Positioning of the first Met tRNA on the mRNA is a critical step in translation inititiation. Selection of the first correct AUG is achieved by a complex of Met tRNA with a. initiation factors
b. initiation factors plus poly A binding protein
c. small subunit ribosome d.small subunit ribosome plus initiation factors

Answers

Answer 1

The positioning of the first Met tRNA on the mRNA during translation initiation is a crucial step. The correct selection of the first AUG codon is achieved by a complex of Met tRNA with the small subunit ribosome plus initiation factors. This complex ensures the accurate initiation of protein synthesis.

Translation initiation is the process by which protein synthesis begins in cells. It involves the assembly of the ribosome, mRNA, and initiator tRNA at the start codon of the mRNA molecule. The first Met tRNA, carrying the amino acid methionine, plays a crucial role in this process.

To ensure accurate initiation, a complex is formed between the small subunit ribosome, the initiator tRNA, and several initiation factors. These initiation factors help in the proper positioning of the components and facilitate the recognition of the start codon. Among the initiation factors, one important factor is the initiation factor 2 (IF2) that interacts with the initiator tRNA and the small subunit ribosome.

The initiation complex scans the mRNA molecule until it reaches the correct start codon, which is typically AUG. The start codon is recognized by the anticodon of the initiator tRNA, which is base-paired with the AUG codon. The interaction between the Met tRNA and the start codon is facilitated by the small subunit ribosome and the initiation factors. Once the correct start codon is recognized, the large subunit of the ribosome joins the complex, and protein synthesis begins. The initiator tRNA occupies the P-site of the ribosome, ready to receive the next amino acid and initiate the elongation phase of translation.

In conclusion, the positioning of the first Met tRNA on the mRNA during translation initiation is achieved by a complex consisting of the small subunit ribosome plus initiation factors. This complex ensures the accurate selection of the first AUG codon and facilitates the proper initiation of protein synthesis.

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Related Questions

15) UTI's with microbial etiology include: A. cystitus. B. Urethritis C. Leptospirosis D. A and B E. A, B and C 16) The cause of gonorrhea is a member of the genus: A. Borrelia B. treponema C. Neisseria D. Mycobacterium E. plasmodium 17) Which antibody is most import in immediate hypersensitivity reactions: A. IgG B. IgM C. IgA D. ISE 18) Which is true. Of. HPV (papillomavirus) A. Only two strains. Effect humans B. It can cause genital warts C. Less than 1% of women are effected D. No vaccine is available 19). Trichomonal. Vaginitis is caused by: A. Yeast B. Bacteria C. Protozoan D. Chlamydia E. A virus 20) Lyme disease A. Is highly contagious B. Early symptoms include rash and flu like symptoms etiology D. Mosquito vector C. Viral

Answers

UTIs with microbial etiology include cystitis and urethritis. The cause of gonorrhea is a member of the genus Neisseria. The most important antibody in immediate hypersensitivity reactions is IgE.

UTIs (urinary tract infections) with microbial etiology commonly involve cystitis (inflammation of the bladder) and urethritis (inflammation of the urethra). These infections are often caused by bacterial pathogens.

Gonorrhea is caused by a member of the genus Neisseria, specifically Neisseria gonorrhoeae, a sexually transmitted bacterium.

In immediate hypersensitivity reactions, the most important antibody involved is IgE. IgE antibodies are responsible for triggering allergic reactions and are associated with conditions like asthma and allergic rhinitis.

HPV (human papillomavirus) is a sexually transmitted infection that can cause genital warts and is also associated with certain types of cancer. There are several strains of HPV that affect humans, not just two, and there is a vaccine available to protect against certain high-risk strains.

Trichomonal vaginitis, also known as trichomoniasis, is caused by a protozoan parasite called Trichomonas vaginalis.

Lyme disease is primarily transmitted through the bite of infected black-legged ticks. It is not highly contagious between humans. Early symptoms of Lyme disease often include a characteristic rash called erythema migrans, along with flu-like symptoms.

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The following are stages of glucose oxidation, except. O electron transport system oxidative phosphorylation O Krebs cycle O glycolysis O all of the

Answers

Glucose oxidation is the metabolic process by which glucose is oxidized to produce ATP energy that can be used by the cells for carrying out their activities.

The process of glucose oxidation takes place in three stages, namely glycolysis, Krebs cycle, and electron transport system, which are discussed below.

Glycolysis:

It is the first stage of glucose oxidation that takes place in the cytoplasm of the cell.

In this process, one glucose molecule is oxidized to form two molecules of pyruvic acid.

Moreover, two molecules of ATP energy are produced in this process.

This process can take place in both aerobic and anaerobic conditions.

Krebs Cycle:

It is the second stage of glucose oxidation, also known as the citric acid cycle.

In this stage, the two molecules of pyruvic acid produced during glycolysis are further oxidized to produce energy.

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Chloroplasts, mitochondria, and bacteria have 70S ribosomes in common. O True False

Answers

70S ribosomes are a type of ribosome found in prokaryotic cells, including bacteria, as well as in certain organelles of eukaryotic cells, such as mitochondria and chloroplasts.

The term "70S" refers to the sedimentation coefficient of the ribosome, which is a measure of its size and density.

The 70S ribosome consists of two subunits: a larger 50S subunit and a smaller 30S subunit. True. Chloroplasts, mitochondria, and bacteria share the characteristic of having 70S ribosomes.

These ribosomes are smaller than the 80S ribosomes found in eukaryotic cells.

The presence of 70S ribosomes in these organelles and bacteria suggests a common evolutionary origin and supports the endosymbiotic theory.

Which proposes that mitochondria and chloroplasts were once free-living bacteria that were engulfed by ancestral eukaryotic cells. The conservation of 70S ribosomes among these organisms highlights their shared ancestry and functional similarities.

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What provides the energy to the ATP Synthase for the formation of ATP? (Select all that apply) a. Proton Flow b. Electron Flow c. Phosphoryl Transfer Potential d. Voltage potential e. Oxidation strength of the synthase

Answers

The energy required to form ATP in the ATP Synthase is provided by Proton Flow and Voltage potential. These two factors cause a conformational change in the structure of ATP Synthase, which results in the formation of ATP molecules (option a and d).

ATP Synthase is an enzyme complex that converts ADP to ATP. The energy required for the formation of ATP is obtained from the electron transport chain and oxidative phosphorylation. The proton gradient that is established in the inner mitochondrial membrane during the electron transport chain is used to synthesize ATP through ATP Synthase.

The process is known as chemiosmotic coupling and it is the key mechanism behind ATP production in the cell. During the chemiosmotic coupling, protons (H+) are pumped out of the mitochondrial matrix into the intermembrane space. This results in the establishment of a proton gradient across the inner mitochondrial membrane.As the protons move back into the matrix through the ATP Synthase, the energy generated is used to produce ATP. This process is called oxidative phosphorylation and it is a crucial step in cellular respiration. Hence, options (a) and (d) are correct.

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QUESTION 4 A 12 year-old boy, accompanied by his mother, came to the hospital with high fever, headache and diarrhea. The doctor guessed the boy was infected by influenza since the epidemic has started in that area. He was acutely ill, so the doctor judged it was necessary for him to be ventilated. Then, however, they found out that all the 5 ventilators that the department had were currently in use for another 11 year-old girl with influenza and 4 elderly people in chronic care. The girl has just came in yesterday but seems to be start recovering, but her doctors request her to be ventilated for one more night at least. Also, the 4 elderly people cannot live without a ventilator. In addition, the epidemic of influenza is expected to get more serious within a week or so, and it is expected to be an increase of influenza patients, and some of them, especially elderly people and young children would need a ventilator. (a) (b) Highlight some of the conflicts and problems in this case. [C4] [SP1, SP2, SP4,SP5, SP6] [10 marks] Do you think one of five ventilators in use should be connected to the boy? If yes, in what occasion or with what reason will it happen? If no, what are reasons and should those reasons be based on ethics, rational or something else? [C5] [SP1, SP2, SP4,SP5, SP6] [15 marks]

Answers

As seen in the passage, there are some conflicts and problems that arise in the case; they are: The hospital has a limited number of ventilators for patients, which means there is a limited capacity to care for people who need it.

The problem becomes worse during an epidemic because there will be more people who require ventilators, especially elderly and young children.

Therefore, the hospital is not equipped to deal with all of these patients. The boy is very sick and needs a ventilator, but all the ventilators are currently in use for the girl and four elderly patients. This is a conflict because the boy's life is at risk, but there is no ventilator available for him.

The doctors have to make a decision about who gets the ventilator, which can be challenging because it involves ethical considerations. The girl has just come in, but her doctors request her to be ventilated for one more night at least, which creates a conflict because it means that the boy may have to wait longer to receive treatment.

Opinion on whether one of five ventilators in use should be connected to the boy: The boy is very sick, and he needs a ventilator. However, all the ventilators are in use for the girl and the four elderly patients, so the doctors have to make a decision about what to do.

The question is whether one of the five ventilators in use should be connected to the boy. In my opinion, the answer is yes, but with some conditions. The boy is acutely ill and requires immediate treatment, and if he doesn't get a ventilator, he might not survive.

The girl has been receiving treatment for a while and is recovering, so her need for the ventilator is less critical. Therefore, I believe that the doctors should disconnect the girl from the ventilator and give it to the boy. However, this decision should be based on ethics and rationality and not solely on the basis of need. The doctors must consider the long-term effects of their decision on all the patients involved.

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What are the five principal reactions that occurred during
primodial nucleosynthesis?
Name all the types of stable nuclei that remained after
primordial nucleosynthesis had finished.
At what proportio

Answers

At the end of primordial nucleosynthesis, the universe was composed of approximately 75% hydrogen, 24% helium, and trace amounts of lithium and other elements.

During primordial nucleosynthesis, the five principal reactions that occurred are as follows:Proton-proton chain reaction: This reaction occurs when protons fuse with one another to form a helium nucleus.Alpha process: It is a sequence of nuclear reactions that produce helium-4 from hydrogen. This process involves the capture of helium nuclei to heavier elements. The alpha process is most efficient at producing elements with even numbers of protons, particularly helium, carbon, and oxygen.Beta decay: It is a process by which an unstable atomic nucleus loses energy by emitting an electron or a positron.

The unstable nucleus changes into a stable nucleus by emitting either a negatively charged electron (beta-minus decay) or a positively charged positron (beta-plus decay).Neutron capture: It is a process in which a neutron is added to a nucleus to produce a heavier nucleus. Neutron capture is important for the formation of heavier elements beyond iron.Nuclear fusion: It is a process by which multiple atomic nuclei join together to form a heavier nucleus. This is the process by which stars produce energy.The types of stable nuclei that remained after primordial nucleosynthesis had finished are as follows:Hydrogen-1, Helium-3, Helium-4, Lithium-6, Lithium-7, Beryllium-7.At the end of primordial nucleosynthesis, the universe was composed of approximately 75% hydrogen, 24% helium, and trace amounts of lithium and other elements.

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In photosynthesis, carbon dioxide is "fixed" in ____.
A. the light-dependent reactions
B. the Carbon cycle
C. the light-independent reactions
D. the Krebs cycle

Answers

The correct answer is C. the light-independent reactions, where carbon dioxide is fixed and converted into organic compounds during photosynthesis.

The process of carbon dioxide fixation refers to the conversion of atmospheric carbon dioxide into organic compounds during photosynthesis. This occurs during the light-independent reactions, also known as the Calvin cycle or the dark reactions. These reactions take place in the stroma of chloroplasts, specifically in the chloroplasts of plant cells.

During the light-independent reactions, carbon dioxide molecules are combined with molecules derived from the light-dependent reactions, such as ATP and NADPH. The key enzyme involved in carbon dioxide fixation is called RuBisCO (Ribulose-1,5-bisphosphate carboxylase/oxygenase). RuBisCO catalyzes the incorporation of carbon dioxide into an organic molecule called ribulose-1,5-bisphosphate (RuBP), which then goes through a series of reactions to produce glucose and other organic compounds.

In contrast, the light-dependent reactions, which occur in the thylakoid membranes of chloroplasts, involve the absorption of light energy and the generation of ATP and NADPH. These energy-rich molecules produced in the light-dependent reactions are subsequently used in the light-independent reactions to drive the carbon dioxide fixation and synthesis of organic molecules.

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27. What does Lugol's test for and a + color? + 28. What does Biuret test for and a + color? + 29. What does benedicts test for and a + color? +

Answers

Lugol's test is used to test for the presence of starch. A positive result is indicated by a dark blue or purple color.Biuret test is used to test for the presence of proteins. A positive result is indicated by a violet or purple color.Benedict's test is used to test for the presence of reducing sugars.

Lugol's test is used to detect the presence of starch in a solution. The test is performed by adding a few drops of Lugol's iodine solution to the solution in question. If the solution turns dark blue or purple, the presence of starch is confirmed.

Biuret test, on the other hand, is used to test for the presence of proteins in a solution. When Biuret reagent is added to a protein solution, the solution turns violet or purple in color. The intensity of the color is proportional to the amount of protein present in the solution.

Benedict's test is used to detect the presence of reducing sugars in a solution. When Benedict's solution is added to a reducing sugar solution and heated, a red, yellow, or green color is formed, depending on the amount of reducing sugar present. The more intense the color, the more reducing sugar is present.

In summary:Lugol's test is used to test for the presence of starch. A positive result is indicated by a dark blue or purple color.Biuret test is used to test for the presence of proteins. A positive result is indicated by a violet or purple color.Benedict's test is used to test for the presence of reducing sugars. A positive result is indicated by a red, yellow, or green color.

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Why taxonomic nomenclature is important? It provides the unified language for communication about biological diversity. It reflects evolutionary relatedness of taxa. Scientific names often capture important characteristics of the animals. It documents the history of science. All of the above.

Answers

Taxonomic nomenclature is important because it provides a standardized language for communication, represents evolutionary relationships, captures important characteristics, and documents the history of scientific discoveries. So, All of the above is the correct choice.

Taxonomic nomenclature is important for several reasons:

It provides a unified language for communication about biological diversity: By assigning unique scientific names to organisms, taxonomic nomenclature allows researchers, scientists, and other professionals to communicate and exchange information accurately and precisely. This ensures clarity and avoids confusion that may arise from using different common names for the same species.It reflects evolutionary relatedness of taxa: Taxonomic nomenclature is based on the principles of evolutionary relationships. Organisms with similar characteristics and shared ancestry are grouped together into taxa (such as genus, family, order, etc.), and their scientific names reflect their evolutionary relationships. This helps in understanding the evolutionary history and biological relationships between different organisms.Scientific names often capture important characteristics of the animals: Scientific names are often chosen to describe important characteristics of the organisms they represent. These names can provide insights into the morphology, behavior, habitat, or other significant features of the species. This additional information enhances our understanding of the organism beyond its common name.It documents the history of science: Taxonomic nomenclature has a long history and has evolved over time. The use of scientific names allows us to trace the development of scientific knowledge, discoveries, and advancements in the field of taxonomy. The history of taxonomic naming provides valuable insights into the progression of scientific understanding and serves as a record of scientific exploration.

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2. Enterobius vermicularis is infective in___ form and causes ____
a. larval; pinworm
b. egg; hookworm
c. egg; pinworm d.larval; hookworm 3. The reproductive structure of Taenia is a a.hook b.proglottid c. scolex d.heterocyst
4. Trichinella spiralis is transmitted by
a. ingestion of a cyst b. ingestion of a larva
c. ingestion of an egg d.a vector 5. Which type of sample would be used to aid in diagnosis of a Clonorchis infection? a. Fecal smear b.Sputum sample
c. Skin scraping d.Blood sample

Answers

Enterobius vermicularis is infective in the egg form and causes pinworm infection. The reproductive structure of Taenia is the proglottid. Trichinella spiralis is transmitted by ingestion of a larva. A fecal smear would be used to aid in the diagnosis of a Clonorchis infection.

Enterobius vermicularis is infective in the egg form and causes pinworm infection. The eggs of Enterobius vermicularis are ingested, usually through contaminated food, water, or by direct contact with infected individuals. Once inside the body, the eggs hatch in the small intestine, and the larvae migrate to the large intestine, where they mature into adult worms. The adult female worms then migrate to the perianal area to lay their eggs, leading to itching and discomfort.

The reproductive structure of Taenia, a genus of parasitic tapeworms, is the proglottid. Proglottids are segments that make up the body of a tapeworm and contain both male and female reproductive organs. Each proglottid is capable of producing eggs, which are then released into the environment through the feces of the infected host. The proglottids can detach from the tapeworm's body and be passed in the feces, enabling the tapeworm to spread and infect new hosts.

Trichinella spiralis, a parasitic roundworm, is transmitted by the ingestion of a larva. The larvae of Trichinella are encysted in the muscle tissue of infected animals, typically pigs or other mammals. When these infected meat products are consumed by humans, the larvae are released in the digestive system, where they mature into adult worms. The female worms then produce larvae that migrate to muscle tissue, causing a condition known as trichinellosis.

To aid in the diagnosis of a Clonorchis infection, a fecal smear would be used. Clonorchis sinensis is a parasitic liver fluke that infects humans through the consumption of raw or undercooked freshwater fish containing the infectious larvae. The adult flukes reside in the bile ducts of the liver. The presence of Clonorchis eggs in a fecal smear can indicate an infection, as the adult flukes release eggs into the feces. Other diagnostic methods may include serological tests or imaging techniques to visualize the flukes in the bile ducts.

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1. Are there drugs that interfere with the ETC? Yes. For example barbiturates and the pesticide Rotenone block electron transfer from complex I to CoQ. Hence, NADH cannot be used. What would the consequence be on protons pumped and ATP produced? Calculate now how many protons could be pumped and ATPs synthesized with NADH's contribution out of the picture.

Answers

The above situation would result in the pumping of 4 protons and will result in ATP synthesis. In the absence of NADH's contribution, 44 protons would be pumped.

In the electron transport chain (ETC), NADH is an important electron carrier that donates electrons to complex I, which leads to the pumping of protons across the inner mitochondrial membrane and ultimately contributes to ATP synthesis. If NADH's contribution is blocked by substances like barbiturates and Rotenone, it would have consequences on the number of protons pumped and ATP produced.

NADH is responsible for donating electrons to complex I, resulting in the pumping of 4 protons (H+) from the matrix to the intermembrane space. These protons contribute to the proton gradient, which drives ATP synthesis.

If NADH's contribution is eliminated, those 4 protons pumped per NADH would not occur. However, other sources of electron input into the ETC, such as FADH2 from the citric acid cycle, can still contribute. FADH2 donates electrons to complex II, bypassing complex I and reducing the number of protons pumped.

FADH2, on average, donates electrons at complex II, resulting in the pumping of 2 protons (H+) from the matrix to the intermembrane space. This means that for each FADH2 molecule, 2 protons are pumped.

To calculate the potential proton pumping and ATP synthesis when NADH's contribution is absent, we need to know the relative ratios of NADH and FADH2 in the electron transport chain during normal conditions. The ratio typically considered is 10 NADH: 2 FADH2.

So, without NADH's contribution, if we consider the contribution of FADH2:

- Protons pumped: (10 NADH * 4 protons) + (2 FADH2 * 2 protons) = 40 protons + 4 protons = 44 protons

- ATP synthesized: (10 NADH * 3 ATP) + (2 FADH2 * 2 ATP) = 30 ATP + 4 ATP = 34 ATP

Therefore, in the absence of NADH's contribution, approximately 44 protons would be pumped, contributing to the proton gradient, and approximately 34 ATP molecules would be synthesized in the electron transport chain.

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what is the total amount of ATP generated in cellular
respiration?
please help quickly with very very short explination!

Answers

A total of up to 36 molecules of ATP can be consisted from just one molecule of glucose in the work of cellular respiration.

What is the ATP?

The total amount of ATP create in basic breathing changes depending on the particular road complicated. Electron transport from the particles of NADH and FADH2 from glycolysis, the revolution of pyruvate, and the Krebs cycle generates as many as 32 more ATP particles.

Therefore,  In general, through the complete disintegration of individual particle of hydrogen, the net result of ATP is 36 to 38 particles in prokaryotes and 30 to 32 fragments in eukaryotes.

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In human fibroblasts, the "loss" of Rb and p53 by a DNA tumor virus, and reactivation of hTERT will lead to which of the following? a. Tumorigenic phenotype b. Morphological transformation c. Immortalization d. quiescence e. crisis

Answers

The loss of Rb and p53 by a DNA tumor virus, and reactivation of hTERT will lead to immortalization. So, option C is accurate.

When human fibroblasts experience the loss of Rb and p53, which are tumor suppressor proteins, and the reactivation of hTERT (human telomerase reverse transcriptase), the cells undergo a process called immortalization. This means that the cells acquire the ability to divide indefinitely, bypassing the usual cellular senescence mechanisms. Rb and p53 are key regulators of the cell cycle and are responsible for suppressing abnormal cell growth and promoting cell cycle arrest or apoptosis in response to DNA damage or other stressors. The loss of their function eliminates these control mechanisms, while the reactivation of hTERT prevents the progressive shortening of telomeres, which are protective caps at the ends of chromosomes that shorten with each cell division. Consequently, the combination of Rb and p53 loss and hTERT reactivation leads to cellular immortalization, a critical step in the development of a tumorigenic phenotype.

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Acetyl-CoA is an important intermediate that participates (either as an input, an output, or an intermediate) in all of the below processes EXCEPT O Photorespiration O the Citric Acid Cycle B-oxidation cycle Acetyl-CoA participates in all these processes O Glyoxylate cycle Determination of an enzyme or pathway Q10 provides information on O a method to compare two alternative enzymes or pathways at a single temperature O gas solubility in response to temperature O the relative thermal motivation of a biochemical pathway a O the temperature sensitivity of an enzyme or pathway O the temperature switch point between C3 and CAM photosynthesis

Answers

Acetyl-CoA is an important intermediate that participates in all of the processes mentioned except gas solubility in response to temperature.

Option (F) is correct.

Acetyl-CoA is a central molecule in cellular metabolism. It is involved in various biochemical processes, including the ones mentioned:

A) Photorespiration: Acetyl-CoA participates in photorespiration as an input in the glycolate pathway, which helps plants recover carbon during inefficient photosynthesis.

B) The Citric Acid Cycle: Acetyl-CoA enters the citric acid cycle, also known as the Krebs cycle, where it undergoes a series of reactions to generate energy-rich molecules such as ATP.

C) β-oxidation cycle: Acetyl-CoA is produced as an output during the breakdown of fatty acids in the β-oxidation cycle, which occurs in mitochondria.

D) Glyoxylate cycle: Acetyl-CoA serves as an intermediate in the glyoxylate cycle, allowing certain microorganisms and plants to convert acetyl-CoA into carbohydrates.

E) Determination of an enzyme or pathway Q10: Acetyl-CoA can participate in the determination of the temperature sensitivity of an enzyme or pathway using the Q10 value, which describes the rate of change with temperature.

However, F) Gas solubility in response to temperature does not involve Acetyl-CoA directly. It refers to the solubility of gases, such as oxygen or carbon dioxide, in liquids and is influenced by factors like temperature and pressure.

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Complete question is:

Acetyl-CoA is an important intermediate that participates (either as an input, an output, or an intermediate) in all of the below processes EXCEPT:

A) Photorespiration

B) The Citric Acid Cycle

C) β-oxidation cycle

D) Glyoxylate cycle

E) Determination of an enzyme or pathway Q10 provides information on

F) Gas solubility in response to temperature

G) The relative thermal motivation of a biochemical pathway

H) The temperature sensitivity of an enzyme or pathway

I) The temperature switch point between C3 and CAM photosynthesis

With regard to the lac operon, which of the following is false under conditions of low (or no) glucose? a) Lactose is not present b) The repressor is bound to the operator c) Lactose is not bound to the repressor d) RNA polymerase can bind to the promoter

Answers

The false statement under conditions of low (or no) glucose with regard to the lac operon is: a) Lactose is not present. In the lac operon, low (or no) glucose conditions induce the lac operon to be active, leading to the expression of genes involved in lactose metabolism.

Lactose, which is the inducer molecule, is typically present under these conditions and plays a crucial role in regulating the lac operon. Lactose binds to the repressor protein, causing it to be released from the operator region, thereby allowing RNA polymerase to bind to the promoter and initiate gene transcription.

The presence of lactose is necessary for the operon to be fully induced and for the expression of the lac genes. Therefore, statement a) is false.

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"True/False Organismal complexity (how complex an organism is)
is not correlated with genome length but is
correlated with the number of protein coding genes
Group of answer choices
a.True
b.False"

Answers

b. False

Organismal complexity is generally correlated with genome length and not necessarily with the number of protein-coding genes alone. While the number of protein-coding genes contributes to an organism's complexity, it is not the sole determining factor.

Genome length encompasses protein-coding genes and non-coding regions, regulatory elements, repetitive sequences, and other genetic components that contribute to the overall complexity of an organism. Therefore, genome length is a more comprehensive measure of organismal complexity than just the number of protein-coding genes.

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Design one simple experiment to find out whether your protein of interest is over-expressed in E. coli. Given the DNA sequence and three restriction enzymes (Hindill, Psti and BamHI), write out the se

Answers

To find out whether the protein of interest is over-expressed in E. coli, we need to carry out a simple experiment called Western Blot. This experiment involves the use of antibodies to detect the protein of interest. The steps involved in this experiment are given below:

Step 1: Protein Extraction - The protein of interest must be extracted from E. coli cells.

Step 2: Protein Quantification - The concentration of the extracted protein must be determined.

Step 3: Protein Separation - The extracted protein must be separated by SDS-PAGE (Sodium Dodecyl Sulfate Polyacrylamide Gel Electrophoresis).

Step 4: Western Blotting - The separated protein must be transferred onto a nitrocellulose membrane and blocked using non-specific protein.

Step 5: Primary Antibody Incubation - The primary antibody (which recognizes the protein of interest) is incubated with the membrane.

Step 6: Secondary Antibody Incubation - The secondary antibody (which recognizes the primary antibody) is incubated with the membrane.

Step 7: Detection - The protein of interest is detected using a substrate that reacts with the secondary antibody.

Western Blot is an effective method for detecting whether a protein of interest is over-expressed in E. coli. This method allows us to detect and quantify the protein of interest using specific antibodies.

Western Blot is a widely used method for detecting proteins in a sample. This method is based on the use of antibodies to detect the protein of interest. The steps involved in Western Blot are Protein Extraction, Protein Quantification, Protein Separation, Western Blotting, Primary Antibody Incubation, Secondary Antibody Incubation, and Detection. Each of these steps is important for the success of the experiment.In the first step, Protein Extraction, the protein of interest must be extracted from E. coli cells.

This step involves the use of lysis buffer and sonication to break the cells and release the protein. The extracted protein must then be purified using methods such as column chromatography or ammonium sulfate precipitation.In the second step, Protein Quantification, the concentration of the extracted protein must be determined. This step is important because it allows us to know how much protein we are working with.

Protein Quantification can be done using methods such as Bradford Assay or UV Spectroscopy.In the third step, Protein Separation, the extracted protein must be separated by SDS-PAGE. SDS-PAGE is a method that separates proteins based on their size.

The separated proteins are then transferred onto a nitrocellulose membrane.In the fourth step, Western Blotting, the separated protein is transferred onto a nitrocellulose membrane and blocked using non-specific protein. This step is important because it prevents non-specific binding of the primary antibody.

In the fifth step, Primary Antibody Incubation, the primary antibody (which recognizes the protein of interest) is incubated with the membrane. The primary antibody binds to the protein of interest and allows us to detect it.In the sixth step, Secondary Antibody Incubation, the secondary antibody (which recognizes the primary antibody) is incubated with the membrane.

The secondary antibody binds to the primary antibody and allows us to detect the protein of interest.In the seventh step, Detection, the protein of interest is detected using a substrate that reacts with the secondary antibody. This reaction produces a signal that can be detected using methods such as Chemiluminescence or Fluorescence.

Western Blot is an effective method for detecting whether a protein of interest is over-expressed in E. coli. This method allows us to detect and quantify the protein of interest using specific antibodies. The steps involved in this experiment are Protein Extraction, Protein Quantification, Protein Separation, Western Blotting, Primary Antibody Incubation, Secondary Antibody Incubation, and Detection.

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Which is a main blocking antibody in Immunologic Intervention for Type-I hypersensitivity reaction (desensitization method)? Selected Answer: IgE Answers: IgE IgA IgG IgD IgM .

Answers

The correct answer os IgE.

IgE is the main blocking antibody involved in immunologic intervention for Type-I hypersensitivity reactions during desensitization methods. IgE antibodies are responsible for triggering allergic reactions by binding to allergens and activating mast cells and basophils. Desensitization aims to reduce the hypersensitivity by gradually exposing the individual to increasing doses of the allergen, leading to the production of blocking IgG antibodies that compete with IgE for binding to the allergen, thereby preventing allergic reactions.

In Type-I hypersensitivity reactions, the immune system responds to harmless substances, called allergens, by producing an excessive amount of IgE antibodies. These IgE antibodies bind to the surface of mast cells and basophils, which are rich in histamine. When the individual is re-exposed to the allergen, the allergen binds to the IgE antibodies on the mast cells and basophils, triggering the release of histamine and other inflammatory mediators. This process leads to the symptoms of an allergic reaction, such as itching, swelling, and respiratory difficulties.

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Serotonin has been shown to be sufficient to cause the development of the gregarious form of the migratory desert locust. What predictions must have been tested to arrive at this conclusion? (provide 1 prediction). If serotonin provides a phenotypic change, is this a proximate or ultimate explanation and explain why? Note: A prediction can be tested to see if a hypothesis is true.

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One prediction that must have been tested to arrive at the conclusion that serotonin is sufficient to cause the development of the gregarious form of the migratory desert locust is whether or not an increase in serotonin levels leads to the development of gregarious behavior.

One prediction that must have been tested to arrive at the conclusion that serotonin is sufficient to cause the development of the gregarious form of the migratory desert locust is whether or not an increase in serotonin levels leads to the development of gregarious behavior. This hypothesis can be tested by manipulating the serotonin levels of locusts and observing their resulting behavior. If the locusts become more gregarious when their serotonin levels are increased, then the hypothesis would be supported and serotonin would be shown to be a sufficient cause for gregarization. If serotonin provides a phenotypic change, it is a proximate explanation.

Proximate explanations focus on the immediate causes of behavior or phenotype, such as the physiological mechanisms underlying the behavior. In this case, serotonin is the immediate cause of the locust's gregarious behavior.

Ultimate explanations, on the other hand, focus on the evolutionary or adaptive significance of a behavior or phenotype. While serotonin may have an ultimate explanation in terms of its evolutionary history and the selective pressures that favored the development of gregarious behavior in locusts, the fact that serotonin causes this behavior is a proximate explanation.

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Which one is the correct hierarchical sequence of the auditory stimulus processing? (Some intermediate structures may be omitted.)
a) Vesibulocochlear nerve - Inferior Colliculus - Cochlear Nuclei - Medial Geniculate nucleus - Primary Auditory cortex.
b) Cranial nerve VIII - Cochlear Nuclei – Medial Geniculate nucleus - Inferior Colliculus - Primary Auditory cortex.
c) Cranial nerve V - Cochlear Nuclei – Inferior Colliculus - Medial Geniculate nucleus - Primary Auditory cortex.
d) Hair cells – Spiral ganglion cells – Cochlear Nuclei – Inferior Colliculus - Medial Geniculate nucleus - Primary Auditory cortex.

Answers

The correct hierarchical sequence of the auditory stimulus processing is (b) Cranial nerve VIII - Cochlear Nuclei – Medial Geniculate nucleus - Inferior Colliculus - Primary Auditory cortex. Here is an explanation for each of the structures:

Auditory stimulus processing is the step-by-step process that sound waves undergo as they travel from the ear to the brain for interpretation. The structures involved in this process are as follows:

Cranial nerve VIII (CN VIII) or Vestibulocochlear nerve: This is the nerve responsible for transmitting sound information from the ear to the brain.

Cochlear Nuclei: These are two small clusters of cells located in the brainstem. They receive and process sound information from the cochlea.

Medial Geniculate Nucleus: This is a group of nuclei in the thalamus that act as the main relay center for auditory information processing.

Inferior Colliculus: This is a midbrain structure that receives and integrates auditory information from both ears.

Primary Auditory Cortex: This is the first cortical region in the temporal lobe responsible for processing auditory information from the thalamus.

The correct sequence, therefore, is Cranial nerve VIII - Cochlear Nuclei – Medial Geniculate nucleus - Inferior Colliculus - Primary Auditory cortex.

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Fibrous proteins Fibrous and globular proteins Globular proteins Answer Bank primarily function as structural proteins includes collagen includes hemoglobin exhibit secondary protein structure somewhat spherical in shape rod-like in shape insoluble in water some function as enzymes polymers of amino acids soluble in water

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Fibrous proteins are rod-shaped and insoluble in water. These types of proteins primarily function as structural proteins in living organisms, such as keratin, collagen, and elastin.

Globular proteins, on the other hand, are soluble in water and have a somewhat spherical shape. Hemoglobin and enzymes are examples of globular proteins that exhibit a secondary protein structure.
Fibrous proteins and globular proteins are two of the major protein categories. Structural proteins, such as collagen, keratin, and elastin, are primarily made up of fibrous proteins. The insoluble characteristic of fibrous proteins gives them a significant structural advantage because they can withstand and resist external forces. The tight packing of the amino acid residues, as well as their repeated sequences, results in the fibrous protein's overall shape and strength. On the other hand, globular proteins, such as hemoglobin and enzymes, are folded into a compact, spherical shape, making them soluble in water. This shape allows globular proteins to interact with water molecules while still maintaining their 3D structure. As a result, globular proteins are involved in various biochemical reactions in living organisms. As enzymes, they can catalyze metabolic reactions, whereas, as transporters, they can shuttle molecules around the body.

Fibrous proteins are usually involved in providing structural support to cells, tissues, and organs. The strength and resistance of these proteins come from their repeating sequences and the tight packing of amino acid residues. Globular proteins, on the other hand, are involved in a variety of biochemical functions, including enzymatic reactions and transportation of molecules. The protein's compact, spherical shape allows for interactions with water molecules while maintaining its 3D structure.

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State one possible hypothesis that can explain the global distribution of lactase persistence (lactose tolerance) and lactase nonpersistance (lactose intolerance). Be sure to include the following keywords in your explanation; selection, fitness, survival.

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The natural selection, fitness hypothesis suggests the global distribution of lactase persistence and non persistence may have arisen an adaptive response to availability or absence of dairy farming practices.

One possible hypothesis to explain the global distribution of lactase persistence (lactose tolerance) and lactase nonpersistence (lactose intolerance) is the "natural selection and fitness" hypothesis. This hypothesis suggests that lactase persistence may have been positively selected for in populations that traditionally relied on dairy consumption as a significant source of nutrients, while lactase non persistence may have been advantageous in populations with limited or no history of dairy farming.

In regions where dairy farming has been prevalent for thousands of years, individuals with the genetic mutation that allows for lactase persistence would have had a survival advantage. The ability to digest lactose, the sugar present in milk, would have provided a valuable source of nutrition, especially during times of scarcity or limited food resources. This increased fitness and survival among lactase-persistent individuals would have led to a higher prevalence of the lactase persistence trait in these populations over generations.

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Urine with fixed specific gravity is a distinctive feature of acute renal failure. Select one: True False

Answers

False, urine with fixed specific gravity is not a distinctive feature of acute renal failure.

Explanation:Urine with a fixed specific gravity is when the kidney is unable to concentrate or dilute urine in response to changes in water intake.

The specific gravity of urine can be used to detect kidney disease or injury.

In acute renal failure, the kidneys are unable to filter waste products from the blood effectively, resulting in an accumulation of toxins in the bloodstream.

This leads to a variety of symptoms and may be caused by a number of factors including injury, infection, or medication.

A decrease in urine output or anuria, a significant increase in blood pressure, electrolyte imbalances, and accumulation of nitrogenous waste products in the blood can all be signs of acute renal failure. Urine with a fixed specific gravity is not a distinctive feature of acute renal failure.

Therefore, the statement "Urine with fixed specific gravity is a distinctive feature of acute renal failure" is false.

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Beyond confused with figuring out these unknowns organisms.
I think the more I research, the more I get confused.
Please help.
Organism A Organism B
Gram Reaction Positive Negative
Morphology Bacilli Bacilli
Arrangement Chains/Pairs Chains/Pairs
Catalase Positive Positive
EMB growth Clear colonies, red agar, non-lactose fermenting
MAC growth Clear to pink colonies, non-lactose fermenting
PEA growth Growth present
TSA growth High growth High growth
MSA growth (Halophile/Halotolerant or Not) No growth No growth
Coagulase Negative Negative
Oxidase Negative Negative
Indole Negative Positive
Motile Non Positive?
Nitrate Positive Positive
Mannitol Broth Positive Negative
Glucose Broth Positive Positive
Lactose Broth Negative Negative
Sucrose Broth Negative Negative
Urea Positive Negative
Methyl Red (MR) Negative Negative
Voges-Proskauer (VP) Negative Negative
Simmon's citrate Positive Positive
Starch Negative Negative
Bacitracin Sensitive Acid-Fast Yes Spore Forming No

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Organism A is a Gram-positive, catalase-positive, non-lactose fermenting, and positive for nitrate, urea, and Simmon's citrate. Organism B is Gram-negative, catalase-positive, non-lactose fermenting, and positive for indole.

The provided information presents a comparison of various biochemical characteristics between Organism A and Organism B. These characteristics help in differentiating and identifying the organisms.

Organism A is Gram-positive, meaning it retains the crystal violet stain in the Gram staining process. It is catalase-positive, indicating the presence of the catalase enzyme that breaks down hydrogen peroxide. It does not ferment lactose, as evidenced by the negative growth on EMB (eosin methylene blue) and MAC (MacConkey agar) media. It is positive for nitrate reduction, urea hydrolysis, and Simmon's citrate utilization. Additionally, Organism A is motile, suggesting the presence of flagella for movement.

On the other hand, Organism B is Gram-negative, meaning it loses the crystal violet stain in the Gram staining process. It is catalase-positive like Organism A. It also does not ferment lactose, as indicated by the non-lactose fermenting growth on EMB and MAC media. Organism B is positive for indole production, which is a byproduct of tryptophan metabolism. It is non-motile, suggesting the absence of flagella.

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1 point If glucagon production stopped which of the following would be observed? Enhanced production of bile Increased absorption of glucose from the small intestine Increased release of lipids from the liver Decreased release of glucose from the liver and muscle cells Increased production of insulin * 1 point Chronic untreated type 2 diabetes can lead to which of the following conditions? High blood glucose High blood pressure leading to glomerular damage Obesity The excretion of hyper-concentrated urine Hypoglycemia

Answers

If glucagon production stopped, the following would be observed: Decreased release of glucose from the liver and muscle cells. Glucagon is a hormone produced by alpha cells of the pancreas and is involved in regulating glucose homeostasis.

Glucagon regulates glucose production in the liver, and its effects are opposite to insulin. When glucagon is produced, it inhibits insulin production and causes an increase in glucose production in the liver.

Hence, if glucagon production stopped, there would be a decrease in glucose production from the liver and muscle cells. This would result in the inability of the body to maintain blood glucose levels.

Chronic untreated type 2 diabetes can lead to high blood glucose. This condition is characterized by the inability of the body to regulate glucose levels. In type 2 diabetes, insulin production is affected, which leads to an inability to manage glucose levels.

The result of this is high blood glucose levels that can lead to further complications if left untreated. Some of the complications of untreated type 2 diabetes include kidney damage, nerve damage, cardiovascular disease, and vision problems.

Therefore, it is important to manage diabetes effectively to avoid these complications.

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Achondroplasia is caused by mutations in the Fibroblast growth factor receptor 3 gene. It is a disorder of bone growth that prevents the changing of cartilage to bone. O Statement 1 is correct. Statement 2 is incorrect Both statements are incorrect Statement 1 is incorrect. Statement 1 is correct. Both statements are correct Neurofibromatosis 1 is considered an autosomal dominant disorder because the gene is located on the long arm of chromosome 17. It is caused by microdeletion at the long arm of chromosome 17 band 11 sub-band 2 involving the NF1 gene. Both statements are incorrect O Both statements are correct O Statement 1 is correct. Statement 2 is incorrect O Statement 1 is incorrect, statement 2 is correct Genetic disorder is a disease that is caused by an abnormality in an individual's DNA. Range from a small mutation in DNA or addition or subtraction of an entire chromosome or set of chromosomes. O Both statements are correct Statement 1 is correct. Statement 2 is incorrect O Statement 1 is incorrect, statement 2 is correct O Both statements are incorrect.

Answers

The correct option is "Statement 1 is correct, Statement 2 is incorrect."Genetic disorders are diseases caused by abnormalities in an individual's DNA.

They can range from a small mutation in DNA to the addition or subtraction of an entire chromosome or set of chromosomes.Achondroplasia is a disorder of bone growth that prevents the changing of cartilage to bone. It is caused by mutations in the Fibroblast growth factor receptor 3 gene.

Statement 1 is correct about Achondroplasia.Neurofibromatosis 1 is caused by microdeletion at the long arm of chromosome 17 band 11 sub-band 2 involving the NF1 gene. Neurofibromatosis 1 is considered an autosomal dominant disorder because the gene is located on the long arm of chromosome 17. Statement 2 is incorrect about Neurofibromatosis 1.

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What is the body mass index? a. an index of body fat relative to height b. a measure of aerobic fitness relative to body weight c. an index of body weight relative to height d. a measure of blood glucose relative to body weight

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The body mass index (BMI) is an index of body weight relative to height. It is a numerical value calculated by dividing an individual's weight in kilograms by the square of their height in meters (BMI = weight (kg) / height^2 (m^2)). The correct answer is option c.

The body mass index serves as a tool to assess whether an individual's weight falls within a healthy range based on their height.

It is widely used as a screening tool to evaluate weight status and potential health risks associated with underweight, normal weight, overweight, and obesity.

BMI is useful because it provides a quick and simple measure to categorize individuals into different weight categories. These categories are commonly defined as follows:

Underweight: BMI less than 18.5

Normal weight: BMI between 18.5 and 24.9

Overweight: BMI between 25.0 and 29.9

Obesity: BMI 30.0 and above

It's important to note that the BMI is an indicator of body weight relative to height and does not directly measure body fat percentage or other factors related to health.

While BMI can be a useful initial screening tool, it may not provide a complete assessment of an individual's health status. Other factors such as body composition, muscle mass, and distribution of fat can influence overall health.

For instance, individuals with higher muscle mass may have a higher BMI even if they have a lower percentage of body fat. Additionally, BMI does not take into account differences in body shape or fat distribution, which can affect health risks.

For a more comprehensive evaluation of an individual's health, additional measurements and assessments, such as body fat percentage, waist circumference, and overall health indicators, may be necessary.

In summary, the body mass index (BMI) is an index of body weight relative to height. It is used as a quick and simple screening tool to assess weight status and potential health risks associated with underweight, normal weight, overweight, and obesity.

While BMI provides a useful initial measure, it is important to consider other factors, such as body composition and overall health indicators, for a comprehensive assessment of an individual's health.

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The following DNA sequences were used to generate a contig from a genome sequencing project.
ttcagattttccccg
gctaaagctccgaa
gccattaacgcc
tttagcatactacggcgtta
aaaaccggggaaaat
tccgaatcggtcattcaga
Examine the fully assembled double strand sequence. Counting bases starting at 1 for the 5'-most base of each strand, at what position is the first place where a base the same distance from each end matches? (For example if the sequence reads 5'-CACGG... from one end and 5'-GTCGA... from the other end, then the first match occurs at position 3.)

Answers

The first place where a base the same distance from each end matches in the fully assembled double strand sequence is at position 9. This is because the first base in the 5'-most strand (ttcaga) matches the ninth base in the 3'-most strand (tcagtt).

To find the first match, we can start at the 5'-most end of the sequence and count bases until we find a match with the 3'-most end of the sequence. In this case, the first match occurs at position 9.

It is important to note that this is only the first match in the sequence. There may be other matches that occur later in the sequence.

Here is a diagram of the fully assembled double strand sequence, with the first match highlighted:

5'-ttcagattttccccg-3'

| |

3'-tcagttccgaatcgg-5'

The highlighted bases are the first match in the sequence.

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Pericardial effusion: Please define and describe this diagnosis.
Please name 4 possible causes for this diagnosis. 1-2
paragraph.

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Pericardial effusion is the accumulation of fluid around the heart, which can impair its functioning. It can be caused by factors such as inflammation, heart attack, cancer, and kidney failure.

Pericardial effusion refers to the accumulation of fluid in the pericardial sac, the double-layered membrane that surrounds the heart. It can exert pressure on the heart, impairing its ability to pump blood effectively.

Pericardial effusion can be caused by various factors. Four possible causes include:

Inflammation: Inflammation of the pericardium, known as pericarditis, can lead to pericardial effusion. It may occur due to viral or bacterial infections, autoimmune disorders, or certain medications.

Heart attack: Myocardial infarction (heart attack) can cause damage to the heart muscle, leading to pericardial effusion.

Cancer: Certain types of cancer, such as lung cancer or breast cancer, can metastasize to the pericardium and result in fluid accumulation.

Kidney failure: In some cases, kidney failure can cause an imbalance in fluid levels, leading to pericardial effusion.

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Three Identical Strangers (2018) Two of the brothers were reported to show behaviors indicating emotional distress shortly after they were adopted at 6 months. What upsetting behavior did they display?

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In the documentary "Three Identical Strangers" (2018), two of the adopted brothers displayed behaviors indicating emotional distress shortly after their adoption at 6 months.

The specific upsetting behavior they exhibited was "separation anxiety." Separation anxiety refers to a condition where individuals, often children, experience excessive fear or distress when separated from their primary caregivers or attachment figures. It is characterized by clinginess, distress, crying, and a strong desire to be in close proximity to their caregivers. The brothers' display of separation anxiety indicated their emotional turmoil and the challenges they faced in adjusting to their new environment after being separated from their biological family.

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Which of the following is a fluorescent agent used to stain DNA for visualization following agarose gel electrophoresis? a. GelRed O b. GFP Oc Fluorescein O d. Methylene blue At steady state, 5 kg/s of saturated water vapor at p1 = 1 bar enters a Direct Contact Heat Exchanger and mixes with 5 kg/s of liquid water entering at T2 = 25C, p2 = 1 bar. A two-phase liquidvapor mixture exits at p3 = 1 bar. Neglect heat transfer with the surroundings and the effects of motion and gravity. Let To = 30C, po = 1 bar. Our Sun has a peak emission wavelength of about 500 nm and a radius of about 700,000 km. Your dark-adapted eye has a pupil diameter of about 7 mm and can detect light intensity down to about 1.5 x 10-11 W/m2. Assume the emissivity of the Sun is equal to 1.First, given these numbers, what is the surface temperature of the Sun in Kelvin to 3 significant digits?What is the power output of the Sun in moles of watts? 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How would you perform an extensive sequence comparison?Select one alternative:1- Perform a database search for homologous sequences after introducing several modifications to my sequence to make a better fit2- I would combine a strategy based on all the three other statements3- Perform a database search for homologous sequences by using resourses like Blosum-624- Perform a database search for homologous sequences by using resourses like Basic Local Alignment Search Tool (BLAST) How has the mobility of people and goods changed the health and well-being of nations across the globe?Has this expansion improved or hindered global health?What is your suggestion for addressing these health concerns? Suppose the economy begins in long run equilibrium. Then, suppose the U.S. dollar appreciates. a. Which curve will shift and in which direction? A. AD will shift Left B. SAS will shift Left OC. SAS will shift Right OD. AD will shift Right b. What will happen to Y, P, and the unemployment rate? A. Y will decrease, P will increase, and the unemployment rate will increase. B. Y will increase, P will increase, and the unemployment rate will decrease. OC. Y will increase, P will decrease, and the unemployment rate will decrease. O D. Y will decrease, P will decrease, and the unemployment rate will increase . c. Is the new equilibrium above, below, or equal to full employment? O A. Above full employment OB. Equal to full employment OC. Below full employment d. Is the unemployment rate above, below, or equal to the Natural Rate of Unemployment? O A. Below NRU O B. Equal to NRU O C. Above NRU e. Is the economy experiencing any added inflation; and if so, is it cost-push or demand-pull inflation? A. Yes, demand-pull inflation. B. Yes, cost-push inflation. O C. N/A the economy is not experiencing any added inflation. f. If the Fed uses monetary policy to correct the economy (assume that they neutralize the shock), which curve will shift and in which direction? A. AD shifts Right B. SAS shifts Left O C. AD shifts Left O D. SAS shifts Right g. What are 3 monetary policies the Fed could use to correct the economy? A. Decrease taxes, increase government spending, and increase transfer payments B. Decrease the discount rate, decrease the IOR rate, and decrease the ON RRP rate. OC. Increase taxes, decrease government spending, and decrease transfer payments D. Increase the discount rate, increase the IOR rate, and increase the ON RRP rate. Prompts Submitted Answers Choose a match A encodes a mutant RNA that is able to respond to a termination codon and insert an amino acid at that site Gr Choose a match encodes a tRNA that has been mutated to recognize a different codon. By inserting a different amino acid at a mutant codon the tRNA suppresses the effect of the original mutation Choose a match ERNA synthetases use to correct insertion of an incorrect amino acid after the addition reaction has been completed Chemical proofreading A mechanism that depends on incorrect events proceeding more slowly than correct events so that incorrect events are reversed before a subunit is added to a polymeric chain The reduced specificity at the wobble position of a codon A decreases the effects of a mutation at this position. B increases the effects of a mutation at this position. has no impact on a mutation's effect at this position. >M12-LCMT-F_D02.ab1AGGGGTCCCTTGACCAC-------- >M13-LCMT-F_E02.ab1AGGGGTCCCTTGACCAC. >M14-LCMT-F_Fe2.ab1AGGAGTCCCTTGACCAC >M15-LCMT-F_Ge2.ab1AGGGGTCCCTTGACCAC- >M12-LCMT-F De2.ab1---- >M16-LCMT F_H02.ab1AGGGGTCCCTTGACCACNNCCNTTC >M13-LCMT-F_E02.ab1-. >M14-LCMT-F_F02.ab1--- >M15-LCMT-F_G02.ab1>M16-LCMT-F_H02.ab1CTCA T T 1 EE -400 425 >M12-LCMT-F_D02.ab1AGGGGTCCCTTGACCAC-------- >M13-LCMT-F_E02.ab1AGGGGTCCCTTGACCAC. >M14-LCMT-F_Fe2.ab1AGGAGTCCCTTGACCAC >M15-LCMT-F_Ge2.ab1AGGGGTCCCTTGACCAC- >M12-LCMT-F De2.ab1---- >M16-LCMT F_H02.ab1AGGGGTCCCTTGACCACNNCCNTTC >M13-LCMT-F_E02.ab1-.>M14-LCMT-F_F02.ab1--- >M15-LCMT-F_G02.ab1 >M16-LCMT-F_H02.ab1CTCA T T 1 EE -400 425 >M12-LCMT-F_D02.ab1AGGGGTCCCTTGACCAC-------- >M13-LCMT-F_E02.ab1AGGGGTCCCTTGACCAC. >M14-LCMT-F_Fe2.ab1AGGAGTCCCTTGACCAC >M15-LCMT-F_Ge2.ab1AGGGGTCCCTTGACCAC- >M12-LCMT-F De2.ab1---- >M16-LCMT-F_H02.ab1AGGGGTCCCTTGACCACNNCCNTTC >M13-LCMT-F_E02.ab1-.>M14-LCMT-F_F02.ab1--- >M15-LCMT-F_G02.ab1 >M16-LCMT-F_H02.ab1CTCA T T 1 EE -400 425 >M12-LCMT-F_D02.ab1AGGGGTCCCTTGACCAC-------- >M13-LCMT-F_E02.ab1AGGGGTCCCTTGACCAC. >M14-LCMT-F_Fe2.ab1AGGAGTCCCTTGACCAC >M15-LCMT-F_Ge2.ab1AGGGGTCCCTTGACCAC- >M12-LCMT-F De2.ab1---->M16-LCMT-F_H02.ab1AGGGGTCCCTTGACCACNNCCNTTC >M13-LCMT-F_E02.ab1-.>M14-LCMT-F_F02.ab1--->M15-LCMT-F_G02.ab1 >M16-LCMT-F_H02.ab1CTCA T T 1 EE -400 425>M12-LCMT-F_D02.ab1AGGGGTCCCTTGACCAC-------- >M13-LCMT-F_E02.ab1AGGGGTCCCTTGACCAC. >M14-LCMT-F_Fe2.ab1AGGAGTCCCTTGACCAC >M15-LCMT-F_Ge2.ab1AGGGGTCCCTTGACCAC- >M12-LCMT-F De2.ab1---- >M16-LCMT-F_H02.ab1AGGGGTCCCTTGACCACNNCCNTTC >M13-LCMT-F_E02.ab1-. >M14-LCMT-F_F02.ab1--->M15-LCMT-F_G02.ab1 >M16-LCMT-F_H02.ab1CTCA T T 1 EE -400 425please write down the differences between the five DNA sequences in each of these images. 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What are some ethical, medical, economic, and societal issueswith in vitro transcription/ translation (NOT IVF (in vitrofertilization))? Include your references please! Help to determine the specifications (unstretched length and spring constant k) for the elastic cord to be used at a bungee-jumping facility. Participants are to jump from a platform 45m above the ground. When they rebound, they must avoid an obstacle that extends 5m below the point at which they jump.Establish reasonable safety limits for the minimum distance by which participants must avoid the ground and obstacle whilst accounting for different weights for each participant(you may specify the maximum allowable weight for participant). A. 0.25 cm B. 0.5 cm C. 2 cm D. 4 cm Questions 9 and 10 are related to the context below. A. A particle is moving in a two dimensional plane and the position is given by F= (4t-10)i + (8t - 5t)j 9. 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