Answer: 1/64 would be normal
Explanation:
Let the woman phenotype be Yy, since a carrier is usually heterozygous;
Also let the husband phenotype be Yy
Then Yy seperate to yield two alleles "Y" and "y"
So, the cross of the two alleles:
"Y", "y" X "Y", "y"
F1 will be YY, Yy, Yy and yy
From the crossing,
- the normal phenotype are YY,
-the carriers are Yy, Yy
- while yy is recessive.
Since, the probability that one of their offspring is normal is 1/4; for three children, then 1/4 x 1/4 x 1/4 = 1/64
Thus, probability that all three children are normal is 1/64
The region of the transcript from the 5’ cap to the nucleode just upstream of the start codon is called the 5’ untranslated region (5’UTR) because it is part of the transcript that isnot translated. How long (in ribonucleotides) is the 5’UTR
In prokaryotes the 5' UTR is 3-10 nucleotides.
In Eukaryotes the 5'UTR is 100 to many thousand nucleotides long.
Explanation:
Leader sequence or 5' UTR starts at transcription site and ends at the initiation codon just one nucleotide away from it.
It is present in mRNA.
These are GC rich and form secondary structure, helps in protein synthesis.
Shine Dalgarno sequence in prokaryotes is an example of 5'UTR.
It acts as an entry point of ribosome.
The length of the 5' untranslated region (5' UTR) varies for different mRNAs and is not a fixed value. It requires specific gene information to determine the exact length in ribonucleotides.
Explanation:The length of the 5' untranslated region (5' UTR) differs among mRNAs and is not a fixed value. This region stretches from the 5' cap to the nucleotide just upstream of the start codon. The 5' UTR is a crucial part of the mRNA as it plays roles in translation regulation and mRNA stability. Its length can influence the efficiency with which a ribosome binds and initiates translation, which can affect protein synthesis.
Proteins known as RNA-binding proteins (RBPs) can also bind to the 5' UTR, affecting the stability and lifespan of the mRNA molecule. Understanding the exact number of ribonucleotides in a specific 5' UTR would require sequencing data or detailed annotation from a specific gene. Without such specific information, the exact length in ribonucleotides cannot be determined.
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Belinda's food choices are lacto-vegetarian. Since vegetarians require 1.8 times more iron to make up for the low bioavailability of their diet, what would her daily iron requirement be?
Belinda's daily iron requirement would be 1.8 times the standard Recommended Dietary Allowance due to the lower bioavailability of iron in her lacto-vegetarian diet. This adjusted amount ensures she receives sufficient iron despite the dietary restrictions.
To determine Belinda's daily iron requirement, we need to consider that the bioavailability of iron in a vegetarian diet is lower compared to diets that include animal products. According to the information provided, the Recommended Dietary Allowance (RDA) for iron is 1.8 times higher for those following a lacto-vegetarian diet due to the low bioavailability of iron from plant sources.
The RDA for iron varies by age and gender. For instance, adolescent boys and adult men typically require less iron than adolescent girls and women of childbearing age, who may need more iron to compensate for loss during menstruation or to support a pregnancy.
Given that Belinda follows a lacto-vegetarian diet, if the standard RDA for her demographic is X mg of iron per day, her requirement would be 1.8X mg/day to account for the lower bioavailability of plant-based iron sources. For example, if the standard RDA for a woman in her age group is 18 mg/day, Belinda would need approximately 32.4 mg/day (18 x 1.8 = 32.4 mg/day).
Use the terms in the answers below to fill in the blanks in the following sentences. "An investigator is studying mutants in methionine synthesis. The _________ mutants are unable to ATP sulfurylase, also known as ______. This protein is the product of the _________ gene. "
Question is incomplete i have added full question in ask for detail section.
Answer:
Option a. met3, Met3p, MET3 is correct answer
"An investigator is studying mutants in methionine synthesis. The _met3_ mutants are unable to ATP sulfurylase, also known as _Met3p_. This protein is the product of the _MET3_gene. "
Explanation:
MET3 encodes ATP sulfurylase, which is a catalyst of first step of the sulfur assimilation pathway. This pathway results in the formation of hydrogen sulfide which is a precursor in the biosynthesis of cysteine, homocysteine, and methionine.
Source: National Center for Biotechnology Information, U.S. National Library of Medicine
Final answer:
The met mutants cannot synthesize methionine due to a non-functional enzyme ATP sulfurylase, product of the MET3 gene. Scientific methods like the use of differential media aid in the study of these mutants and their genetic makeup, which informs on methionine biosynthetic pathways.
Explanation:
An investigator is studying mutants in methionine synthesis. The met mutants are unable to synthesize methionine due to a defective ATP sulfurylase, also known as Met3p. This protein is the product of the MET3 gene. Through research on yeast strains that are unable to synthesize essential sulfur-containing amino acids due to inactivations within the biosynthetic pathway, scientists are able to use a genetic screen method to distinguish which MET genes are missing and thus understand the metabolic pathways for Met and Cys in more detail.
By using selective media containing various sulfur sources and differential media, such as BiGGY agar, these mutants could be characterized based on their growth properties. This method is essential to predict how mutations in genes involved in Met and Cys synthesis will affect the concentrations of metabolites in the pathway.
which of the followings are true about the pentose pathway? A. Pentose pathway generates NADH B. Pentose pathway generates 5-carbon monosaccharide C. Pentose pathway generates GAP D. Pentose pathway generates 4-carbon and 7 carbon monosaccharides
Answer:
C. Pentose pathway generates 5-carbon monosaccharide
D. Pentose pathway generates 4-carbon and 7 carbon monosaccharides
Explanation:
The pentose phosphate pathway (PPP) is an alternative pathway involved in the oxidation of glucose. The major products of the pentose phosphate pathway are 5- carbon monosaccharides and NADPH. Examples of the 5- carbon monosaccharides with phosphate attachments are ribulose-5-phosphate and xylulose-5-phosphate.
In addition to the 5 carbon monosaccharides, 4- carbon monosaccharides with phosphate attachment like erthyrose-4-phosphate and 7 carbon monosaccharides like sedoheptulose-7-phosphate are produced within the pathway.
The PPP doesn't generate NADH and GAP.
What are the 5 steps in the immune system
Answer:
hi!
Explanation:
The main parts of the immune system are: white blood cells, antibodies, the complement system, the lymphatic system, the spleen, the thymus, and the bone marrow.
Which of the following is/are true?
A. Sympatric speciation is best described as a random event that disrupts the allele frequencies in a population
B. Redheaded females in a population will only breed with red-headed males. The yellow-headed females will only breed with yellow-headed males. The red and yellow populations live on separate continents and rarely met in nature. The red and yellow individuals could breed and produce fertile offspring, but they normally do not.
This would be an example of sympatric speciation
C. Sympatric speciation does not require geographic isolation
D. Redheaded females in a population will only breed with red headed males. The yellow-headed females will only breed with yellow-headed males. This population lives in the same geographic area The red and yellow individuals could breed and produce fertile offspring, but they normally do not.
This would be an example of sympatric speciation.
E. Sympatric speciation can be due to sexual (mate) selection
Answer: Only Options A, C and E are correct
A) Sympatric speciation is best described as a random event that disrupts the allele frequencies in a population
C) Sympatric speciation does not require geographic isolation.
E) Sympatric speciation can be due to sexual(mate) selection
Explanation:
Sympatric speciation is a random or naturally occurring event whereby organisms of the same species:
> live in the same territory or nearby territories ( i.e not living in isolation)
> DO NOT interbreed, but select a sexual mate from a much diverse territory which results in an uneven gene flow or disruption of alleles among the population of same species of the parents organisms.
Coral reefs:
(A) may occur as barrier, fringing, or atoll formations.
(B) protect coastlines from erosion.
(C) are found in shallow, nutrient poor, tropical waters.
(D) are the most diverse of all marine environments.
(E) All of these choices are correct
Answer:
(E) All of these choices are correct
Explanation:
Coral reefs are some of the most diverse ecosystems in the world. It is produced by Coral polyps and may occur as barrier, fringing, or atoll formations.
Coral reefs protect coastlines from storms and erosion, it is a rich source of job for local communities, and provides avenues for recreation. It is rich in food nutrients and can be used medicinally too.
It exhibits mutualism with photosynthetic algae called zooxanthellae. It supplies shelter and safety and materials needed by zooxanthellae for photosynthesis while it obtains nutrients and oxygen and waste removal mechanism from it.
When 5110 first appeared, the Grants did not know how to classify him.
Where did he come from? Was he a new species of finch, or was he an infertile hybrid of known species? If birds prefer to mate with other birds from their own species, what would 5110's fate be on Daphne where there were no other birds like him?
Answer:
When a new infertile hybrid appears in an environment, it feels like an anomaly in the environment and there are no possible measures to classify it as a new specie or a member of a specie just by a physical examination. When 5110 subject appears in that environment it is possible to classify it as a piece of a terminated branch of only one member as an anomaly of the closely related species.
His fate is solely relied on better care of himself for a good life span cause there are no other measures to save him. Futhermore added he is also not a proper member of a specie so no mating oppurtunities at all.
Answer:
Genetic testing was carried out on 5110 and the grants concluded that he was a hybrid. He has G. fortis and G. scandens genes.
See below further explanation
Explanation:
When 5110 first appeared, the Grants did not know how to classify him.
Where did he come from? Was he a new species of finch, or was he an infertile hybrid of known species? If birds prefer to mate with other birds from their own species, what would 5110's fate be on Daphne where there were no other birds
Genetic testing was carried out on 5110 and the grants concluded that he was a hybrid. He has G. fortis and G. scandens genes.
He was probably from a G fortis parent and hybrid G scandens .
crossbreeding is when The same species of different features/genetic make up are made to mate with the same of different features/genetic makeup and the product can have an enhanced quality
5110 will be a big bird, if there is an interbreeding between the hybrid G Scandens and G. fortis
Consider the data on cliff swallow mortality. Why is this an example of directional selection?Overall genetic variation decreased.Selection did not occur, because no reproduction occurred (just survival).Individuals with intermediate phenotypes survived best.The average trait value changed in one direction (in this case, larger size).
Answer:
The average trait value changed in one direction.(In this case, larger size)
Explanation:
In evolution a natural selection can be disruptive, directional or stabilizing
In stabilizing no extreme trait is favored hence provides intermediate values .
Disruptive selection both extreme traits are favored over the intermediate trait.
Directional, the enviroment will favor the survival of one trait hence a change in direction either towards the left or the right.
In the case of swallow cliff mortality, selection favored the larger size.
Theories are used for: A. Testing hypotheses B. Investigating phenomenon C. Validating existing knowledge D. All of the above
Answer: D. All of the above
Explanation:
A scientific theory includes the explanation for the different aspects of cause of natural processes or events. These explanations are repeated tested and verified by utilizing the scientific methodologies. The scientific methodologies includes the making an observation, constructing a scientific question, drawing hypothesis, conducting experimentation, analyzing the data or results, and concluding the results.
A theory can be used for testing the hypothesis, investigating the cause of the natural phenomena, and give evidences for validating the knowledge of existing theories.
A scientist at the University of Iowa uses a microscope to observe cells in the brain known as microglia. He makes observations about their structure, location, and activity. The scientist ntually observes the cells undergo a sudden and radical shift in their structure/shape and their motility (ability to move). He asks himself questions about what is causing this shift in behaviors and begins to design an experiment to determine the answer. Briefly describe how the scientist practiced both the exploration and testing aspects of scientific inquiry.
Answer:
Scientific inquiry can be illustrated as a procedure by which one can comprehend and extend the present information. It incorporates deducing of a hypothesis by knowing the current scenario, arranging materials and procedures, performing the experiment, assessing the data attained and providing unbiased outcomes. It is very essential that one goes through each o the steps diligently as leaving even one can modify the anticipated outcomes.
In the given case, the scientist was witnessing microglia. He observed their activity, structure, and location. This is termed as exploration as he is utilizing the accessible resources to extend his knowledge regarding a specific topic. After that, he witnesses that they are shifting briskly, which makes him question the phenomenon.
This can be a previously familiarized information, which he is just witnessing or something which was all new to him. Thus, to know more regarding this or to answer the specific question, he decides to develop an experiment. This is considered as the testing aspect of the scientific inquiry as he is arranging the materials and will determine the reason for this movement methodically.
What is one method or line of reasoning that scientists could NOT use when testing the hypothesis that dolphins and Ichthyosaurs form a monophyletic group?
Answer:
They cannot use the reasoning, "Dolphins and Ichthyosaurs form a paraphyletic group."
Explanation:
This is because they always form a monophyletic goup. Both of the species have specific morphological similarities that are enough to consider them in monophyletic group instead of a divergent paraphyletic.
Mitochondria and chloroplasts are thought to have evolved from free-living aerobic bacteria that were engulfed by an ancestral anaerobic cell and established a mutually beneficial (symbiotic) relationship with it. Which of the following statements is NOT true about these organelles? a. They are similar in size to small bacteria.b. They have their own circular genomic DNA.c. They have their own ribosomes.d. They have their own transfer RNAS.e. They are found in all eukaryotes.
Answer:
They are found in all eukaryotic
Explanation:
Animal cell lack chloroplast
The statement 'They are found in all eukaryotes' is not true about mitochondria and chloroplasts. While mitochondria are found in most eukaryotes, chloroplasts are only present in plants and some algae.
Explanation:The statement that is NOT true about mitochondria and chloroplasts is 'They are found in all eukaryotes'. Both mitochondria and chloroplasts are thought to have evolved from free-living bacteria that were engulfed by an ancestral cell, and they formed a symbiotic relationship. They are similar in size to small bacteria, contain their own circular genomic DNA, have their own ribosomes, and transfer RNAs. These characteristics provide compelling evidence for the theory of endosymbiosis. However, it is not accurate to say that both organelles are found in all eukaryotes. While mitochondria are indeed found in almost all eukaryotes, chloroplasts are specific to plants and some algae, not present in other eukaryotes like animals or fungi.
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Gas diffusion is dependent upon the partial pressure of gasses.
Explain the movement of the following gasses between lung alveoli and blood capillaries as a function of their relative partial pressures in each location.
1. Oxygen
2. Carbon Dioxide
Diffusion is helping these gases to do the gas exchange at alveoli, and blood capillaries. And diffusion depends on partial pressure because it is the pressure produced by the gas in a mixture when it occupied by its own volume.
Explanation:
In a body, the gas exchange will occur in two places, one is lungs where oxygen will take and carbon dioxide will release into the respiratory membrane. The second place the tissues where the oxygen is released and the carbon dioxide is taking. In external respiration, exchanging gases with the external environment and this occurs in the alveoli of the lungs. In the internal respiration, change of gases will happen with the internal environment and occurs in the tissues.The exchange of gases is happened because of the simple diffusion. Diffusion is the process at which the gas-particle will move from high concentration to low concentration. there is no energy required to move carbon dioxide or oxygen between the membranes. Because they will create a pressure gradient which allows them to diffuse between the membranes.Final answer:
Gas diffusion is dependent on the partial pressure of gases. Oxygen moves from alveoli to blood capillaries due to the pressure gradient, while carbon dioxide moves from capillaries to alveoli.
Explanation:
The behavior of gases can be explained by the principles of Dalton's law and Henry's law, both of which describe aspects of gas exchange. Dalton's law states that each specific gas in a mixture of gases exerts force (its partial pressure) independently of the other gases in the mixture. Henry's law states that the amount of a specific gas that dissolves in a liquid is a function of its partial pressure. The greater the partial pressure of a gas, the more of that gas will dissolve in a liquid, as the gas moves toward equilibrium.
Gas molecules move down a pressure gradient; in other words, gas moves from a region of high pressure to a region of low pressure. The partial pressure of oxygen is high in the alveoli and low in the blood of the pulmonary capillaries. As a result, oxygen diffuses across the respiratory membrane from the alveoli into the blood. In contrast, the partial pressure of carbon dioxide is high in the pulmonary capillaries and low in the alveoli. Therefore, carbon dioxide diffuses across the respiratory membrane from the blood into the alveoli. The amount of oxygen and carbon dioxide that diffuses across the respiratory membrane is similar.
Mitosis and meiosis always differ in regard to the presence of a. chromatids. b. homologs. c. bivalents. d. centromeres. e. spindles.
Answer:
c. bivalents.
Explanation:
In Meiosis a bivalents are formed during the first stage of meiosis which is the prophase. The bivalent consist of a paired chromosome and four chromatids i.e two chromosomes in a tetrad. one chromosome comes from each parent.
In mitosis, a bivalent is not formed. There is the presence of chromatids, homologs, centromeres and spindles during mitosis and meiosis.
Mitosis and meiosis always differ in regard to the presence of bivalents (Option c).
A chromosome is a specific linear chain of genetic material (DNA), which is transmitted as a unit during cell division.A bivalent refer to two homo-logous chromosomes that exchange genetic material during recombination.A bivalent occurs only in meiosis where the interchange of genetic material between homo-logous chromosomes occurs.In conclusion, mitosis and meiosis always differ in regard to the presence of bivalents (Option c).
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Claim: An individual virus docks on the surface of a cell, infects it, hijacks the cellular machinery inside, and replicates itself, sometimes thousands of times. Justification: Based on what you learned about size, scale, and the component parts of a virus, justify with scientific reasoning how a virus is able to accomplish this.
Answer:
Virus is a ultramicroscopic pathogenic particle that consist of nucleic acid (DNA or RNA) enclosed in a protein coat and reproduces by binary fission.
Explanation:
Virus is a ultramicroscopic pathogenic particle capable of passing through bacteriological filters;
- consist of nucleic acid (DNA or RNA) enclosed in a protein coat
- capable of replicating within a living host only and spreading from cell to cell
- infects cells of bacteria, plants and animals
- hijack the metabolism of the infected cell, though they cannot carry out metabolism themselves.
- they reproduce by binary fission i.e without the formation of gametes
The claim is justified based on the understanding of virus structure and the cellular processes it exploits. Virus is able to accomplish by an extremely small infectious agent.
Viruses infect cells by a process known as viral entry. This begins with the docking of the virus to specific receptors on the host cell's surface. The virus's capsid proteins are precisely shaped to bind to these receptors, much like a key fits into a lock. Once bound, the virus can enter the cell through various mechanisms, such as endocytosis or membrane fusion.
After entering the cell, the virus releases its genetic material, which can be either DNA or RNA. This genetic material then exploits the host cell's machinery to replicate and produce viral components. The virus hijacks the host's cellular processes by using its own genetic instructions to direct the synthesis of viral proteins and the replication of its nucleic acid.
The replication process can result in the production of many new virus particles, or virions. The number of virions produced can vary widely, from a few to thousands, depending on the type of virus and the conditions within the host cell. Eventually, the newly formed virions are released from the host cell, often causing the cell to lyse (burst) and die, and the cycle repeats as these new viruses go on to infect other cells.
A virus is an extremely small infectious agent, typically consisting of a nucleic acid molecule in a protein coat called a capsid, and in some cases, an outer lipid envelope. The size of a virus is on the order of tens to hundreds of nanometers, which is several orders of magnitude smaller than a typical eukaryotic cell, which is about 10 to 100 micrometers in diameter.
Which of the following chemical reactions could be used to distinguish between a polyunsaturated vegetable oil and a petroleum oil containing a mixture of saturated and unsaturated hydrocarbons? A) addition of bromine in carbon tetrachloride B) ozonolysis C) hydrogenation D) lipidification E) saponification
Answer:
Option C
Explanation:
The main difference between a polysaturated and unsaturated fatty acid is double bond which is absent in the polysaturated fatty acid and is present in the unsaturated fatty acid. In a hydrogenation process, hydrogen molecule saturates the double or triple bonds in the presence of catalyst in an unsaturated fatty acid. Hence, in this way the hydrogenation can distinguish between the polysaturated and unsaturated hydrocarbon.
Hence, option C is correct
To differentiate between polyunsaturated vegetable oil and petroleum oil, the addition of bromine in carbon tetrachloride would cause decolorization in the presence of double bonds found in unsaturated oils. Hydrogenation could also show a change, as vegetable oils would solidify after converting unsaturated fatty acids to saturated ones, while petroleum oils would remain largely unchanged.
Explanation:To distinguish between a polyunsaturated vegetable oil and a petroleum oil containing a mixture of saturated and unsaturated hydrocarbons, the addition of bromine in carbon tetrachloride (A) would be an effective method. When bromine is added to compounds containing double bonds, such as those present in unsaturated oils, the bromine reacts and decolorizes because it adds across the double bonds. Since polyunsaturated vegetable oils have multiple sites of unsaturation (double bonds), they would react with bromine and cause a loss of the reddish-brown color of bromine in carbon tetrachloride. On the other hand, petroleum oils, which also contain saturated hydrocarbons, would not cause the bromine solution to decolorize as much or at all if they are mostly saturated.
Hydrogenation (C) is another reaction that could differentiate between the two oils, as this involves the addition of hydrogen to double bonds, converting them to single bonds. Polyunsaturated vegetable oils would show a change in physical state after hydrogenation due to the conversion of unsaturated fatty acids to saturated fatty acids. This would result in an increase in melting point and, potentially, the oils solidifying. In contrast, petroleum oils that are already saturated would not undergo a significant change during hydrogenation.
Acetylcholinesterase (AChE) is a protein that catalyzes the conversion of acetylcholine to acetate and choline. When the concentration of AChE in an aqueous solution is held constant, the rate of the reaction catalyzed by AChE increases with increasing concentrations of substrate. At low concentrations of acetylcholine, a small increase in the substrate concentration results in a large increase in the reaction rate. At high concentrations of acetylcholine, however, a large increase in the substrate concentration results in only a small increase in the reaction rate.
Which of the following statements correctly explains the observed effect of the acetylcholine concentration on the rate of the enzyme-catalyzed reaction?
A.The active site of AChE is specific for acetylcholine, and only one substrate molecule can occupy the active site at a time.
B. AChE begins converting product into substrate as the acetylcholine concentration changes from low to high.
C. The AChE protein becomes denatured as the acetylcholine concentration changes from low to high.
D. The substrate specificity of AChE changes as the acetylcholine concentration changes from low to high
Answer:
The answer is A.The active site of AChE is specific for acetylcholine, and only one substrate molecule can occupy the active site at a time.
Explanation:
Accetylcholineesterase (AChE) is a hydrolytic enzymes that hydrolyses the neurotransmitter acetylcholine into acetate and choline.AChE works at the synaptic connections, facilitating the transmission of nerve impulses by breaking down acetylcholine.Effect of Substrate on Acetylcholineesterase Activity:
At constant enzyme concentration, an increase in the substrate increases enzyme activity. At low substrate concentration, the enzyme activity also increases.At very high substrate concentration, the rate of catalysis increases up to a certain point after which in increase in rate is observed. This occurs because all the enzyme molecules are saturated with the substrate. Therefore, since, the enzyme can house only one substrate molecule at a time, the rate of catalysis becomes constant and does not rise.The statement 'the active site of AChE is specific for acetylcholine, and only one substrate can occupy the active site at a time' explains the effect of the acetylcholine on the rate of the reaction (Option A).
Acetylcholine (ACh) is a neurotransmitter required for the normal functioning of the parasympathetic nervous system (PNS).Moreover, Acetylcholinesterase (AChE) is an enzyme that catalyzes the breaking down of ACh into acetic acid and choline.AChE binds specifically to ACh at the enzyme's active site in order to hydrolyze it into acetic acid and choline.The relationship between enzyme activity is often directly proportional to the concentration of substrate, which is a consequence of the specificity of the binding of the enzyme by its corresponding substrate.In conclusion, the statement 'the active site of AChE is specific for acetylcholine, and only one substrate can occupy the active site at a time' explains the effect of the acetylcholine on the rate of the reaction (Option A).
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Several factors influence the rate of diffusion and among these factors are temperature, ____________ , electrical currents, and molecular size. For example, as temperature ____________ , the rate of diffusion increases.
Answer: States of matter; increases
Several factors influence the rate of diffusion and among these factors are temperature, STATES OF MATTER, electrical currents, and molecular size. For example, as temperature INCREASES, the rate of diffusion increases.
Explanation:
Factors influencing the rate of diffusion include:
- Temperature: High temperature increases the speed at which molecules move. Thus, as temperature increases, the rate of diffusion increases
- States of matter: Diffusion varies within the three states of matter. The diffusion of gases is much faster than that of liquids and solids, because the gas molecules are freer and therefore faster than the rest.
- Molecular size: The smaller the molecules, the faster the rate of diffusion while the larger the molecules, the slower the rate of diffusion.
Other factors include electrical currents.
"Pancreatic juices aid digestion and absorption by"__________. a. working against homeostasis b. releasing bicarbonate to neutralize gastric acidity c. secreting salivary enzymes d. producing bile e. secreting cholecystokinin
Answer:
The correct answer is - option B.
Explanation:
Pancreatic juice is the secretion of fluid from the pancreas that contain different enzymes that help in digestion and absorption of food and nutrients. The pancreatic juice is alkaline in nature as trypsinogen, amylase, nucleases and more that helps in the digestion of fat, protein, and carbohydrate.
These enzymes and juice help in the acidity of the juices by releasing the bicarbonate in the which acts as the buffer to the acidity and neutralize it. Neutralizing the gastric acidity helps in maintaining pH to the level for enzymes to act and absorption of the nutrients.
Thus, the correct answer is - option B.
Pancreatic juices aid digestion and absorption by releasing bicarbonate to neutralize gastric acidity. Therefore option B is correct.
Pancreatic juices aid digestion and absorption by releasing bicarbonate to neutralize gastric acidity.
The pancreas produces digestive enzymes, including proteases, lipases, and amylases, which are essential for breaking down proteins, fats, and carbohydrates in the small intestine.
To ensure optimal enzyme activity, the pancreas secretes bicarbonate ions to neutralize the acidic chyme from the stomach, creating a more suitable environment for the enzymes to function effectively.
This process is crucial for efficient digestion and absorption of nutrients in the small intestine.
The coordination of bicarbonate release with enzyme secretion is regulated by hormones like secretin and cholecystokinin, ensuring proper digestive function.
Therefore option B is correct.
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Match the items.
a. specialized cells that carry out specialized functions of the organ
b. the process by which tissue stem cells mature into specialized cells with characteristics of the cells they replace
c. the process of cell division to increase the number of cells
d. support cells
1. parenchymal cells
2. differentiation
3. stroma
4. proliferation
Answer:
the process by which tissue stem cells mature into specialized cells with characteristics of the cells they replace - Differentiation
the process of cell division to increase the number of cells - Proliferation
specialized cells that carry out specialized functions of the organ - Parenchymal cells
support cells - Stroma
Explanation:
Consider the following definition: "A cell is the smallest unit of an organism capable of independent functioning, composed of a membrane, enclosing a nucleus, cytoplasm and inanimate matter." Which of the following phrases best criticized this definition?
Human cells die apart from the body that supports them
Answer: The phrase 'composed of a membrane' best critized the definition
Explanation:
Not all cells are composed of membranes. Prokaryotic cells are found in unicellular organisms, they are with a single chromosome, no nuclear envelope, no membrane-bounded organelles.
Thus, the absence of membrane in the organelles of Prokaryotes like bacteria and cyanobacteria, invalidates and also critized the above definition.
Answer: The phrase composed of cell membrane criticized the definition.
Explanation:
Not all cells are composed of membrane . Eukaryotes cell are composed of membrane while prokaryotic cells are no composed of membrane.
Prokaryotes are unicellular organism that lack internal membrane bound structures and organelles. They lack nucleus and have single chromosome located in the nucleoid area of the cell But Eukaryotes have nucleus.
Zane, a 26-year-old male, came upon a car accident and immediately started to help the victims, who were pinned in their car. There was blood all over the scene and Zane acted without having any personal protective equipment. While helping the victims, he cut his arms in several places on the sharp metal and shards of glass. A few weeks later, he developed flu-like symptoms that persisted for several days. He went to his clinic and tested negative for influenza. Because of his recent exposure, Zane’s doctor was suspicious of blood borne pathogens. Which of the following disorders are blood-borne diseases?a. HIV infection b. Hepatitis c. Cirrhosis d. Leukemia
Answer:
According to the approach, in which Zane had several injuries and had contact with blood from the victims of the accident, it is likely that he has contracted a disease such as hepatitis (option b), a blood-borne disease like HIV infection.
Explanation:
Both hepatitis B and HIV infection are considered blood-borne disease, as transmission occurs through contact with a patient's blood.
In the case of Zane, you may have acquired hepatitis B if his wounds came into contact with contaminated blood, in addition to:
Acute hepatitis B infection may begin a few weeks after infection. Initial symptoms of fever, general discomfort, and weakness are similar to a cold.Zane can't have the other diseases because:
HIV infection, a blood-borne disease, takes a long time to give symptoms. Cirrhosis is a chronic degenerative disease of the liver, a product of the constant and prolonged intake of alcohol, in addition to being a consequence of chronic liver infections. Cirrhosis it's not transmitted by blood.Leukemia is a leukocyte disorder that consists of unusual increase of these cells in the blood, and it's not contagious.In conclusion, Hepatitis (type B) and HIV infection are blood-borne diseases, while cirrhosis and leukemia don't.
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Three alleles control the ABO blood types. IA and IB are codominant genes, so the combination of IAIB produces the AB blood type. The third allele Iois recessive to the other two alleles. Indicate which of these parents could produce the given child.O A X AB produce B child.O A X O produce A child.O A X B produce O child.O A X AB produce O child.O A X AB produce B child.O B X B produce O child.O AB X AB produce A child.
Answer:
The ABO blood group represents the phenomena of Co dominance and the multiple alleles.
1. The cross between O A X AB results in the formation of progeny with the genotype OA ( A blood group), OB ( blood group), AA ( A blood group) and AB ( AB blood group). Their cross results in the B progeny.
2. The cross between O A X OO results in the progeny with A blood group ( AO, AO ) and ( OO, OO) and O blood group. Their cross results in the A progeny.
3. The cross between O A X B produce the children with genotype A, B, AB and O if the parent is OB.
4. The cross between O A X AB produces the child with the genotype OA, AA ( A blood group) OB ( B blood group) and AB (AB blood group). No child with O blood type is produced by the parents.
5. The cross between O B X B produce the children with genotype OO, OB,OB and BB if the parent is OB.
6. The cross between AB X AB results in the progeny with genotype AA ( A blood group) , AB, AB (AB blood group) and BB (B blood group). Their cross results in the progeny with A blood group.
The Km of your favorite enzyme that operates by normal Michaelis-Menten kinetics is 10-6 M. You start off with a substrate concentration of 10-3 M and an enzyme concentration of 10-9 M. kcat is 10sec-1 . After a short time you have reached steady state.
a)__________ is the concentration of free enzyme.
b)__________ is the concentration of the ES complex.
Please provide an explanation. I am a bit confused.
Answer key says answers are:
a)0
b) 10^-9M
Answer: (a) 0 no free enzyme left
(b) 10^-9M
Explanation:
ANSWER:
Given that
Kcat = 10 sec-1
Km = 10^-6 M
[S] = 10^-3 M
[Enzyme] = 10^-9 M
The reaction follows the following path-
Enzyme (E) + Substrate (S) <=> ES complex -> E + Product (P)
According to the improved model of Michaelis-Menten kinetics, upon addition of substrate and the enzyme, instead of dynamic equilibrium, a steady state is reached. The time taken is very less, almost instantaneously (since Kcat is much higher than the concentrations we are dealing with (10 per second! Whereas we are dealing with concentrations as low as 10-9).
In this steady state, the Enzyme and substrate instead of existing individually, exist as an Enzyme-Substrate complex, or ES complex.
Physically, Km is a measure of how well substrate complexes with an enzyme, i.e. It's binding affinity.
You can imagine this as if 1 unit of the substrate can bind to "Km" units of Enzyme. For the give conditions, 1 M of the substrate requires 10^-6 M enzyme for complete binding. So, 10^-3 M of the substrate will require 10^-3 x 10^-6 = 10^-9 M of the enzyme, which is the exact amount of enzyme added to the reaction mixture.
So it is safe to assume that when the steady state is reached, all of the enzyme is bound to the available substrate producing the ES complex with the concentration equal to the limiting reactant, i.e. the enzyme = 10^-9 M
Hence, there will be no free enzyme left after the short duration of the reaction. And the concentration of the ES complex will be 10^-9 M
Which is the correct order of egg layer development? Group of answer choices chitin, vitelline, lipid vitelline, lipid, chitin lipid, vitelline, chitin vitelline, chitin, lipid
Answer:
1. Vitelline (The outer most layer)
2. Chitin (The inner layer)
Final answer:
For egg layer development, the correct order is vitelline, chitin, and lipid.
Explanation:
The correct order of prenatal development is zygote, embryo, fetus. This sequence starts with the zygote, which is the initial cell formed when a sperm cell fertilizes an egg cell. Following fertilization, the zygote begins a series of divisions and transforms into an embryo, where basic structures and organ systems begin to form. Eventually, the embryo develops into a fetus, during which time the tissues and organs that formed earlier begin to mature. This stage continues until birth.
In the context of embryonic development, there are three germ layers: the ectoderm, mesoderm, and endoderm. These layers form during a process called gastrulation and will eventually differentiate into all the organs and tissues of the body. The ectoderm gives rise to structures such as the skin and nervous system, the mesoderm forms components like the muscles and circulatory system, and the endoderm develops into internal structures such as the gut and lungs.
Regarding egg layer development, the correct composition of an egg cell layers is vitelline, chitin, lipid. The vitelline layer is immediately outside the egg plasma membrane, followed by a protective chitin layer, and finally, the lipid-rich yolk constitutes a major portion of the egg's content, providing nourishment for the developing embryo.
"In order to determine if the amygdala is needed for rats to learn to avoid an electric shock, Kevin plans to inject a neurotoxin into the amygdala. Which technique is Kevin using?
Answer:
Lesion technique
Explanation:
A lesion is a damage on a region of a tissue or organ.
The lesion method is a technique where a lesion or some damage is deliberately inflicted on the brain causing damage to the neurons.
The main intent being to know the specific function of a certain part of the brain.
These can be done surgically or chemically.
In our case, Kevin wants to know if the amygdala is needed for rats to learn to avoid an electric shock.
He injects a neurotoxin into the amygdala, this cause a lesion on the amygdala.
From these he can get to know the function of the amygdala.
Kevin is using the D. lesion technique to inject a neurotoxin into the amygdala to study its role in fear learning in rats.
The method he is utilizing is known as an injury. Lesioning includes purposefully harming explicit pieces of the mind to concentrate on their capability by noticing the impacts of the harm on conduct or physiology. For this situation, Kevin is disturbing the capability of the amygdala to notice its part in the learning and dread reaction to an electric shock. Studies, similar to those including patient SM or examination with rhesus monkeys, have shown that sores in the amygdala bring about diminished dread reaction, demonstrating its basic job in handling dread and aversive learning.Complete question:
"In order to determine if the amygdala is needed for rats to learn to avoid an electric shock, Kevin plans to inject a neurotoxin into the amygdala. Which technique is Kevin using?
a. MRI
b. fMRI
c. PET
d. lesion
When the energy flow through an ecosystem is measured at each trophic level, approximately __________ of the available energy passes from one level to the next.
Answer:
5-20%
Explanation:
Only approximately 5-20% of energy is passed from the one trophic level to the next trophic level and 10% law says only 10% energy is transferred. The rest of the energy is lost as heat and some energy is lost as undigested food.
So due to this at each level, most of the energy is lost and very less energy is available for the higher trophic level. This is the reason the number of organisms at higher trophic levels is very less and the trophic level is limited to 4 trophic levels.
Due to the constant random motion of its atoms and molecules, a substance will exhibit net movement from a region where it has a higher concentration to a region where it has a lower concentration.
This net movement is called ___________.
Answer: The net movement is called diffusion.
Explanation:
Diffusion is the net movement of substance from region of higher concentration to a region of lower concentration.
The movement is due to constant and random motion characteristics of atoms ,ions and molecules due to kinetic energy. It continues until the concentration of substance is uniform.
The net movement of particles from an area of higher concentration to an area of lower concentration is called as diffusion.
Diffusion is a fundamental process driven by the constant random motion of atoms and molecules in a substance. This motion, often referred to as thermal motion, results from the kinetic energy of particles. In a region with a higher concentration of particles, there is a greater likelihood of collisions and interactions among the particles.
Diffusion occurs because particles tend to spread out and distribute themselves evenly, seeking a state of equilibrium. This means that over time, substances will naturally move from areas of higher concentration to areas of lower concentration until they reach a state of dynamic equilibrium, where there is still motion but no net change in concentration.
Diffusion plays a crucial role in various biological, chemical, and physical processes, from the exchange of gases in cells to the mixing of substances in a solution, and it is governed by the second law of thermodynamics.
To know more about Diffusion here
https://brainly.com/question/24746577
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Proteins are constantly being synthesized in a living cell. Why doesn't the number of protein molecules become too great for the cell to contain, leading to cell destruction?
Answer:
Both eukaryotes and prokaryotes have gene regulation strategies. By gene regulation, the amount of protein is maintained in the living cell.
In prokaryotes, operons are present which are controlled by a regulatory gene that regulates the synthesis of protein in prokaryotes. In eukaryotes, repressor and transcription activators regulate the expression of gene and protein formation.
So by gene regulation cell maintains the number of protein molecules to become too great for a cell. Without gene regulation the survival of any organism is not possible.