2.which of the following statements about glycolysis is wrong?
All the intermediates in glycolysis are phosphorylated
The sugar is phosphorylated twice during the preparation phase and both times the phosphate donor is an ATP
All the ATP molecules generated during the payoff phase are through substrate-level phosphorylation
The total energy yield from glycolysis is 2 Atp per glucose (4 ATP from the payoff phase - 2 ATp during the preparation phase), all things considered.
a. What metabolic fate(s) exist for glucose-6-phosphate?
(Check All That Apply)
It can enter the pentose phosphate pathway.
It can be used to synthesize glycogen.
It can be broken down through glycolysis.
The phosphate can be removed so that the sugar can be released into the bloodstream
b. What metabolic fate(s) exist for fructose-1,6-bisphosphate?
(Check All That Apply)
It can enter the pentose phosphate pathway.
It can be used to synthesize glycogen.
It can be broken down through glycolysis.
The phosphate can be removed so that the sugar can be released into the bloodstream

Answers

Answer 1

The incorrect statement about glycolysis is that all the ATP molecules generated during the payoff phase are through substrate-level phosphorylation.

The correct statement is that one ATP molecule is generated through substrate-level phosphorylation, while the remaining two ATP molecules are generated through oxidative phosphorylation.

The incorrect statement in the given options is that all the ATP molecules generated during the payoff phase of glycolysis are through substrate-level phosphorylation.

Substrate-level phosphorylation refers to the direct transfer of a phosphate group from a high-energy molecule to ADP to form ATP. However, in glycolysis, the final step of the pathway involves the conversion of phosphoenolpyruvate (PEP) to pyruvate, which generates one ATP molecule through substrate-level phosphorylation.

The other two ATP molecules in the payoff phase are produced through oxidative phosphorylation, where the high-energy electrons generated during glycolysis are transferred to the electron transport chain in the mitochondria, leading to the synthesis of ATP.

Regarding the metabolic fates of glucose-6-phosphate, it can undergo multiple pathways. It can enter the pentose phosphate pathway, where it is converted to ribose-5-phosphate, a precursor for nucleotide synthesis, or it can generate NADPH, an important reducing agent.

Glucose-6-phosphate can also be used to synthesize glycogen through the process of glycogenesis. Additionally, it can be further metabolized through glycolysis to generate energy.

The phosphate group attached to glucose-6-phosphate can also be removed by enzymes, allowing the release of glucose into the bloodstream.

As for fructose-1,6-bisphosphate, its metabolic fates include entering the pentose phosphate pathway, where it can be used to generate ribose-5-phosphate or NADPH.

It can also be utilized for glycogen synthesis through glycogenesis. Moreover, fructose-1,6-bisphosphate serves as a key intermediate in glycolysis and is broken down further to generate energy.

The phosphate group can be removed, leading to the release of fructose into the bloodstream. In summary, the incorrect statement is that all ATP molecules generated during the payoff phase of glycolysis are through substrate-level phosphorylation.

In reality, only one ATP molecule is produced through substrate-level phosphorylation, while the other two ATP molecules are generated through oxidative phosphorylation.

Glucose-6-phosphate can enter the pentose phosphate pathway, synthesize glycogen, undergo glycolysis, or have its phosphate group removed for the release of glucose.

Fructose-1,6-bisphosphate can enter the pentose phosphate pathway, be used for glycogen synthesis, undergo glycolysis, or have its phosphate group removed for the release of fructose.

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Related Questions

Question 1
What is the osmotic fragility test and what does it assess?
How does the flow cytometric osmotic fragility test determine hereditary spherocytosis?
What is osmotic gradient ektacytometry and how can it be used to diagnose inherited RBC membrane disorders? Be sure to include a discussion around what the terms Omin, Elmax and Ohyp are and how they can be used to determine hereditary spherocytosis, hereditary elliptocytosis and Southeast Asian ovalocytosis (pictures may assist you here).

Answers

Osmotic fragility test is a laboratory test that is used to determine the ability of erythrocytes (red blood cells) to swell or shrink depending on the osmotic environment.

This test is important in the diagnosis of hemolytic anemias as it assesses the integrity of the RBC membrane. What is the osmotic fragility test? The osmotic fragility test assesses the rate at which red blood cells break down (hemolysis) under different degrees of saline (salt) concentration. It is a diagnostic test that is performed on a blood sample to identify and evaluate various hemolytic conditions.

The test is based on the fact that red blood cells undergo hemolysis when they are placed in hypotonic solutions that cause them to swell and eventually burst. How does flow cytometric osmotic fragility test determine hereditary spherocytosis? The flow cytometric osmotic fragility test determines the degree of osmotic fragility of red blood cells.

The test helps to determine the degree of hemolysis in hereditary spherocytosis patients and can also help in the diagnosis of other forms of hemolytic anemia. In this test, the red blood cells are exposed to varying degrees of osmotic pressure and the degree of hemolysis is measured.

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a. Describe the 'gain of function' experiments performed with highly pathogenic avian influenza virus H5N1. (5 marks) b. What are three reasons people have provided as to why these experiments should not have been performed. c. Would it be scientifically valid to perform similar experiments for SARS-CoV-2?

Answers

It is not scientifically valid to perform similar experiments with SARS-CoV-2 as it poses a risk of accidental release, dual-use concerns, and ethical concerns. SARS-CoV-2 is a highly infectious virus that has already caused a global pandemic.

a. Gain of function experiments are experiments where researchers increase the transmissibility or virulence of pathogens to understand how they work and how they can better prepare for and prevent outbreaks. Highly pathogenic avian influenza virus H5N1 (HPAI H5N1) is a deadly influenza virus that has shown evidence of human-to-human transmission. Gain of function experiments have been performed with HPAI H5N1 to study its behavior and characteristics. The experiments have been carried out to identify genetic changes that allow the virus to become more transmissible and/or more virulent. The researchers were able to identify specific genetic changes that allow the virus to spread more easily and quickly between birds. However, the experiments have also raised concerns about the potential for accidental release of the virus and the potential for misuse.

b. Three reasons why gain of function experiments with HPAI H5N1 should not have been performed include:1. Safety concerns: The experiments were conducted in high-level biosafety laboratories, but there is always the potential for accidental release or escape of the virus. If the virus were to escape, it could cause a pandemic, and it could be difficult to contain.2. Dual-use concerns: Dual-use concerns refer to the potential for the research to be used for harmful purposes.

c. It is not scientifically valid to perform similar experiments with SARS-CoV-2 as it poses a risk of accidental release, dual-use concerns, and ethical concerns. SARS-CoV-2 is a highly infectious virus that has already caused a global pandemic. Performing gain of function experiments with this virus could make it even more infectious or more lethal. The risks associated with these experiments are significant, and the potential benefits are uncertain. Instead, scientists should focus on studying the virus and developing vaccines and treatments to prevent and treat COVID-19.

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Which of the following is true about chimpanzee tool use?
Chimpanzees use tools mostly for acquiring food
Chimpanzees depend on tool use for survival, as it is widespread across Africa
Although there is evidence it exists, chimpanzee material culture has not been observed
Whereas most primates make tools, chimpanzees do not

Answers

The answer that is true about the use of tools by chimpanzees is that chimpanzees use tools mostly for acquiring food.

Chimpanzees are one of the few primates who use tools extensively. They use tools mostly for acquiring food. Researchers have documented chimpanzees using sticks to extract termites and ants from their nests. Additionally, they have been known to use stones to crack open nuts.

Tool use among chimpanzees is so common that it's regarded as a cultural trait among some populations. There is evidence that chimpanzee material culture exists, however, it is often difficult to observe in the wild.

Some researchers believe that tool use in chimpanzees is so widespread that it can be considered a survival trait, although this is still being debated.

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Papineau argues that the ability to form long-term intentions is one of the features that distinguishes humans from other animals. a.True b.False

Answers

The statement "Papineau argues that the ability to form long-term intentions is one of the features that distinguishes humans from other animals" is True.

What are long-term intentions?

The future-oriented intentions that the individuals have and that guide them to realize their long-term plans and goals are known as long-term intentions. Long-term plans necessitate a certain level of mental proficiency, such as the ability to think ahead, engage in goal-directed behavior, and act accordingly.

Papineau is a Canadian philosopher who is known for his work on the philosophy of mind, philosophy of science, and metaphysics. He argues that the ability to form long-term intentions is one of the features that distinguishes humans from other animals.

Papineau argues that one of the essential things that differentiate humans from other animals is the ability to plan for the future and to act accordingly. He argues that this ability is closely linked to the ability to form long-term intentions.

Other animals may make short-term plans or have immediate intentions, but they don't have the ability to think ahead and plan for the future like humans do. Therefore, the given statement is true.

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Question 31 (1 point) Alex stopped his pack-a-day smoking habit last year. He now says he can taste his food much better and everything seems more flavorful. Which of the following structures is allowing Alex to specifically taste his food? O A) Taste pore OB) Basal cells O C) Papilla OD) Gustatory cell O E) Support cells Question 38 (1 point) Which of the following articulates directly with the skull? OA) A) Axis B) Atlas OC) Vertebral prominens OD) Coccyx O E) Transverse foramen Question 39 (1 point) Which of the following is TRUE? A) Gray matter is thickest in the cervical and lumbar regions because of the cervical and lumbar enlargements B) Ascending and descending tracts are located peripherally in the gray matter OC) The ratio of gray to white matter is the same throughout of the spinal cord D) Gray matter is located peripherally while the white matter is located centrally E) There are 7 cervical spinal nerves

Answers

Alex stopped his pack-a-day smoking habit last year, and he now says he can taste his food much better and everything seems more flavorful. Gustatory cells are the structures.

These are specialized cells found in taste buds that detect and respond to taste stimuli. Gustatory cells are primarily found in the papillae. The tiny bumps on the tongue's surface. The cells transmit signals to the brain about the presence and intensity of different tastes.

The papillae on the tongue's surface are also responsible for the sensation of texture in foods and drinks.The Atlas articulates directly with the skull. The Atlas is the first cervical vertebra, and it is so named because it holds up the skull, just as the mythical Atlas supported the heavens on his shoulders.

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Select all correct answers. In the streak-plate technique A. a wire loop is used to streak the mixed culture on an agar plate B. the wire loop is sterilized between each of the series of streaks C. melted agar is poured into a petri plate D.streaks are made in series to cover as much of the agar surface as possible E. serial dilutions are made

Answers

The A, B, and D. The streak-plate technique is used to isolate pure cultures of microorganisms by using a wire loop to streak a mixed culture onto an agar plate. The procedure includes the following steps:Step 1: Sterilize the wire loop by heating it until it glows red in a flame.

Step 2: Collect a small amount of the mixed culture on the wire loop.Step 3: Streak the mixed culture on the agar plate using the wire loop, starting from the center and working outwards in a series of streaks.Step 4: Sterilize the wire loop by heating it in a flame between each of the series of streaks.Step 5: Turn the plate 90 degrees and repeat the streaking process using the same wire loop, starting from the end of the first series of streaks.Step 6: Repeat the process a third and fourth time, always sterilizing the wire loop between each series of streaks.

This results in a series of streaks that cover the surface of the agar plate and gradually dilute the concentration of the mixed culture.Step 7: Incubate the plate at the appropriate temperature and observe the growth of colonies. By using the streak-plate technique, it is possible to isolate pure cultures of microorganisms from mixed cultures and study their characteristics.

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Central Dogma Problem Solving. In the given strand, do the transcription and then translation to determine the polypeptide product, or if there is no product. Note: the starting codon is AUG for Methionine (Met). (5 points each, 10 points total) a. 3'ATGCTGCAAGCGTCGGATGAGCTAGACTGCAGTCGATGACCGAGCCGTAGCTAGS
b. 3'GCAACGATGGGTACCACGTGGACTGAGGACTCCTCACTTAGS

Answers

The central dogma is a principle of molecular biology that states that the information present in nucleic acids is converted into the sequence of amino acids of proteins through a process of gene expression.

According to the given problem, the transcription and translation of the given strands will determine the polypeptide product or if there is no product. a) Transcription: 5'-AUG CUG CAA GCG UCG GAU GAG CUA GAC UGC AGU CGA UGA CCG AGC CGU AGC UAG-3'Translation: AUG - Met; CUG - Leu; CAA - Gln; GCG - Ala; UCG - Ser; GAU - Asp; GAG - Glu; CUA - Leu; UGC - Cys; AGU - Ser; CGA - Arg; UGA - Stop. The polypeptide product would be Methionine-Leucine-Glutamine-Alanine-Serine-Aspartic acid-Glutamic acid-Leucine-Cysteine-Serine-Arginine.

b) Transcription: 5'-GCA ACG AUG GGU ACC ACG UGG ACU GAG GAC UCC UCA CUU AG-3'Translation: AUG - Met; GGU - Gly; ACC - Thr; ACG - Thr; UGG - Trp; ACU - Thr; GAG - Glu; GAC - Asp; UCC - Ser; UCA - Ser; CUU - Leu; AG - Stop. The polypeptide product would be Methionine-Glycine-Threonine-Threonine-Tryptophan-Threonine-Glutamic acid-Aspartic acid-Serine-Serine-Leucine-Stop.

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Discuss how Na+-K+ pump in nerve cells facillitate transmission of electrical impulse to and from your brain

Answers

The Na+-K+ pump is a type of active transport that is responsible for maintaining the gradient of the two vital ions in the nerve cells, Na+ and K+.

It moves three Na+ ions out of the cell and two K+ ions inside the cell against their respective gradients. It requires the use of ATP as an energy source. The Na+-K+ pump is important for the transmission of electrical impulses in nerve cells.The transmission of electrical impulses in nerve cells is facilitated by the Na+-K+ pump, which keeps the concentration gradient of Na+ and K+ ions balanced across the plasma membrane. Electrical impulses in nerve cells are conveyed by changes in the membrane potential of neurons, which are made up of ion channels that allow ions to flow across the plasma membrane.The Na+-K+ pump is essential for maintaining the ionic balance that generates the resting membrane potential in the nerve cells. It helps to establish the electrochemical gradient for Na+ and K+ ions that contribute to the generation of action potentials. When an electrical impulse is sent from the brain to the peripheral nerve, the depolarization of the neuron's membrane allows Na+ ions to enter the cell. This generates a positive charge that spreads to the adjacent regions of the neuron. The propagation of the action potential along the axon of the neuron is facilitated by the Na+-K+ pump, which ensures that the ionic balance is maintained.Overall, the Na+-K+ pump plays a crucial role in the transmission of electrical impulses to and from the brain. It helps to establish the resting membrane potential of the nerve cells and contributes to the generation of action potentials that enable the transmission of electrical impulses along the neurons.

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From the point of view of evolution, why is sexual reproduction
an advantage for the continuity of species?

Answers

Sexual reproduction is beneficial for the continuity of species from the point of view of evolution because it promotes genetic variation among offspring. During sexual reproduction, gametes (sperm and egg cells) combine to produce offspring with unique combinations of genetic material from both parents.

As a result, each offspring is genetically distinct from its parents and siblings. This genetic diversity allows for increased adaptability to changing environmental conditions and better chances of survival, increasing the continuity of the species.

Furthermore, the genetic diversity that arises from sexual reproduction allows for the selection and propagation of advantageous traits and the elimination of harmful ones. Over time, this can lead to the evolution of new species that are better adapted to their environments.

In contrast, asexual reproduction produces genetically identical offspring that lack the variability necessary for natural selection and adaptation. Thus, sexual reproduction is an essential evolutionary advantage for the continuity of species.

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What and where are the different types of neurons that directly
innervate the muscles

Answers

The different types of neurons that directly innervate muscles are called motor neurons. They are located in the spinal cord and the brainstem.

Motor neurons are specialized nerve cells that transmit signals from the central nervous system (CNS) to the muscles, enabling voluntary movement and muscle control. There are two main types of motor neurons: upper motor neurons and lower motor neurons. Upper motor neurons are located in the motor cortex of the brain and transmit signals down to the lower motor neurons. They form connections with lower motor neurons in the brainstem and spinal cord. Lower motor neurons, also known as alpha motor neurons, are located in the ventral horn of the spinal cord. They directly innervate skeletal muscles, sending signals from the spinal cord to the muscles to initiate muscle contractions. The synapses formed between lower motor neurons and muscle fibers are called neuromuscular junctions. At these junctions, neurotransmitters, such as acetylcholine, are released by the motor neurons, binding to receptors on the muscle fibers and triggering muscle contraction. In summary, motor neurons, specifically lower motor neurons, are responsible for directly innervating muscles and initiating muscle contractions. They are located in the spinal cord and play a crucial role in the voluntary control of movement.

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HELP PLEASE
2. What are the five (5) types of ELISA that are mentioned in the video?
3. What are two proteins that can be used to block the surface of the well to prevent the antibody from binding directly to the bottom of the well?
4. How are the primary antibodies generated? 5. Give two (2) reasons why Indirect ELISA is advantageous over Direct ELISA?

Answers

ELISA utilizes various assay types, blocking proteins, immunized primary antibodies, and indirect ELISA for signal amplification and versatility. Therefore :

(2) The five common types of ELISA mentioned are direct, indirect, sandwich, competitive, and reverse ELISA.

(3) Bovine Serum Albumin (BSA) and casein are commonly used proteins to block the well surface in ELISA.

(4) Primary antibodies are generated through immunization of animals and collecting and purifying the antibodies from their blood.

(5) Indirect ELISA is advantageous due to signal amplification and its versatility in using the same secondary antibody with different primary antibodies.

(2.) The five types of ELISA mentioned in the video can vary depending on the specific context, but here are five common types:

a. Direct ELISA: This involves directly labeling the primary antibody with an enzyme, allowing for the detection of the antigen without the need for secondary antibodies.

b. Indirect ELISA: This uses a primary antibody to bind to the antigen, followed by a secondary antibody labeled with an enzyme that recognizes and binds to the primary antibody.

c. Sandwich ELISA: This involves capturing the target antigen between two antibodies—the capture antibody coated on the plate and the detection antibody labeled with an enzyme.

d. Competitive ELISA: In this type, a known labeled antigen competes with the unlabeled antigen present in the sample for binding to a limited amount of specific antibodies.

e. Reverse ELISA: This is used to detect the presence of specific antibodies in a sample by immobilizing the antigen instead of the antibody.

(3.) Two proteins commonly used to block the surface of the well in ELISA to prevent non-specific binding of antibodies are:

a. Bovine Serum Albumin (BSA): BSA is often used to coat the well surface because it helps to block any non-specific binding sites, reducing background noise.

b. Casein: Casein is another protein that can be used as a blocking agent to prevent non-specific binding. It forms a protective layer on the well surface, minimizing unwanted interactions.

(4.) Primary antibodies are generated through immunization. Animals, such as rabbits or mice, are injected with the specific antigen of interest. The immune system of the animal recognizes the antigen as foreign and produces antibodies to neutralize it. These antibodies can then be collected from the animal's blood and purified for use as primary antibodies in ELISA or other immunoassays.

(5.) Two reasons why Indirect ELISA is advantageous over Direct ELISA are:

a. Signal amplification: Indirect ELISA allows for signal amplification because multiple secondary antibodies can bind to a single primary antibody. This amplifies the detection signal, increasing sensitivity.

b. Versatility: Indirect ELISA is more versatile because the same secondary antibody can be used with different primary antibodies. This saves time and resources as only the primary antibody needs to be changed, while the secondary antibody can remain constant.

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Select two viral infections of adults with potentially serious health consequences and compare and contrast them in terms of virus structure, mode of spread, disease characterization and possible preventative measures

Answers

Two viral infections that can have serious health consequences in adults are influenza (flu) and human immunodeficiency virus (HIV).

Influenza, caused by the influenza virus, is a respiratory infection that primarily affects the nose, throat, and lungs. The influenza virus belongs to the Orthomyxoviridae family and has a segmented RNA genome surrounded by an envelope. It is spread through respiratory droplets when an infected person coughs or sneezes.

Influenza is characterized by symptoms such as high fever, cough, sore throat, muscle aches, fatigue, and headache. It can lead to severe complications, particularly in older adults and those with underlying health conditions.

To prevent influenza, annual vaccination is recommended, as well as practicing good respiratory hygiene, such as covering the mouth and nose when coughing or sneezing, and frequent handwashing.

On the other hand, HIV is a viral infection caused by the human immunodeficiency virus. HIV belongs to the Retroviridae family and has an RNA genome and an envelope. It is primarily transmitted through unprotected sexual intercourse, sharing contaminated needles, or from mother to child during childbirth or breastfeeding. Unlike influenza, HIV primarily affects the immune system, specifically targeting CD4 T-cells.

This leads to a gradual weakening of the immune system, making individuals more susceptible to opportunistic infections and cancers. HIV infection progresses to acquired immunodeficiency syndrome (AIDS) if left untreated. Prevention measures for HIV include practicing safe sex, using sterile needles, and implementing strategies such as pre-exposure prophylaxis (PrEP) for high-risk individuals and antiretroviral therapy (ART) for individuals living with HIV.

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The dilator muscles of the iris are responsive to which of the
following:
A.
Parasympathetic nervous system
B.
Sympathetic nervous system
C.
Bright light
D.
Lack of light
E.
B

Answers

The dilator muscles of the iris (colored part of the eye that controls the size of the pupil) are responsive to the sympathetic nervous system.

The iris is the colored part of the eye that controls the size of the pupil, which is the opening through which light enters the eye. The muscles responsible for dilating the pupil are called the dilator muscles. The activity of these muscles is regulated by the autonomic nervous system.

The sympathetic nervous system, which is a branch of the autonomic nervous system, controls the dilation of the pupil. When the sympathetic nerves are activated, they cause the dilator muscles of the iris to contract, resulting in the enlargement of the pupil. This response is known as mydriasis.

On the other hand, the parasympathetic nervous system, also a branch of the autonomic nervous system, controls the constriction of the pupil. When the parasympathetic nerves are activated, they cause the circular muscles of the iris, called the sphincter muscles, to contract, resulting in the narrowing of the pupil. This response is known as miosis.

Bright light and lack of light are environmental stimuli that can indirectly influence the activity of the iris muscles. Bright light causes the pupils to constrict as a protective mechanism to limit the amount of light entering the eye, while in darkness, the pupils dilate to allow more light in for better vision. However, the direct control of the dilator muscles is mediated by the sympathetic nervous system.

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Rhizomes are?
a. a modified underground plant stem that sends out roots and shoots from seagrass nodes b. a modified underground holdfast that sends out roots and shoots from nodes of macroalgae c. the above-ground portion of seagrasses d. the above-ground portion of marine macroalage

Answers

Rhizomes are modified underground plant stems that serve as a means of vegetative propagation. The correct answer is option a.

They are horizontally oriented and grow underground, producing roots and shoots from their nodes. Rhizomes are commonly found in various plant species and serve multiple purposes. They enable plants to spread horizontally, allowing for the colonization of new areas and the formation of extensive clonal colonies.

Rhizomes also store nutrients and energy reserves that aid in the plant's survival and regrowth. Examples of plants that utilize rhizomes include bamboo, ginger, and iris. Through their ability to produce roots and shoots from nodes, rhizomes play a vital role in the growth, reproduction, and expansion of plant populations.

The correct answer is option a.

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A 27-year old male seen in the family practice office is found to have an elevated PT, with a normal APTT. Platelet count is 220,000/microliter. Bleeding time is 6 minutes. Which of the following factor deficiencies is suggested? O A. V OB. VII OC. VIII OD.X The following laboratory date were obtained from a 14-year old male with a history of abnormal bleeding: • PT: 13 seconds • APTT: 98 seconds • Factor VIII Activity: markedly decreased • Platelet Count 153,000 • Bleeding Time: 7 minutes • Platelet Aggregation . ADP: normal • EPl: normal . Collagen: normal Ristocetin: normal Which of the following disorders does this patient most likely have? A. hemophilia A B. von Willebrand's disease C. hemophilia B D.factor VII deficiency A citrated plasma specimen was collect at 7:00 am and prothrombin time results were released. At 3:00 pm, the physician called the lab and requested that an APTT be performed on the same sample. The technician should reject this request due to which of the following? A. the APTT will be prolonged due to increased glass contact factor OB. the APTT will be decreased due to the release of platelet factors OC. the APTT will be prolonged due to the loss of factor V and/or VIII OD. the APTT will be prolonged due to the loss of factor VII

Answers

A 27-year-old male seen in the family practice office is found to have an elevated PT, with a normal APTT. Platelet count is 220,000/microliter. Bleeding time is 6 minutes.

The most likely factor deficiencies suggested are Factor VII deficiency (D) or Factor X deficiency (OD).Factor VII and Factor X are both factors within the extrinsic pathway. Both are dependent on Vitamin K. Intrinsic pathways rely on Factors VIII, IX, XI, and XII, all of which are dependent on Hageman Factor or Factor XII.

The given laboratory data of a 14-year-old male with a history of abnormal bleeding suggests Von Willebrand's disease. In patients with Von Willebrand's disease, the primary symptoms are usually those of a mucous membrane type, which includes easy bruising, epistaxis, and menorrhagia.

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I need you to show me step by step how to do the problem
In a population of 100 individuals, 49 percent are of the NN blood type. What percentage is expected to be MN assuming Hardy-Weinberg equilibrium conditions?
In a population of 100 individuals, 49 percent are of the NN blood type. What percentage is expected to be MN assuming Hardy-Weinberg equilibrium conditions?
51 percent
21 percent
42 percent
9 percent
There is insufficient information to answer this question.

Answers

the best answer would be the closest one, which is 42 percent.

The expected percentage of the MN blood type in a population with 100 individuals and a 49 percent frequency of the NN blood type is 42 percent. This is the answer that fits the multiple-choice options provided. To show how this answer was derived, the Hardy-Weinberg equation can be used.

Hardy-Weinberg equationp² + 2pq + q² = 1, where:

p² = frequency of the homozygous dominant genotype (NN)

2pq = frequency of the heterozygous genotype (MN)

q² = frequency of the homozygous recessive genotype (MM)

p = frequency of the dominant allele (N)

q = frequency of the recessive allele (M)

To solve for the frequency of the MN genotype, 2pq must be calculated. First, the frequency of the NN genotype can be determined as:

p² = 0.49 (given)

Then the frequency of the q allele can be found by taking the square root of q²:

q² = 1 - p²q² = 1 - 0.49q = √(1 - 0.49)q = 0.63

Finally, the frequency of the MN genotype can be calculated as:

2pq = 2 × 0.63 × 0.51pq ≈ 0.64, or 64%

However, the question specifies that the population is in Hardy-Weinberg equilibrium conditions, meaning that the frequency of each allele and genotype is not changing over time.

This means that the percentage of MN individuals in the population will be equal to the frequency of the MN genotype calculated above, which is approximately 64%. Since this option is not among the answer choices, the best answer would be the closest one, which is 42 percent.

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Damage to the fusiform gyrus leads to a condition in which people are unable to recognize familiar faces (sometimes even their own), called

Answers

The pathogen or antigen's entry into a Peyer's patch via a M cell, a series of events that lead to the generation of pathogen/antigen-specific IgA antibodies in the effector compartment of a mucosal tissue can be summarised as follows:

1. Antigen uptake: An M cell in the mucosal epithelium of the intestinal lining is where the pathogen or antigen enters the Peyer's patch. M cells are specialised cells that move antigens from the intestine's lumen to the lymphoid tissue beneath.

2. Antigen presentation: Once inside the Peyer's patch, specialised antigen-presenting cells known as dendritic cells (DCs) take the antigens up. In the Peyer's patch, T cells get the antigens from DCs after being processed.

3. T cell activation: The given antigens stimulate CD4+ T cells, which arethe most common type of T cells.

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True or False: A clear temporal relationship between exposure
and disease is an advantage of cross sectional studies.
Group of answer choices
A. True
B. False

Answers

The statement "a clear temporal relationship between exposure and disease is an advantage of cross sectional studies" is false.

A clear temporal relationship between exposure and disease is not an advantage of cross-sectional studies. Cross-sectional studies are observational studies that assess the relationship between exposure and disease at a specific point in time. They are designed to gather data on exposure and disease prevalence simultaneously, but they do not establish a temporal sequence between exposure and disease.

In cross-sectional studies, researchers collect data from a population or sample at a single time point, without following the participants over time. Therefore, they cannot determine the temporal sequence of events, such as whether the exposure preceded the disease or vice versa. Cross-sectional studies are mainly used to estimate disease prevalence, examine associations between exposure and disease, and generate hypotheses for further research.

To establish a clear temporal relationship between exposure and disease, longitudinal studies or experimental studies such as randomized controlled trials (RCTs) are typically conducted. Longitudinal studies follow participants over an extended period, allowing for the assessment of exposure status before the development of the disease outcome.

RCTs, on the other hand, involve random allocation of participants to different exposure groups, allowing researchers to observe the effects of exposure on disease development over time.

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Which list is the correct list for the following results: endospore former, positive acid-fast stain, and gram negative bacilli? a. Bocillus subtilis, Mycrobacterium smegmatis, and Escherichia coli b. Bacillus subtilis, Mycobacterium smegmatis, and Escherichia coli Mycobacterium smegmatis, Bacillus subtilis, and Escherichia coli d. Bacillus subtilis, Mycobacterium smegmatis, and Escherichia coli alldelar hair

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The list that represents the correct list for the following results: endospore former, positive acid-fast stain, and gram negative bacilli is option c. Mycobacterium smegmatis, Bacillus subtilis, and Escherichia coli. Hence option C is correct.

Endospores are a dormant and non-reproductive form of bacteria that withstands environmental pressure in the Bacillus and Clostridium genera. They can stay dormant in soil, air, and water for years before they experience favorable conditions to germinate again.Positive acid-fast stainThis result is shown by a few species of bacteria, like Mycobacterium, which have an extra-thick cell wall that can resist stain decolorization by an acid-alcohol mixture following staining with basic dyes such as methylene blue. It also implies that it cannot be identified by the Gram stain procedure.

A gram-negative bacillus is a type of bacteria that is commonly found in the human body and is often responsible for infections. Bacteria in the bacillus genus are long and thin, with a rod-like form. They are gram-negative, meaning they do not retain the crystal violet stain and appear pink or red in the Gram staining procedure. Gram-negative bacilli are a category of bacteria that cause a variety of diseases.

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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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Fish have specialized organs known as gills that allow them to obtain oxygen from water. The gills in fish are designed to increase oxygen uptake efficiency and minimize blood pressure. This is because gill capillaries in fish are fragile, and high blood pressure could result in rupture, causing the fish to suffocate.

The oxygen delivery to fish is affected by the low blood pressure that is required to preserve the fragile capillaries in the gills. The lower blood pressure in fish leads to a lower oxygen supply to the tissues, which affects the metabolic rate of fish.The metabolic rate of fish is the rate at which the fish utilizes oxygen and nutrients to produce energy for physiological processes such as growth, reproduction, and movement. Therefore, fish with lower oxygen supply have lower metabolic rates and are usually less active compared to fish with higher oxygen supply.Besides, low oxygen supply in fish could lead to changes in behavior, such as a decrease in feeding, which can lead to a decline in growth and survival.

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2. Name the equilbrium potential(s) that mainly determine the resting membrane potential of a cell? (1 point) 3. How does increasing K+ concentration in the affect (a) the equilibrium potential of K+

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The resting membrane potential of a cell is primarily determined by the equilibrium potentials of potassium . Increasing the concentration of  in the extracellular fluid will affect the equilibrium potential of .

The resting membrane potential of a cell is the electrical potential difference across the cell membrane when the cell is at rest. It is mainly determined by the equilibrium potentials of various ions. In most cells, including neurons, the resting membrane potential is primarily influenced by the equilibrium potential of potassium . This is because the membrane is more permeable to  compared to other ions.

The equilibrium potential of  is determined by the Nernst equation, which takes into account the concentration gradient of  across the membrane. An increase in extracellular  concentration will raise the concentration gradient, resulting in a higher equilibrium potential of K+. In other words, the resting membrane potential of the cell will shift towards a more positive value.

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6 1 point Choose the following options which indicate pleiotropy: A mutant allele at one locus X creates mice with brown fur, while an allele at locus Y creates mice with red eye color. When mice are

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The options that indicate pleiotropy in this scenario are: "A mutant allele at one locus X creates mice with brown fur" and "an allele at locus Y creates mice with red eye color."

Pleiotropy refers to a genetic phenomenon where a single gene or allele influences multiple, seemingly unrelated traits or phenotypes. In the given scenario, the following options indicate pleiotropy:

"A mutant allele at one locus X creates mice with brown fur."This suggests that a mutation at locus X affects both the color of the mouse's fur and potentially other traits."An allele at locus Y creates mice with red eye color."This indicates that an allele at locus Y influences the color of the mouse's eyes, which is a distinct trait from the fur color affected by locus X.

By having different alleles at these loci (X and Y), the mice exhibit different phenotypes for both fur color and eye color. This demonstrates the concept of pleiotropy, where a single gene or allele can have multiple effects on the organism's traits.

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If you could make chemicals that can prevent transcription regulators from functioning and you want to stop root growth, then which transcription regulator would you inhibit with a chemical? O WUS CLV3 BRC1 WOX5

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Transcription regulators are proteins that control gene expression by regulating the transcription of genes. If a chemical that can prevent transcription regulators from functioning is made and is used to stop root growth, then the transcription regulator that would be inhibited with this chemical is WOX5.

WOX5 (WUSCHEL-RELATED HOMEOBOX 5) is a transcription factor that plays a vital role in the growth of plant roots. WOX5 acts as a transcriptional regulator and binds to the DNA to activate or inhibit gene expression. WOX5 is expressed in the quiescent center (QC), which is a group of cells located at the tip of plant roots.

The QC is responsible for maintaining the stem cell population in the root and is essential for root growth. WOX5 plays a critical role in root growth by regulating the differentiation of stem cells into specific cell types. If the function of WOX5 is inhibited, then the differentiation of stem cells is affected, and root growth is stopped.

Therefore, to stop root growth, a chemical that can prevent the functioning of transcription regulators should be developed to inhibit WOX5.

Answer: To stop root growth, the transcription regulator that would be inhibited with a chemical is WOX5.

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Immunological memory consists of memory B cells that secrete IgM only. memory Th2 cells only. memory phagocytes. both Memory B cells and memory T cells of all types. Treg cells.

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Immunological memory comprises memory B cells that secrete only IgM and memory T cells of all types, including Th2 cells and Treg cells. Additionally, memory phagocytes play a role in immunological memory.

Immunological memory is a crucial aspect of the adaptive immune system. It allows the immune system to recognize and respond more effectively to previously encountered pathogens or antigens. Memory B cells are a type of B lymphocyte that have been activated by an antigen and have differentiated into plasma cells or memory cells.

These memory B cells produce and secrete antibodies, with IgM being the primary antibody class secreted. On the other hand, memory T cells are T lymphocytes that have encountered an antigen and undergone clonal expansion and differentiation. Memory T cells include various types, such as Th2 cells (helper T cells that assist B cells in antibody production) and Treg cells (regulatory T cells that suppress immune responses).

In addition to memory B and T cells, memory phagocytes, such as macrophages and dendritic cells, play a role in immunological memory by efficiently recognizing and eliminating previously encountered pathogens.

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Match the muscle with it's function correctly. Muscles of the Anterior Trunk Pectoralis Major Serratus Anterior External intercostals Internal intercostals Rectus abdominis External abdominal oblique

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Muscles of the Anterior Trunk: Pectoralis Major, Rectus Abdominis and External Abdominal Oblique.

Muscles of the Chest Wall: Serratus Anterior, External Intercostals and Internal Intercostals.

Muscles of the Anterior Trunk:

Pectoralis Major - Function: Adducts and flexes the arm at the shoulder joint.Rectus Abdominis - Function: Flexes the trunk and compresses the abdomen.External Abdominal Oblique - Function: Compresses the abdomen and aids in trunk rotation.

Muscles of the Chest Wall:

Serratus Anterior - Function: Protracts and stabilizes the scapula, allowing for forward-reaching movements of the arm.External Intercostals - Function: Elevates the ribcage during inspiration, aiding in inhalation.Internal Intercostals - Function: Depresses the ribcage during forced exhalation, aiding in exhalation.

The specific function of each muscle may vary slightly depending on the movement and position of the body.

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One of the following is an example of the application of BCI:
Group of answer choices
a.Speech synthesizer
b.Wheelchair control
c.Memory enhancement
d.All of the above

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Brain-Computer Interface (BCI) technology enables direct communication between the brain and an external device, allowing individuals to control and interact with their environment using their brain signals. BCI has various applications, including speech synthesizers, wheelchair control, and memory enhancement. Thus, The correct answer is d. All of the above.

a. Speech synthesizer: BCI can be utilized to help individuals with speech impairments communicate by converting their brain activity into synthesized speech. By detecting and interpreting specific brain signals related to speech intentions, BCI systems can generate spoken words or sentences, providing a means of communication for individuals who have lost the ability to speak.

b. Wheelchair control: BCI technology can be employed to assist individuals with severe physical disabilities in controlling motorized wheelchairs. By monitoring the user's brain signals, BCI systems can translate their intentions into wheelchair commands, enabling them to navigate and move independently.

c. Memory enhancement: BCI research explores the potential of using brain signals to enhance memory and cognitive functioning. By stimulating specific regions of the brain or decoding neural patterns associated with memory encoding and retrieval, BCI applications aim to improve memory performance in individuals with memory impairments or cognitive disorders.

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Head & Neck Q54. The motor function of the facial nerve can be tested by asking the patient to: A) Clench his teeth. B) Open his mouth. C) Shrug his shoulders. D) Close his eyes. E) Protrude his tongu

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The motor function of the facial nerve can be tested by asking the patient to close his eyes.

The facial nerve, also known as cranial nerve VII, is responsible for controlling the muscles of facial expression. Testing the motor function of the facial nerve involves assessing the patient's ability to perform specific facial movements.

Among the options provided, the action of closing the eyes is the most relevant for testing the motor function of the facial nerve. The facial nerve innervates the muscles involved in eyelid closure, such as the orbicularis oculi muscle. Asking the patient to close their eyes allows the examiner to observe the symmetry and strength of the eyelid closure, which are indicative of proper facial nerve function.

While the other options listed (clenching teeth, opening mouth, shrugging shoulders, and protruding tongue) involve various muscle movements, they are not directly related to the motor function of the facial nerve. These actions are controlled by other cranial nerves or muscle groups.

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Which of the following is NOT a function of the plasma
membrane?
Group of answer choices
It regulates which substances can enter or leave the cell.
It receives information from outside the cell and tr

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Ansmits signals to the cell's interior. It provides structural support and shape to the cell. It synthesizes proteins for cellular processes.

The plasma membrane, also known as the cell membrane, is a vital component of all living cells. It is a selectively permeable barrier that surrounds the cell, separating its internal environment from the external environment. The primary function of the plasma membrane is to regulate the movement of substances into and out of the cell. It controls the entry and exit of ions, molecules, and nutrients, ensuring the maintenance of proper internal conditions necessary for cell function. Additionally, the plasma membrane is involved in cell signaling, as it receives external signals and transmits them to the cell's interior, allowing the cell to respond to its surroundings. The plasma membrane also plays a role in cell adhesion, cell recognition, and maintaining the cell's structural integrity.

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Drosophila sex determination involves the regulation of alternative RNA splicing by the sex-lethal (Sxl), transformer (tra), and doublesex (dsx) genes. Match each effect on Drosophila sexual development with the gene deletion that would cause it. Sxl deletion tra deletion dsx deletion Answer Bank male-specific splicing of tra yields male traits in females absence of male-determining regulatory protein yields female traits in males absence of female-determining regulatory protein yields male traits in females male-specific splicing of dux yields male traits in females

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The effects on Drosophila sexual development with the gene deletion that would cause it are Sxl deletion, tra deletion and dsx deletion.

The following effects on Drosophila sexual development with the gene deletion that would cause it are:

Sxl deletion:

absence of female-determining regulatory protein yields male traits in females.

tra deletion:

absence of male-determining regulatory protein yields female traits in males.

dsx deletion:

male-specific splicing of dsx yields male traits in females.

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With respect to the levels of organization of the human body, organs would fall between Select one: a. organ systems and atoms b. atoms and cells c. organelles and organ systems d. cells and tissues e

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The correct answer is c. organelles and organ systems.

Organs fall between the organelles and organ systems in the hierarchy of the levels of organization of the human body.

In the levels of organization of the human body, organs are structures composed of two or more different types of tissues that work together to perform specific functions. Organs are part of the third level of organization, falling between organelles (such as mitochondria or nuclei within cells) and organ systems (such as the cardiovascular system or respiratory system).

Atoms are the basic building blocks of matter and are not specific to the human body alone.

Cells are the smallest functional units of life and are the building blocks of tissues.

Tissues are groups of cells that work together to perform a particular function.

Organs are structures composed of different types of tissues that work together to perform specific functions.

Organ systems are groups of organs that work together to carry out a particular set of functions in the body.

The organism is the highest level of organization, representing the entire individual.

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A ground-water flow study was performed near your home in the Coachella Valley. A tracer dye was injected into a well 500 feet north of the Whitewater River. The tracer dye was detected in the river exactly 100 days after it was injected a. What is the general directions of ground water flow? b. What is the ground water velocity in feet per day? c. What is the ground-water velocity in feet per hour? 14. There has been a contaminant spill of a mile from your home. If the groundwater is flowing at the same rate as your answer from 13b. How many days would it take for the contaminants to reach your homes well? (1 miles = 5280 ft) Haemoglobin G Makassar is similar to HbS in that Glutamate is replaced at position 6 of each chain by Alanine. What would you expect the electrophoretic pattern for this Hb? And this mutation does not cause sickling of the haemoglobin protein. Speculate on why this may be the case. Name three animal phyla and describe the uniquecharacteristics which cause these groups to be different from theothers.SHORT ANSWER / SIMPLE Transformation of E. coli with the pUC-Factor X plasmid was undertaken following a similar protocol to that of BIOL10412, but using the volumes and concentrations of reagents given below:- 200 l transformation solution (CaCl2) added to E. coli- 20 l plasmid DNA added to the competent cells (DNA plasmid concentration 12.5 g ml-1)- 600 l LB broth added following the heat shock - 100 l of the transformation mixture plated on each LB/LB+amp plate- Average of 185 colonies grown on each LB+amp plate after 24 hours- Lawn of bacteria on LB plate (no ampicillin) after 24 hoursQ1.3 Calculate (showing your working) the transformation efficiency of this experiment in units of transformants g-1 plasmid DNA. (5 marks) If, instead of Eq. (4-70), we choose the Falkner-Skan similarity variable 11 = y(\U\/vx) /, the Falkner-Skan equation becomesf"' + 2/(m + 1)ff" + m(f - 1) = 0 subject to the same boundary conditions Eq. (4-72). Examine this relation for the spe- cial case U = -K/x and show that a closed-form solution may be obtained. Air enters the compressor of a gas turbine at 100 kPa and 300 K with a volume flow rate of 5.81 m/s. The compressor pressure ratio is 10 and its isentropic efficiency is 85%. At the inlet to the turbine, the pressure is 950 kPa and the temperature is 1400 K. The turbine has an isentropic efficiency of 88% and the exit pressure is 100 kPa. On the basis of an air-standard analysis, what is the thermal efficiency of the cycle in percent? 1. An impedance coil with an impedance of (5 + j8) is connected in series with a capacitive reactance X and this series combination is connected in parallel with a resistor R. If the total impedance of the circuit is (4 + j0) , find the value of the resistance of the resistor.2. A capacitance C is connected in series with a parallel combination of a 2 k resistor and a 2 mH coil inductor. Find the value of C in order for the overall power factor of the circuit be equal to unity at 20 kHz.NEED HELP PLEASE. THANK YOU A study has shown that the probability distribution of X, the number of customers in line (including the one being served, if any) at a checkout counter in a department store, is given by P(X= 0) = 0.30, P(X= 1) = 0.25, P(X= 2) = 0.20, P(X= 3) = 0.20, and P(X 2 4) = 0.05. Consider a newly arriving customer to the checkout line. Round your answers to two decimal places, if necessary. a. What is the probability that this customer will not have to wait behind anyone? b. What is the probability that this customer will have to wait behind at least one customer? c. 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The wall is maintained at temperature of 200C and the fins dissipate heat by convection into the ambient air at 40C with h = 50 W/m.(a) determine the fin efficiency.(b) determine the area-weighted fin efficiency.(c) Determine the heat loss per square meter of wall surface. everal mutants are isolated, all of which require compound G for growth. The compounds (A to E) in the biosynthetic pathway to G are known, but their order in the pathway is not known. Each compound is tested for its ability to support the growth of each mutant (1 to 5). In the following table, a plus sign indicates growth and a minus sign indicates no growth. What is the order of compounds A to E in the pathway? Compound tested A B C D E G Mutant 1 - - - + - +2 - + - + - + 3 - - - - - + 4 - + + + - + 5 + + + + - + a. E-A-B-C-D-Gb. B-A-E-D-C-G c. A-B-C-D-E-G d. E-A-C-B-D-G e. B-A-E-C-D-G Analysis of variance showed significant differences among cultivars in 1% probability for Number of rows in-ear, Number of seeds per row, 100-seeds weight, Harvest index, Seed yield, and 5% probability for Biological yield (Table 1), which demonstrated the existence of variation among cultivars studied in this research. The highest coefficient of variation (CV) was shown by harvest index and the least values were shown by developmental characteristics such as seed weight and to Number of rows in-ear. Irrigation treatment had a significant influence on all traits, too (Table 1). Several studies have shown that seed yield and yield components of maize, were markedly affected by irrigation treatments (Rivera-Hernandez et al., 2010., Moser et al., 2006 Cakir.. 2004) Effect of cultivar was significant on all traits in the error level of 1% expect for biological yield that for this trait was significant in error level of 5% (Table 1). Mostafavi et al. (2011), in a similar experiment on the effects of drought stress on Maize hybrids, stated variety was significantly affected either by the yield parameters. The Highest Number of rows in-ear (NRE) was achieved with control and had significant differences between other treatments. The lowest NRE is related to 150 mm levels of evaporation. KSC720 cultivar has highest NRE and had significant differences with KSC- N84-01 and KSC 708GTbut had no significant differences with KSC720. The lowest NRE is related to KSC 708GT (Table 2). Rivera-Hernandez et al. (2010) reported that although significant differences were observed among irrigation treatments for a variable number of rows per ear, this was the least affected by the rise in soil moisture tension. This suggests that the number of rows per ear is more influenced by heredity factors than by crop management. The Highest Number of seeds per row (NSR) was achieved with control and had significant differences between other treatments. The lowest NSR is related to 150 mm levels of evaporation and KSC720. the cultivar has the highest NSR with significant differences from other cultivars and the lowest NSR related to KSC 708GT (Table 2). Moser et al. (2006) reported that pre-anthesis drought significantly reduced the number of kernels per row. The highest 100 seed weight was achieved in control and has significantly different from other treatments, but the lowest 100 seed weight is related to 150 mm levels of evaporation. The results show that the highest 100 seed weight was from the KSC720 cultivar and other cultivars had significant differences together (Table 2). Zenislimer et al. (1995) stated that the drought effect on the number of grains per and 100-grain weight, grain yield was reduced. 1. 2 points The product of two imaginary values is an imaginary value. O a. True O b. False 2. 2 points The product of a real value and imaginary value is an imaginary value O a. True O b. 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