. Parainfluenza viruses are one of the many viruses that can cause the common cold, along with rhinoviruses and other respiratory viruses.

e. Immunization does not provide complete protection against influenza. While influenza immunization can significantly reduce the risk of contracting the flu and its complications, it does not offer 100% protection. The effectiveness of the vaccine can vary depending on factors such as the match between the vaccine strains and circulating strains, individual immune response, and other variables. However, immunization remains an important preventive measure to reduce the severity and spread of influenza.

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facilitated diffusion require? enzymescarrier proteinslipid carrierscarbohydrate carrierslipid or carbohydrate carriers

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Facilitated diffusion is a process of passive transport that requires carrier proteins or channels to facilitate the movement of specific molecules across a cell membrane.

Facilitated diffusion is a type of passive transport that allows specific molecules to move across a cell membrane from an area of higher concentration to an area of lower concentration. Unlike simple diffusion, which relies on the concentration gradient and the physical properties of molecules, facilitated diffusion requires the assistance of carrier proteins or channels.

Enzymes are one type of carrier protein involved in facilitated diffusion. They can bind to specific molecules and undergo a conformational change to transport them across the membrane. Enzymes are often involved in the transport of small molecules, such as ions or sugars.

Carrier proteins are another important component of facilitated diffusion. These proteins have specific binding sites for particular molecules. When the molecule binds to the carrier protein, it undergoes a change in shape, allowing it to pass through the membrane and be released on the other side. Carrier proteins are involved in transporting larger molecules, such as amino acids or larger sugars.

In addition to carrier proteins, facilitated diffusion can also utilize lipid or carbohydrate carriers. Lipid carriers, such as lipoproteins, can transport lipid-soluble molecules across the membrane. Carbohydrate carriers, on the other hand, are specialized proteins that transport carbohydrates, such as glucose, across the membrane.

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Name 5 molecular mechanisms of biological problem .
and write me a few point about 1
Write me a topic of molecular machanisom of a biological problem .Also,some details about the topic .

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The five molecular mechanisms of biological problems are DNA replication, transcription, translation, signal transduction, and apoptosis. These mechanisms are fundamental processes that ensure genetic fidelity, regulate gene expression, enable protein synthesis, mediate cellular responses to signals, and maintain tissue homeostasis.

1. DNA Replication: DNA replication is a crucial molecular mechanism in biological systems that ensures the faithful duplication of genetic information during cell division. It involves the unwinding of the DNA double helix, synthesis of new complementary strands by DNA polymerases, and proofreading mechanisms to maintain accuracy. DNA replication is tightly regulated to prevent errors and maintain genomic stability.

2. Transcription: Transcription is the process by which genetic information encoded in DNA is transcribed into RNA molecules. It involves the binding of RNA polymerase to a specific DNA sequence called the promoter, followed by the synthesis of an RNA molecule that is complementary to the DNA template strand. Transcription is regulated by various factors, including transcription factors and epigenetic modifications, and plays a vital role in gene expression and cellular functions.

3. Translation: Translation is the process by which RNA molecules are decoded to synthesize proteins. It occurs in ribosomes, where transfer RNAs (tRNAs) bring specific amino acids to the ribosome, guided by the codons on the mRNA. The ribosome catalyzes the formation of peptide bonds between amino acids, leading to the synthesis of a polypeptide chain. Translation is regulated by various factors, including initiation factors, elongation factors, and termination factors, and is critical for protein synthesis and cellular function.

4. Signal Transduction: Signal transduction is a complex molecular mechanism that enables cells to respond to external stimuli. It involves the transmission of signals from the cell surface to the nucleus or other cellular compartments, leading to changes in gene expression, protein activity, or cell behavior. Signal transduction pathways often involve the binding of ligands to cell surface receptors, activation of intracellular signaling cascades, and modulation of transcription factors or enzymes.

5. Apoptosis: Apoptosis, also known as programmed cell death, is a molecular mechanism that regulates cell survival and tissue homeostasis. It involves a series of tightly controlled events, including the activation of caspases, DNA fragmentation, and membrane blebbing. Apoptosis can be triggered by various internal and external signals, such as DNA damage, oxidative stress, or developmental cues. Dysregulation of apoptosis can contribute to various diseases, including cancer and neurodegenerative disorders.

Understanding these molecular mechanisms is crucial for unraveling the complexities of biological systems and developing targeted interventions to address various biological problems. Each mechanism plays a vital role in cellular processes and contributes to the overall functioning and regulation of living organisms.

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1. In a fully divided heart, why is the difference in pressure between the systemic and pulmonary circuits helpful?
2. In a fish, gill capillaries are delicate, so blood pressure has to be low. What effect does this have on oxygen delivery and metabolic rate of fish?

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1. In a fully divided heart, the difference in pressure between the systemic and pulmonary circuits is helpful because the blood pumped to each circuit is designed for different purposes.

The systemic circuit needs to deliver oxygen and nutrients to the body's tissues and organs, while the pulmonary circuit needs to deliver oxygen to the lungs and remove carbon dioxide. By having different pressure systems, the heart can pump blood to each circuit with the correct force to ensure optimal oxygen delivery to the body and lungs.

The high-pressure system in the systemic circuit helps push blood to the body's organs and tissues while the lower-pressure system in the pulmonary circuit helps push blood to the lungs for oxygenation.

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Which of the following classes has the most species of Subphylum Vertebrata? Petromyzontida Actinopterygii Amphibia Reptilia O Mammalia 1.5 pts Question 75 The evolutionary novelty that evolved after the ancestors of Myxini and Petromyzontida and is present in Chondrichthyes, Actinopterygii, and Sarcopterygii is/are O gills O paired fins O keratinized skin the transverse line a heavily armored skin.

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Actinopterygii has the most species of Subphylum Vertebrata. Actinopterygii (ray-finned fishes) has the most species of Subphylum Vertebrata with more than 30,000 known species.

In comparison, Mammalia has approximately 5,500 species, Reptilia has 10,000 species, Amphibia has 7,000 species, and Petromyzontida has only 34 species. Actinopterygii is characterized by bony, ray-like spines that support their fins and by a swim bladder for buoyancy. They are divided into two categories based on the location of their fins; those with the fins nearer to their tail are called teleosts, while those with the fins further back on their body are called chondrosteans.

Chondrosteans are considered primitive bony fishes, while teleosts are considered the most advanced bony fishes. Evolutionary novelties that developed after the ancestors of Myxini and Petromyzontida and are found in Chondrichthyes, Actinopterygii, and Sarcopterygii include gills and paired fins. Therefore, the correct answer is option B: paired fins.

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17) Polypolidy led the lilly flower to become two distinct species. This is an example of A) melting that ended the "snowball Earth" period. B) Sympatric speciation C) allopatric speciation D) Directional selection E) origin of multicellular organisms.

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Polypolidy led the Lilly flower to become two distinct species. This is an example of Sympatric speciation. So, option B is accurate.

The scenario described, where polyploidy leads to the formation of two distinct species, is an example of sympatric speciation. Sympatric speciation occurs when new species emerge from a common ancestral species without the physical barrier of geographic isolation. Polyploidy refers to the condition where an organism has multiple sets of chromosomes, often resulting from errors during cell division. In plants, polyploidy can lead to reproductive isolation and the formation of new species within the same geographic area. In the case of the lily flower, the occurrence of polyploidy caused genetic divergence and reproductive barriers between the polyploid individuals and their diploid relatives, leading to the formation of two distinct species.

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Which of the following statements regarding highly efficacious agents is incorrect? abe They bind to the receptor and produce a response abe They must have a high affinity for the receptor abe They favour activation of the receptor abc They produce a large stimulus to the cell upon binding to the receptor abe They may give rise to the phenomenon of "spare receptors"

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The incorrect statement regarding highly efficacious agents is "abc They produce a large stimulus to the cell upon binding to the receptor."

Highly efficacious agents are substances that bind to receptors and produce a response. They must have a high affinity for the receptor, meaning they have a strong binding interaction. They favor activation of the receptor, meaning they promote the activation of downstream signaling pathways. Additionally, they may give rise to the phenomenon of "spare receptors," where even when a small fraction of receptors is occupied by the agonist, it can still produce a maximal response.

Highly efficacious agents do produce a response upon binding to the receptor, but the size of the stimulus or response is not necessarily related to their efficacy. Efficacy refers to the ability of an agent to activate the receptor and initiate a cellular response, but it does not determine the magnitude of the response. The magnitude of the response can be influenced by factors such as the downstream signaling pathways, cellular context, and presence of other modulating factors.

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Question 9 (3 points) Define "carrying capacity". Can the carrying capacity of a population change? Explain. A

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Carrying capacity is the maximum population that a particular ecosystem can sustain over a prolonged period under specific environmental conditions. It varies based on different factors like availability of resources, competition for resources, and other environmental factors. The carrying capacity of a population can change, depending on the changes in the environment.

For instance, if there is a decline in the availability of food or water, the carrying capacity would decrease, and if there is an increase in the availability of food and other resources, the carrying capacity would increase. The carrying capacity of a population can also change due to external factors like natural disasters, diseases, and human activities like deforestation, pollution, hunting, and climate change.

For example, if a forest that supports a particular population is destroyed, the carrying capacity of that ecosystem would decrease, and the population would decline. In conclusion, the carrying capacity of a population can change based on various internal and external factors that affect the ecosystem.

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Zoology experiment: The Predator-prey Interactions Between Zebrafish and Daphnia
1. Six 1-L beakers were filled with aged tap water.
2. To test the effect of light on the survival of Daphnia, the 6 beakers were divided equally into 2 treatments: light & dark. Beakers assigned to the dark treatment were covered w/ aluminum foil.
3. One zebrafish (about 2-3 cm) starved for 24 hours was placed in each beaker.
4. Fifty (50) Daphnia sp. individuals were added in each beaker containing the starved zebrafish. The top of the beakers assigned to the dark treatment were covered with aluminum foil.
5. One hour after, the zebra fish was scooped out & the no. of surviving Daphnia in each set-up were counted.
QUESTIONS:
1. What would be your hypothesis in this experiment?
2. What is your basis for formulating that hypothesis?
3. What do you think will happen to the survival rate of Daphnia when exposed to its predator under well-lit environment? In a completely dark set-up?

Answers

In this experiment, I hypothesize that the presence of a zebrafish predator (e.g. a starved zebrafish) will have a negative impact on the survival rate of Daphnia, which will be greater when exposed to light than in a completely dark set-up.

This is based on the fact that a well-lit environment will facilitate better visibility for the zebrafish, and thus higher predation efficiency. This is in contrast to a completely dark set-up, where the zebrafish will not be able to detect the Daphnia as easily, and so predation efficiency will be lower.

As the presence of zebrafish in the environment will effectively be a top-down control that determines the population size of Daphnia, it is likely that the observed change in the Daphnia’s survival rate will be greater when the zebrafish is experienced in a light environment, as opposed to a dark environment.

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Which statement regarding facultative anaerobes is true?
a. They can survive in the presence or absence of oxygen.
b. They require oxygen to survive.
c. They require the absence of oxygen to survive.
d. They cannot metabolize glucose.
e. They require carbon dioxide to survive.

Answers

Facultative anaerobes can survive in the presence or absence of oxygen.

The correct answer is (a) They can survive in the presence or absence of oxygen. Facultative anaerobes are microorganisms that have the ability to switch between aerobic and anaerobic metabolism based on the availability of oxygen. In the presence of oxygen, they can perform aerobic respiration to generate energy.

However, in the absence of oxygen, they can switch to anaerobic metabolism, such as fermentation, to produce energy. This versatility allows facultative anaerobes to survive and thrive in environments with varying oxygen levels, making them adaptable to different conditions.

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2. Explain why ampicillin acts as an functions in bacteria. antibiotic, and the mechanism whereby the ampi gene [2]

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Ampicillin is an antibiotic that acts by inhibiting bacterial cell wall synthesis. It belongs to the class of antibiotics called penicillins and specifically targets the enzymes involved in the construction of the bacterial cell wall.

The mechanism of action of ampicillin involves interfering with the transpeptidation step of peptidoglycan synthesis. Peptidoglycan is a crucial component of the bacterial cell wall responsible for maintaining its structural integrity. It consists of alternating units of N-acetylglucosamine (NAG) and N-acetylmuramic acid (NAM), cross-linked by short peptide chains. Ampicillin works by binding to and inhibiting the transpeptidase enzymes known as penicillin-binding proteins (PBPs). These enzymes are responsible for catalyzing the cross-linking of the peptide chains in peptidoglycan. In summary, ampicillin acts as an antibiotic by inhibiting bacterial cell wall synthesis through the inhibition of transpeptidase enzymes.

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Exposure of zebrafish nuclei to cytosol isolated from eggs at metaphase of mitosis resulted in phosphorylation of NEP55 and L68 proteins by cyclin-dependent kinase 2. NEP55 is a protein of the inner nuclear membrane, and Les is a protain of the nuclear lamina. What is the most lkely role of phosphorylation of thase proteins in the process of mintois? a. They are incolved in chromosome condensation b. They are involved in migration of centrospmes to coposite sides of the nucleus. c. They are involved in the disassembly of the nuclear envelope
d. They eriafie the anachment of apindle mierecutoules to knetochares

Answers

The phosphorylation of NEP55 and L68 proteins by cyclin-dependent kinase 2 in zebrafish is most likely involved in the disassembly of the nuclear envelope during mitosis.

The process of mitosis involves several key events, including the condensation of chromosomes, the migration of centrosomes to opposite sides of the nucleus, the disassembly of the nuclear envelope, and the attachment of spindle microtubules to kinetochores. Among the given options, the most likely role of the phosphorylation of NEP55 and L68 proteins is in the disassembly of the nuclear envelope.

NEP55 is a protein of the inner nuclear membrane, while L68 is a protein of the nuclear lamina. Phosphorylation of these proteins by cyclin-dependent kinase 2 suggests that they are targeted for modification during mitosis. Phosphorylation events are known to play a crucial role in regulating the disassembly of the nuclear envelope, allowing for the separation of the nuclear contents from the cytoplasm and facilitating chromosome segregation. Therefore, the phosphorylation of NEP55 and L68 proteins is likely involved in the disassembly of the nuclear envelope, which is a critical step in mitotic progression.

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Describe the property of lipids that makes them a better energy source than proteins or carbohydrates. Refer to bond energy in your description.

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Lipids are an excellent source of energy as they are the primary components of cellular membranes and carry out various functions in the human body. Lipids also have the highest energy density of all macronutrients and can generate more energy than carbohydrates or proteins per unit of weight.

Lipids are energy-dense due to the high number of carbon-hydrogen bonds that they contain. They also have lower levels of oxygen compared to carbohydrates and proteins, which means that they can generate more energy per molecule. The reason why lipids have more energy per molecule is that carbon-hydrogen bonds store more energy than oxygen-hydrogen bonds found in carbohydrates and proteins. As a result, when the body breaks down lipids, more energy is released than when carbohydrates and proteins are broken down.Lipids are also insoluble in water, and this property enables them to be stored in adipose tissues.

They can be broken down and released into the bloodstream to provide a long-lasting source of energy when there are no other energy sources available to the body. As a result, lipids can be stored for more extended periods and used by the body as an energy source when carbohydrates and proteins are not available.

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What is the mechanism of action of contraceptive pills? Describe
they interfere the uterine and ovarian cycles. Include: how do they
prevent ovulation? Pls don't copy paste from other chegg answers, I

Answers

Contraceptive pills contain synthetic estrogen and progesterone hormones that prevent ovulation and also alter the cervical mucus and lining of the uterus.

Contraceptive pills are used to prevent pregnancy. It contains synthetic estrogen and progesterone hormones which interfere with the ovarian and uterine cycles in females. It prevents ovulation by inhibiting the production of follicle-stimulating hormone (FSH) and luteinizing hormone (LH), which are responsible for the growth and maturation of follicles in the ovary. By doing so, the ovary does not release an egg, and therefore fertilization does not occur. Also, contraceptive pills thicken the cervical mucus, which makes it difficult for sperm to enter the uterus. If by chance the egg is released, the pills also alter the lining of the uterus, which makes it less receptive to the fertilized egg. Thus, the egg is not implanted, and pregnancy is avoided.Contraceptive pills contain synthetic estrogen and progesterone hormones that prevent ovulation and also alter the cervical mucus and lining of the uterus.

Contraceptive pills are highly effective in preventing pregnancy when taken correctly. It is essential to take them at the same time every day to ensure maximum protection. However, they do not protect against sexually transmitted infections (STIs).

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Why are the shape, orientation and location of the protein encoded by mc1r gene important in the fulfillment of its role?
Using the diagram below, describe the chain of events of protein synthesis of the MC1R protein. Starting from the mc1r gene (point A), indicate the molecules and details of the role of the process involved in each of the numbered steps 1-6.
Using the same diagram, describe the pathway which is triggered at point 7. Include in your answer the molecules and processes involved in each of the numbered steps 7-11.

Answers

The shape, orientation, and location of the protein encoded by the MC1R gene are important for its role because they determine the protein's functionality and interaction with other molecules. The specific shape of the protein allows it to bind to specific molecules, such as melanocyte-stimulating hormone (MSH), and activate signaling pathways involved in pigmentation regulation.

In protein synthesis (steps 1-6), the MC1R gene is transcribed into mRNA (step 1), which is then processed and transported out of the nucleus (step 2). The mRNA binds to ribosomes (step 3), and the ribosome reads the mRNA sequence to synthesize the corresponding amino acids (step 4). These amino acids are linked together to form a polypeptide chain (step 5), which folds into a specific 3D structure to become the MC1R protein (step 6).

In the pathway triggered at point 7, the MC1R protein interacts with MSH (step 7), leading to activation of the cAMP signaling pathway (step 8). This pathway activates enzymes, such as protein kinase A (PKA), which phosphorylate downstream proteins (step 9). Phosphorylated proteins initiate a series of cellular responses, such as the production of melanin, which determines skin and hair pigmentation (step 10). These responses ultimately lead to changes in pigmentation, such as tanning or red hair color (step 11).

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Cystic fibrosis (CF) is a monogenic, recessive disorder caused by a mutation in the CFTR gene. F is the symbol for the normal, dominant allele and f is the symbol for the recessive, CF-causing allele. Another trait, widow's peak, is dominant in humans. W is the "widow's peak" allele and w is the straight hairline allele. Imagine that a woman who has widow's peak, but her father did not, has children with a man who does not have widow's peak. Both the man and the woman are heterozygous at the CFTR locus. Famous actor Gary Cooper and his widow's peak. a. (2 pts) What are the genotypes and phenotypes of the woman and man? b. (2 pts) What are the odds of their having a girl with CF and widow's peak? c. (2 pts) If the couple has two children, what are the odds that they are both boys without CF, but with widow's peak?

Answers

a. The woman has the genotype Ww for widow's peak and Ff for the CFTR gene. Her phenotype is widow's peak (expressing the dominant W allele) and being a carrier for CF (not expressing the recessive f allele).

The man has the genotype ww for a straight hairline and Ff for the CFTR gene. His phenotype is a straight hairline (expressing the recessive w allele) and being a carrier for CF (not expressing the recessive f allele).

b. To determine the odds of having a girl with CF and widow's peak, we need to consider the inheritance of each trait separately.

For CF:

The woman is heterozygous (Ff) and the man is also heterozygous (Ff), which means they both carry the recessive CF-causing allele. The probability of passing on the recessive allele to a child is 1/4 for each parent. Thus, the probability of having a child with CF is (1/4) x (1/4) = 1/16.

For widow's peak:

The woman is heterozygous (Ww) and the man is homozygous recessive (ww). The dominant widow's peak allele (W) is always expressed when present. Therefore, all their children will have a widow's peak.

Combining the probabilities, the odds of having a girl with CF and widow's peak is (1/16) x 1 = 1/16.

c. If the couple has two children, the odds that they are both boys without CF, but with widow's peak can be calculated by considering each trait separately.

For CF:

The probability of having a child without CF is 3/4 for each child since both parents are carriers (Ff). Therefore, the odds of having two boys without CF is (3/4) x (3/4) = 9/16.

For widow's peak:

All their children will have a widow's peak since the woman is heterozygous (Ww). Therefore, the odds of having two boys with a widow's peak is 1 x 1 = 1.

Combining the probabilities, the odds that they have two boys without CF, but with a widow's peak is (9/16) x 1 = 9/16.

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GEFEL I 8 EE E C The structure shown in this image represents which part of a cell? Integral protein Integral protein Endoplasmic membrane Questions Filter (10) Y Pore Channel Polar head (hydrophilic)

Answers

The structure shown in this image below represents the Plasma membrane.

What is the plasma membrane?

Plasma membrane is the outer membrane of a cell. It is a phospholipid bilayer that separates the cell from its environment.

The plasma membrane is responsible for regulating the movement of substances into and out of the cell. It also plays a role in cell signaling and cell adhesion.

Integral proteins are proteins that are embedded in the membrane of a cell. They can be either transmembrane proteins, which extend all the way through the membrane, or peripheral proteins, which are attached to the surface of the membrane.

The above answer is based on the full question below;

The structure shown in this image represents which part of a cell? Pore Channel Integral protein Integral protein Polar head hydrophilic Fatty acid tal (hydrophobic)

A Nucleus

B) Lysosomes

C) Plasma membrane
D) Endoplasmic membrane

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Which of the following statements about venous blood pressure are true? Select ALL correct answers: A) Pressure in veins is much lower than pressure in arteries B) Arterial blood pressure fluctuates as the heart squeezes and relaxes, but venous blood pressure does not C) Valves in veins help prevent blood from flowing backward D) Contraction of skeletal muscles in the legs helps move venous blood back up towards the heart

Answers

All the following statements about venous blood pressure are true. Therefore, the correct options are: A, C, and D. Venous blood pressure is the pressure that is exerted by the circulating blood against the walls of veins, particularly the venae cavae and pulmonary veins.

There are various facts that surround this type of blood pressure and they include:

A. Pressure in veins is much lower than pressure in arteries: this statement is true. This is because the pressure in arteries is higher than that in veins. Arteries have a high-pressure rate that is due to the high volume of blood flowing through them in a short amount of time. Blood vessels act as conduits of blood circulation, and veins' thin walls help in the flow of low-pressure blood flow.

Therefore, statement A is correct.

B. Arterial blood pressure fluctuates as the heart squeezes and relaxes, but venous blood pressure does not: This statement is false. Venous blood pressure fluctuates when the heart squeezes and relaxes, and this is referred to as pulse pressure. However, venous pressure fluctuates more gently than arterial pressure. Therefore, statement B is incorrect.

C. Valves in veins help prevent blood from flowing backward: This statement is true. The venous system has valves that function to prevent the backward flow of blood and assist in the return of blood to the heart. The vein's wall contracts the blood and the valves in the veins block the blood from flowing backward. Therefore, statement C is correct.

D. Contraction of skeletal muscles in the legs helps move venous blood back up towards the heart: This statement is true. Venous blood returns to the heart through muscular contractions. The contraction and relaxation of the muscle's veins move the blood back to the heart, as the valves prevent blood from flowing backward. Therefore, statement D is correct.

The correct options that explain venous blood pressure are A, C, and D.

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If a population reaches the carrying capacity of the environment, O food and other resources will increase O the population will decline rapidly O unrestrained growth will occur O the population size

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If a population reaches the carrying capacity of the environment, the population size will fluctuate around this level (option d).

The carrying capacity of an environment is the maximum number of individuals of a particular species that an environment can support based on the resources available. If the population exceeds this carrying capacity, there may be a decline in resources, leading to a decrease in the population size. In contrast, if the population is below the carrying capacity, there may be room for growth until the carrying capacity is reached.

However, once the population reaches the carrying capacity, it is unlikely to continue to grow at the same rate. The availability of resources may fluctuate due to environmental factors such as weather patterns or natural disasters, causing the population to fluctuate in response. For example, if a drought occurs, there may be a decrease in the availability of water and food, leading to a decline in the population. Similarly, if there is an abundance of resources, the population may increase until it reaches the carrying capacity again.

Overall, once a population reaches the carrying capacity of the environment, the population size will fluctuate around this level due to the availability of resources and other environmental factors. It is important for populations to remain at or below the carrying capacity to ensure the continued health and survival of the species.

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The full question is given below:

If a population reaches the carrying capacity of the environment:

a. unrestrained growth will occur.

b. the population will decline rapidly.

c. food and other resources will increase.

d. the population size will fluctuate around this level.

Spemann organizer is a dorsal structure . Planar cell polarity involves interaction with ECM. Hedgehog and Wingless are both secreted factors that establish the anterior/posterior identity of parasegments in Drosophila.

Answers

The Spemann organizer is a dorsal structure involved in embryonic development. Planar cell polarity is a cellular process that includes interactions with the extracellular matrix (ECM). Hedgehog and Wingless are secreted factors that play roles in establishing the anterior/posterior identity of parasegments in Drosophila.

The Spemann organizer is a region in the developing embryo that plays a critical role in dorsal-ventral patterning. It was first identified by the German embryologist Hans Spemann and his student Hilde Mangold during their experiments with salamander embryos. The Spemann organizer is responsible for inducing the formation of the dorsal mesoderm and axial structures during embryonic development. It secretes various signaling molecules, such as Chordin and Noggin, which inhibit the activity of BMP (Bone Morphogenetic Protein) signaling and promote the development of dorsal structures.

Planar cell polarity refers to the coordinated orientation of cells within a tissue plane. It involves the alignment and polarization of cells along a specific axis, which is essential for the proper organization and function of tissues. Planar cell polarity is regulated by a complex network of signaling pathways and interactions with the extracellular matrix. The ECM provides cues and signals that guide the polarization of cells, influencing their orientation and behavior.

Hedgehog and Wingless are two secreted factors involved in establishing the anterior/posterior identity of parasegments in Drosophila, a commonly studied model organism in developmental biology. Hedgehog signaling pathway plays a key role in patterning the anterior/posterior axis by establishing concentration gradients of Hedgehog protein, which then activates target genes in a concentration-dependent manner. Wingless, also known as Wnt, is another signaling pathway that helps establish the anterior/posterior identity by regulating gene expression in specific regions of the embryo.

In summary, the Spemann organizer is a dorsal structure involved in embryonic development, planar cell polarity involves interactions with the extracellular matrix, and Hedgehog and Wingless are secreted factors that play important roles in establishing the anterior/posterior identity of parasegments in Drosophila.

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Assume the diameter of the field of vision in your microscope is 3 mm under low power. If on Bacillus cell is 3um, how many bacillus cells could fit end to end across the field? How many 20 um yeast cells could fit across the field? Show your work.

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Under low power with a field of vision diameter of 3 mm, approximately 1000 Bacillus cells could fit end to end across the field. This calculation is based on the assumption that the Bacillus cells are each 3 μm in size.

By dividing the diameter of the field (converted to micrometers) by the size of the Bacillus cell, we obtain the number of cells that can fit. In the case of yeast cells measuring 20 μm in size, the same calculation indicates that approximately 150 yeast cells could fit across the field.

It's important to note that these calculations assume a perfect arrangement of cells without any overlap or gaps, which may not be entirely accurate in real-world microscopy.

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Journal Review for: Phylogeny of Gekko from the Northern Philippines, and Description of a New Species from Calayan Island DOI: 10.1670/08-207.1
In terms of the molecular data
1. What type of molecular data was used? Describe the characteristic of the gene region used and how did it contribute to the findings of the study.
2. What algorithms were used in the study and how were they presented? If more than 1 algorithm was used, compare and contrast the results of the algorithms.
In terms of the morphological data
3. Give a brief summary of the pertinent morphological characters that were used in the study. How where they presented?
4. Phylogenetic studies are usually supported by both morphological and molecular data. In the journal assigned, how was the collaboration of morphological and molecular data presented? Did it create conflict or was it able to provide sound inferences?
Separate vs. Combined Analysis
5. Identify the substitution model utilized in the paper.
6. In the phylogenetic tree provided identify the support value presented (PP or BS). Why does it have that particular support value?
7. Did the phylogenetic analysis utilize separate or combined data sets? Explain your answer.

Answers

1. The type of molecular data used in the paper “Phylogeny of Gekko from the Northern Philippines, and Description of a New Species from Calayan Island” is mitochondrial and nuclear genes. The molecular phylogenetic analysis was based on 3469 base pairs of two mitochondrial genes (12S and 16S rRNA) and one nuclear gene (c-mos).

Mitochondrial DNA is generally used in phylogenetic analysis because it is maternally inherited and has a high mutation rate. In contrast, nuclear DNA evolves at a slower rate and is biparentally inherited.
2. In this paper, the maximum parsimony (MP) and Bayesian inference (BI) algorithms were used. MP was presented as a strict consensus tree, and BI was presented as a majority rule consensus tree. MP is a tree-building algorithm that seeks to minimize the total number of evolutionary changes (such as substitutions, insertions, and deletions) required to explain the data. In contrast, BI is a statistical method that estimates the probability of each tree given the data. It is known to be a powerful tool for inferring phylogenies with complex evolutionary models. In this study, the two algorithms produced similar topologies, suggesting that the tree topology is robust.
3. The morphological data used in the study included the number of scales around the midbody, the presence of a preanal pore, the number of precloacal pores, and the length of the fourth toe. These morphological characters were presented as a table that shows the values for each species.
4. In this study, both molecular and morphological data were used to infer the phylogeny of the Gekko species. The phylogenetic tree was based on the combined data set of molecular and morphological data, which was presented as a majority rule consensus tree. The combined analysis provided sound inferences, and there was no conflict between the two datasets.
5. The substitution model utilized in the paper was GTR+I+G. This is a general time reversible model that incorporates the proportion of invariable sites and a gamma distribution of rates across sites.
6. In the phylogenetic tree provided, the support value presented is PP (posterior probability). This particular support value was used because Bayesian inference was used to construct the tree. PP values range from 0 to 1 and indicate the proportion of times that a particular clade is supported by the data.
7. The phylogenetic analysis utilized combined data sets. The authors explained that the combined analysis is a powerful tool that can increase the accuracy and resolution of phylogenetic trees, especially when the datasets are not in conflict with each other.

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What is the risk that Optometry could pose to the public?
What could go wrong?
What dangerous substances/machines/tools/ techniques might be used?

Answers

Optometry, like any healthcare profession, carries certain risks that could potentially pose a threat to the public.

While the overall risk is relatively low, there are some potential concerns that should be addressed. One potential risk is misdiagnosis or incorrect prescriptions. Optometrists play a crucial role in assessing vision health and prescribing corrective measures such as glasses or contact lenses. If there are errors in the examination or prescription process, it could lead to suboptimal vision correction or even exacerbate existing eye conditions.

Another risk involves the improper use of medical instruments or equipment during eye examinations. For instance, incorrect handling or calibration of machines used for measuring intraocular pressure (tonometry) or examining the back of the eye (ophthalmoscopy) could result in inaccurate readings or potential harm to the patient.

Additionally, there is a risk of adverse reactions or complications related to certain substances used in optometric procedures. For instance, during eye examinations, eye drops containing dilating agents are sometimes used to facilitate examination of the retina. While adverse reactions to these eye drops are rare, there is a minimal risk of allergic reactions or other side effects.

It's important to note that optometrists undergo extensive training and follow strict protocols to mitigate these risks and ensure patient safety. Regular audits, quality control measures, and adherence to professional standards help minimize the chances of errors or dangerous situations arising.

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Other Questions
At the exit of an impeller with a backwards angle (82) of 20 the absolute flow velocity is 15 ms with a component of 3.1 m/s in the radio direction. If the rotation speed is 18 m/s, the slip factor will be O 0.870 0.642 O 0.703 O 0.590 O 0.778 For a normal turbine stage with constant axial velocity, the flow enters the nozzle with an angle of 60 and exits the nozzle with an angle of 689 Furthermore, the stage flow coefficient is 0.8. The stage reaction degree is O 0.714 0.675 O 0.792 0.684 O 0.703 A machine has a mass of 130 kg as shown in figure 1. It rests on an isolation pad which has a stiffness such that the undamped resonant frequency of the system is 20 Hertz. The damping ratio of the system is = 0.02. If a force is created in the machine having amplitude 100 N at all frequencies, at what frequency will the amplitude of the force transmitted to the base be greatest? What will be the amplitude of the maximum transmitted force? Neglect gravity. 1. Create the following operations in MATLAB to create signals over time (plot them): a. Rect(t/40).eb. u(t). ec. Cos(100nt) d. os (1000 t). -||2. Find the Fourier transform for the signals of point 1 and plot them. Are the computed transforms the same as those proposed in the theory? Analyze and conclude. Required information The state of stress at a point is x = -9 kpsi, Oy = 11 kpsi, = -19 kpsi, Txy = 6 kpsi, Tyz = 3 kpsi, and Tzx= -19 kpsi. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Determine the principal stresses. The principal normal stress 0 is determined to be___kpsiThe principal normal stress 02 is determined to be___kpsiThe principal normal stress 03 is determined to be___kpsiThe principal shear stress T1/2 is determined to be___kpsiThe principal shear stress T2/3 is determined to be___kpsiThe principal shear stress T1/3 is determined to be ___kpsi Explain in you own words why arteriosclerosis andatherosclerosis can lead to the development of heart diseases(*list what happens with EACH disease?) 7. Which neurons of the autonomic nervous system will slow the heart rate when they fire onto the heart? If input from those neurons is removed, how will the heart rate respond? (2 mark) How is the start codon aligned with the P-site in the prokaryotic initiation complex? O a. The Shine-Dalgarno sequence in the mRNA binds to the 16S rRNA of the 30S ribosomal complex, with the start codon aligning under the P- site. O b. IF-2 binds a GTP and an fMet-tRNA, with the tRNA anticodon base pairing with the start codon in the mRNA. O c. The mRNA is bound by a complex of initiation factors; one that binds the 5' cap, an ATPase/helicase, and a protein that binds to the poly(A)- binding proteins. O d. The 48S complex scans through the mRNA, starting at the 5' cap and reading through until the start codon aligns with the tRNA in the P-site. e. The second codon aligns base-pairs with IF-1 in the A-site. Which of the following is TRUE regarding translation in prokaryotes? O a. Which charged tRNA enters the ribosome complex depends upon the mRNA codon positioned at the base of the A-site. O b. Both RF1 and RF2 recognise all three stop codons. O c. The formation of the peptide bond is catalysed by an enzyme within the 50S subunit. d. Elongation factor G (EF-G) delivers an aminoacyl-tRNA to the A-site. e. The binding of elongation factor Tu (EF-Tu) to the A site displaces the peptidyl-tRNA and stimulates translocation. Clear my choice What must be true for DNA polymerase to work Select one or more: a. There must be a free 3 OH for it to attach nucleotides to. b. New nucleotides must be tri-phosphates c. hydrolysis of the bond between the first and second phosphate drives the polymerization reaction d. Continuous replication doesn't need an RNA primer Okazaki fragments only happen on one of the DNA X strands in a replication bubble (that's a fork going in both directions) 10.11 At f=100MHz, show that silver (=6.1107 S/m,r=1,r=1) is a good conductor, while rubber (=1015 S/m,r=1,r=3.1) is a good insulator. Which of the following "edge effects" is/are often associated with forest fragmentation of the Eastern Deciduous Foret? None of these are associated with this fragmentation. All of these are associated with this fragmentation. Reduction in population sizes of year-round residents that are attracted to habitat edges and nest in cavities due to competition with migrants. Mesopredator release and increased predation (e.g., on ground nests of birds) near forest edges.Increases in most ground-nesting birds that breed in the interior of forest fragments. A reduction in the population size of the Brown-headed Cowbird. What are the missing reagents used in the synthesis of this pharmaceutical intermediate? Draw a tRNA with the anticodon 3ACGUA5 Given wobble, what twodifferent codons could it bind to? Draw each codon on an mRNA,labeling all 5' and 3' ends, the tRNA, and the amino acid itcarries. 14. Explain how Snyder agar is both a selective and differential medium: 15. a. What is one way bacteria use sugar to produce dental caries? b. What type of growth environment do bacteria need to produce acid? What type of metabolism are they doing to produce acid? Which of the following is not a part of the positive feedback that drives the rising phase of the action potential? Select one: ONa+ channel gating Ob voltage-gated channels depolarization Od Na+ channel inactivation Identify whether the structure is part of the conducting division or the respiratory division. conducting division respiratory division trachea larynx nasal cavity primary bronchi respiratory bronchioles pharynx alveolar sacs tertiary bronchi Using named examples of genetic conditions explain the inheritance patterns of:i. a recessive autosomal conditionii. a dominant autosomal conditioniii. a sex-linked conditionYou should use genetic inheritance diagrams. The diagrams should give the genotypes and phenotypes of the parents and F1 zygotes, the gametes produced and the way that the gametes could combine during a monohybrid cross. Find the matrix \( A \) of the linear transformation \( T(f(t))=5 f^{\prime}(t)+8 f(t) \) from \( P_{3} \) to \( P_{3} \) with respect to the standard basis for \( P_{3},\left\{1, t, t^{2}\right\} \). the auditory ossicles transmit and amplify sound waves in the middle ear. in sequence, sound waves pass from: . If the two figures are congruent, which statement is true?A. BCDA FEHGB. ABCD EFGHC. BADC EFGHD. ADCB HGFE The new airport at Chek Lap Kok welcomed its first landing when Government Flying Service's twin engine Beech Super King Air touched down on the South Runway on 20 February 1997. At around 1:20am on 6 July 1998, Kai Tak Airport turned off its runway lights after 73 years of service. (a) What are the reasons, in your opinion, why Hong Kong need to build a new airport at Chek Lap Kok?