Answer:
Phagolysosomal fusion
Explanation:
Macrophages ingest the bacteria through a process called phagocytosis. The phagosome undergoes a series of acidification steps where it fuses with endosomal compartments and encounters bactericidal and antigen processing molecules. Ultimately, it will fuse with the lysosome (phagolysosomal fusion) where it encounters more acidification, hydrolases and bactericidal molecules which may kill the bacteria.
RNA interference (RNAi) is a mechanism of gene silencing that is mediated by the presence of double-stranded RNA. Arrange the steps involved in gene silencing by RNAi.A-Target RNA is degradedB-RNA-induced silencing complex (RISC) binds to short dsRNAC-Long dsRNA is cleaved into short dsRNAD-the sense strand is separated from the antisense strand and degraded.E-antisense RNA pairs with the target RNA
Answer:
C.; B.; D.; E.; A.
Explanation:
C. Long dsRNA is cleaved into short dsRNA.
B. RNA-induced silencing complex (RISC) binds to short dsRNA.
D. The sense strand is separated from the antisense strand and degraded.
E. Antisense RNA pairs with the target RNA.
A. Target RNA is degraded.
RNA interference (RNAi) involves a series of steps starting with the cleavage of long dsRNA into short dsRNA, binding of these fragments to RISC, separation and degradation of sense strand, pairing of antisense RNA with target RNA, and ultimately degradation of the target RNA.
Explanation:RNA interference (RNAi) is a gene silencing mechanism mediated by the presence of double-stranded RNA that interferes with gene expression by binding to mRNA, thus preventing translation and protein synthesis. The steps involved in gene silencing by RNAi are:
Long dsRNA is cleaved into short dsRNA by an enzyme called Dicer.The RNA-induced silencing complex (RISC) binds to the short double-stranded RNA molecules.The sense strand is separated from the antisense strand and degraded.The antisense RNA pairs with the target RNA, also known as antisense RNA pairing.Finally, the target RNA is degraded, completing the gene silencing process.
In the multiple-allele series that determines coat color in rabbits, c+ encodes agouti, cch encodes chinchilla, and ch encodes Himalayan. Dominance within this allelic series is c+> cch > ch. In a cross between c+/ch and cch/ch, what proportion of progeny will be chinchilla?
Final answer:
In the cross between an agouti (c+/ch) and a chinchilla (cch/ch) rabbit, there is a 25% chance that the progeny will display the chinchilla phenotype, owing to the dominance hierarchy of c+> cch > ch.
Explanation:
In the context of coat color in rabbits, controlled by a multiple-allele series, you are asking about the progeny of a cross between an agouti (c+/ch) and a chinchilla (cch/ch) rabbit. Given the dominance hierarchy of c+> cch > ch, we can predict the potential genotypes of the offspring by creating a Punnett square. Since chinchilla (cch) is incompletely dominant over Himalayan (ch), the genotype cch/ch will result in a chinchilla phenotype. When we cross c+/ch with cch/ch, the possible genotypes for the progeny include: c+/cch, c+/ch, cch/ch, and cch/cch. As we can see, cch/ch will result in a chinchilla phenotype due to the dominance of cch over ch. However, any c+ allele will mask the expression of both cch and ch, so only the cch/ch genotype will produce chinchilla-colored offspring. Therefore, in this particular cross, there will be a 1 in 4, or 25% chance, that the progeny will have a chinchilla phenotype.
Give four specific counter examples for why each of the following definitions for a gene are incorrect. For each definition, there may be several possible counter examples. For full credit, each counter example should be distinct. For example, if a friend of yours claimed that "a gene is all the DNA sequences that determine an inherited trait", you could reply: "No, that is incorrect, because multiple genes will often determine an inherited trait."A. "A gene is a sequence of nucleotides from a start codon to a stop codon."B. "A gene is a sequence of nucleotides that codes for a functional protein."C. "A gene is a sequence of nucleotides that codes for an RNA molecule (which may be translated)."D. "A gene is the entire DNA sequence that affects the production of a protein."
Answer:
The correct answer is D A gene is the entire DNA sequence that affects the production of a protein.
Explanation:
A gene cannot be the entire DNA sequence because an entire DNA sequence contain both coding part or exons and non coding parts or introns.
But according to the defination of the gene we know a gene is all the DNA molecules that encodes that primary gene product which may be either RNA or protein.
Gene sequence does not includes non coding parts or introns.
The given definitions of genes are not accurate due to a variety of reasons. For example, not all genes code for proteins, some code for other types of RNA. Also, the concept of one gene-one protein is outdated as many traits are determined by multiple genes interacting with each other.
Explanation:A. The statement 'A gene is a sequence of nucleotides from a start codon to a stop codon' is not entirely correct. Here are some counter examples:
Some genes don't code for proteins and, therefore, lack start and stop codons.Genes contain other sequences like introns, promoters and regulators which are not encompassed between start and stop codons.B. Regarding the statement 'A gene is a sequence of nucleotides that codes for a functional protein', the counter examples are:
Not all genes code for proteins, some genes code for rRNA, tRNA, and other non-coding RNAs.Some genes produce proteins that are never functional. These proteins may be part of the process leading to functional proteins but are not functional themselves.C. For the claim 'A gene is a sequence of nucleotides that codes for an RNA molecule (which may be translated)', counter examples are:
Some genes code for RNA molecules that are not translated into proteins (e.g. rRNA, tRNA, and other non-coding RNAs).Genes contain elements like introns and promoters that never translate into RNA molecules.D. Lastly, the statement 'A gene is the entire DNA sequence that affects the production of a protein' is incorrect. The reasons include:
Not all genes affect protein production. Some genes affect the production of non-protein molecules like tRNA.The concept of gene centrality, or one gene-one protein hypothesis, is outdated. Many complex traits are determined by multiple genes interacting with the environment.Learn more about Gene definitions here:https://brainly.com/question/30767484
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The presence of nitrogen-fixing plants should benefit nearby individuals of other species, because nitrogen is a key nutrient--it is usually in very short supply. But nitrogen--fixing bacteria live inside tight nodules in the host plant's roots, and the host plant has to provide the bacteria with large quantities of sugar. Which of the following statements is correct?
Nitrogen will only become available to other, nearby plants when the nodules or the nitrogen-fixing plant itself dies.
Even when plants are young and nodules are just forming, huge quantities of nitrogen are being produced.
The nitrogen-fixing plant species has so much nitrogen that it readily leaks into the surrounding soil, making it available to nearby plants.
If nitrogen is released from the nodules into the surrounding soil, there is no cost to the host plant.
Answer:
The correct answer is "Nitrogen will only become available to other, nearby plants when the nodules or the nitrogen-fixing plant itself dies".
Explanation:
Nitrogen fixing bacteria maintain a symbiotic relationship with plants. Bacteria performs nitrogen fixation providing the plant with ammonia, while bacteria has a safe place and nutrients to prosper. This relationship is very intimate because it occurs at the root hair of the plant at structures called nodules. Therefore, nitrogen will only become available to other, nearby plants when the nodules or the nitrogen-fixing plant itself dies
Select the statement that is most correct. Select one:
a. The cell bodies of afferent ganglia are located in the spinal cord.
b. Ganglia associated with afferent nerve fibers contain cell bodies of sensory neurons.
c. Ganglia are collections of neuron cell bodies in the spinal cord that are associated with efferent fibers.
d. The dorsal root ganglion is a motor-only structure.
Answer:
b. Ganglia associated with afferent nerve fibers contain cell bodies of sensory neurons.
Explanation:
An oak tree produces two kinds of leaves: large with shallow lobes and narrow with deep lobes. If this dimorphism is based solely on the ecological condition of the amount of sunlight exposure and there is no genetic difference between leaves that express either of the two forms, then:_______.
Answer:
Natural selection is less likely to determine the fitness of one leaf in comparison to the other.
The variation in oak tree leaf forms depending on sunlight exposure, with no genetic difference in the leaves, exemplifies phenotypic plasticity, allowing efficient sunlight capture, critical for the tree's growth and dominance in its environment.
An oak tree produces two kinds of leaves: large with shallow lobes and narrow with deep lobes. If this dimorphism is based solely on the ecological condition of the amount of sunlight exposure and there is no genetic difference between leaves that express either of the two forms, then this phenomenon is an example of phenotypic plasticity. Phenotypic plasticity refers to the ability of an organism to change its phenotype in response to changes in the environment. In the context of oak trees, the variation in leaf shape depending on the sunlight exposure allows the tree to efficiently capture sunlight, which is its primary energy resource. As oak trees develop, they allocate a significant portion of their energy budget to growth and maintaining their large size, which in turn aids them in dominating the competition for sunlight.
Tigers and lions inhabit different areas and different habitats. Tigers live in forested areas in South East Asia while Lions inhabit open savannas in Africa and South West Asia. Lions and tigers never mate in the wild because they are geographically and ecologically isolated. However, in the zoo, they can mate and produce viable offspring. A) Would you classify tigers and lions as the same species using the biological species concept, yes/no and explain what was the main factor in your decision. B) Would you classify tigers and lions as the same species using the ecological species concept, yes/no and explain what was the main factor in your decision.
Answer:
A. In biology, species can be described as organisms which are similar and are able to inter breed and produce fertile offsprings. According to the definition, tigers and lions have the capability to interbreed and produce viable offsprings. Hence, they can be classified as same species.
B. In ecological terms, I will not classify lions and tigers as the same species. Because they are not living in the same areas and belong to different environments.
Metabolic regulation ______a. Always involves changing the amount of an enzyme in the cell b. Is the same thing as metabolic control c. Maintains all reactions in a pathway near the equilibrium of each d. Is the maintenance of homeostasis at a molecular level
Answer:
Metabolic regulation D: Is the maintenance of homeostasis at a molecular level
Explanation:
Homeostasis means resistance to change to maintain a stable and constant internal environment in organisms. Body maintains homeostasis for many reasons like to maintain temperature, controlling pH and glucose levels. If these levels fluctuate, person can get sick.
Metabolic regulation provides an ideal homeostasis mechanism. It is a process by which body takes in nutrients in the form of food and energy is being delivered to us. Metabolic regulation includes regulation of an enzyme in a route of the body by responding either less or more to signals.
Metabolic regulation is the molecular-level maintenance of homeostasis in the cell, achieved through mechanisms such as changing enzyme activity or amounts, and feedback inhibition.
Explanation:Metabolic regulation d. Is the maintenance of homeostasis at a molecular level. Metabolic regulation includes various mechanisms that ensure metabolic pathways operate effectively to meet cellular energy requirements and material needs. These mechanisms include changing the activity or the amount of enzymes involved in these pathways.
Enzymes can be subject to allosteric control, where their activity is influenced by the binding of allosteric effectors, which can be metabolites whose levels reflect the cell's energy status. This is one way in which cells adapt quickly to their metabolic needs. Moreover, some key enzymes in metabolic pathways, particularly those catalyzing reactions far from equilibrium or at early committed steps, are highly regulated to control the entire pathway's output.
Feedback inhibition is a specific regulatory mechanism where a product of a pathway inhibits an enzyme involved in its production, helping to prevent excessive accumulation of the product and wastage of resources. Therefore, it's clear that while changing enzyme amounts can be a part of metabolic regulation, it doesn't always involve this and can operate through other means like allosteric regulation or feedback inhibition.
The DNA in a cell's nucleus encodes proteins that are eventually targeted to every membrane and compartment in the cell, as well as proteins that are targeted for secretion from the cell.
For example, consider these two proteins:
1. Phosphofructokinase (PFK) is an enzyme that functions in the cytoplasm during
glycolysis.
2. Insulin, a protein that regulates blood sugar levels, is secreted from specialized
pancreatic cells.
The question is incomplete. The part of the question after this is: Assume that you can track the cellular locations of these two proteins from the time that translation is complete until the proteins reach their final destinations.
Answer:
PFK: cytoplasm
insulin: ER--> Golgi--> outside cell
Explanation:
The proteins which are made and have to function in the same cell like Phosphofructokinase (PFK) do not have to undergo the modification processes which are required fro transporting a protein. Such kind of proteins are translated in the free cytoplasmic ribosomes and released into the cytoplasm where they start to function.
The proteins like insulin need to be traveled to different cells where they have to function. Such kind of proteins are formed in the ribosomes which have rough Endoplasmic Reticulum attached to them. From here, they travel to the Golgi complex where they are modified and packaged. From the Golgi-complex, these proteins are moved out of the cell.
Other than the economic impact and sociological implications why should physicians be concerned about halitosis?
The correct answer is; it can be a symptom of a medical issue or cause a medical issue if it is not taken care of.
Further Explanation:
Halitosis can cause issues for the people who suffer from bad breath. When a person has bad breath, others do not want to be close to them. It can cause issues in relationships and at work. It can cause people to not keep a job or a significant other.
Many studies have shown that halitosis/bad breath can lead to or be caused from numerous disease of the mouth. Some of the medical issues are;
gingivitis xerostomiaintra-oral neoplasiaPostnasal dripLearn more about halitosis at https://brainly.com/question/1837104
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An interesting observation made by the firefighters was that some parts of the forest seemed to escape the flames. The crown fire would drop down into a ground fire in these areas, and then jump back up into the canopy as the fire burned into adjacent areas. They observed that many of these areas had previously been thinned, which means that some of the trees were cut down and hauled away to be used as timber. In other areas, thinned forests burned with as much severity as the unthinned forest. Your objective is to design an observational study to help answer the question: What was the determining factor for areas that escaped the worst impacts of the fire, compared to areas that were severely burned?
Answer:
According to the previous observational study it is possible to notice that thinning determined in great way the rate of trees that survived the fire.
Explanation:
In general the trees that had been previously cut, were the ones that presented more resistance to the fire, while the trees that had not been thinned were the ones that were severely affected in the spreading conflagration. Reason why in comparison of these two samples, it would be possible to conclude that previously cutting down trees can survive better to the spreading fire.
An observational study can be designed to compare thinned and unthinned areas of the forest to determine the factors contributing to escaping the worst impacts of the fire.
Explanation:An observational study can be designed to help determine the factors that led to some areas of the forest escaping the worst impacts of the fire compared to severely burned areas. The study should focus on comparing areas that were previously thinned and areas that were unthinned. Researchers can collect data on the severity of the fire impacts in each area, measure tree density and species composition, and analyze the fire behavior in these areas. By comparing the two types of forests, the study can help identify whether thinning plays a role in reducing the severity of fire impacts.
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A researcher added a mixture of animal proteins to a physiological bufler solution (pH 7.4) in a test tube incubated at 37"C.
The researcher then added purified pepsin to the mbture, but even after several hours, the proteins were not digested.
Which of the following would explain this result?
a)Pepsin must be converted to its active form, pepsinogen, which digests proteins.
b)The pH was too high
c)Pepsin doesn't digest animal proteins
d)The temperature was too low
Answer:
b)The pH was too high
Explanation:
pepsin is enzyme present in stomach and involves in digestion of proteins. Pepsin dont work at high pH as it normally activates around acidic pH 2. so in stomach, HCL is released from the linings, which will decrease the pH of the stomach and will cause activation of pepsin after the stimulation of food arrival in to the stomach.
Which of the following is an example of heat transfer by convection? Question 2 options: The coils of a toaster become red. A hot wood stove heats an entire room. A teaspoon in a glass of hot tea becomes warm. The handle of a pot that is being heated becomes warm.
Answer:
The correct option is B. A hot wood stove heats an entire room.
Explanation:
Convection can be described as a method of heat transfer in which heat is transferred by a medium like a gas or a fluid.
The warmer air has less density as compared to cooler air. Hence, convection currents are generated which transfer heat.
In the option B, heat is being transferred by the particles of the air. The air near the stove is warmer as compared to the air in other parts of the room. Hence, a convection current is generated which transfers the heat.
Which gland or organ is not strictly classified as an endocrine gland?a) thyroid glandb) adrenal glandc) pancreasd) pineal glande) pituitary gland
Answer:
The correct answer will be option-C
Explanation:
Endocrine glands are the organs of the endocrine system which control the functions of the body by directly secreting the chemical messengers (hormones) into the bloodstream which can act on the distant organs (target organs).
The main endocrine glands of the human body are the pituitary glands present in the brain, pineal gland in the brain, adrenal gland in kidney and the thyroid gland in the neck but the pancreas act as both endocrine and exocrine gland.
Eighty-five percent of the mass of the pancreas secretes digestive enzymes and is considered exocrine whereas only fifteen percent is considered the endocrine as it secretes the insulin hormone.
Thus, Option-C is the correct answer.
Answer:
Pancreas (Ans. C)
Explanation:
Pancreas gland: It is a glandular organ situated at the upper abdomen part of the body and it's behave as a two types of glands in one, like endocrine (hormones producing) and exocrine (producing digestive enzymes).
The pancreas gland secrets two types of hormones one is insulin, and second is glycogen which are helping to control blood sugar levels, function as an endocrine gland.
The pancreas gland excretes the enzymes which are helping in breakdown of lipids, proteins, nucleic acids, and carbohydrates which are present in food, function as an exocrine gland.
So, because of that pancreas gland is not strictly classified as an endocrine gland.
Varying the enzyme- For a one-substrate, enzyme –catalyzed reaction, double –reciprocal plots were determined for three different enzyme concentrations. Which of the three families of curve would you expect to be obtained? Explain.
Answer:
answer is in the image below.
Explanation:
For different enzyme concentrations in a one-substrate, enzyme-catalyzed reaction, the expected double-reciprocal plots are a set of straight lines that intersect the x-axis at the same point (same Km) but intersect the y-axis at different points (different 1/Vmax for different enzyme concentrations). These lines reflect the enzyme specificity and the principle of induced fit.
Explanation:In a one-substrate, enzyme-catalysed reaction, double-reciprocal plots were determined for three different enzyme concentrations. For such reactions, these plots, also known as Lineweaver-Burk plots, are commonly used to investigate different enzyme kinetics. They are typically represented by straight lines in different slopes, each line representing a different concentration of enzyme. The slope of each line is equal to Km/Vmax (where Km is the Michaelis-Menten constant and Vmax is the maximum rate of reaction).
Here, varying the concentration of the enzyme does not change the Km but will increase the Vmax as more enzyme molecules are available to catalyze reactions. Therefore, the slope of the line (Km/Vmax) will decrease as the Vmax increases with increasing enzyme concentration.
In summary, the family of curves expected is a set of lines intercepting the x-axis at the same point (same Km) and the y-axis at different points (different 1/Vmax for different enzyme concentrations). This illustrates the enzyme specificity and the principle of induced fit, which describes the interaction between substrate and enzyme.
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Select the biochemical roles of cholecystokinin, or CCK. stimulates the secretion of sodium bicarbonate by the pancreas stimulates the secretion of digestive enzymes by the pancreas activates pancreatic zymogens to form active enzymes stimulates the secretion of bile salts by the gall bladder
Answer:
The CCK stimulates the secretion of digestive enzymes by the pancreas, and stimulates the secretion of bile salts by the gall bladder.
Explanation:
The cholecystokinin (CCK) is a hormone that has a very important role in the digestive processes.
The CCK is secreted by mucosal epithelial cells in the small intestine, specifically the duodenum segment, and has several functions during the digestive process. The CCK stimulates the release of digestive enzymes from the pancreas, and it signals the gall bladder to contract and pour its contents (bile salts) into the small intestine. Other functions are the inhibition of the gastric emptying, the stimulation of the intestinal motility, the secretion of glucagon (together with the hormone gastrin), and the increase of the pyloric sphincter contraction (together with the hormone secretin).
Colorium is an autosomal dominant trait in Nutonian flies (identical to earth fruit flies in every way). There are two alleles at this locus:F = Dominant allele; flies with this allele cannot observe the color Fuschia; these flies have the Colorium phenotype.+ = Wild type allele; flies who are homozygous for this allele can observe the color Fuschia; these flies do not have the Colorium phenotype.A cross of a true breeding male with Colorium and a true breeding female without Colorium produces F1 offspring that all have the Colorium phenotype. When a complementation cross is performed (with true breeding flies), we expect the following results: A. Half of the F1 offspring will have the Colorium phenotype.B. None of the F1 offspring will have the Colorium phenotype.C. All the F1 offspring will have the Colorium phenotype.D. Half of the female F1 offspring will have the Colorium phenotype.E. Only Fuschia-colored Nutonian flies will have the Colorium phenotype.
Answer:
Half of the F1 offspring will have the Colorium phenotype
Explanation:
The Colorium has an autosomal dominant trait, so we can say the genotype of colorium phenotype is either OO or Oo and the non-colorium is oo.
In the initial cross(F1), the cross between male with colorium (OO or Oo) and female without colorium (oo) will be:
Let use the cross between Oo × oo
we will have: Oo, Oo, oo, oo (check the document below to view the punnet square of the cross)
From the results above, only half of the individuals get colorium, but in the question it is stated that the breeding produces F1 offspring that all have the Colorium phenotype.
Let's look at another cross between OO × oo
we will have: Oo,Oo,Oo,Oo (check the document below to view the punnet square of the cross)
From the above cross, all the F1 generation having colorium phenotype.
This implies that the genotype of colorium phenotype is OO
True breeding implies that the parents are homozygous and not heterozygous . As such If we make a complementation cross with true breeding flies.(i.e a true breeding female without Colorium) and the product of the F1 Ggeneration, we wil have:
the complementary cross between Oo × oo
which are: Oo,Oo,oo,oo (check the document below to view the punnet square of the cross)
Therefore, we conclude that Half of the F1 offspring will have the Colorium phenotype
What is the function of the spliceosome?A-The spliceosome transcribes mitochondrial and chloroplast genes.B-The spliceosome removes the introns and joins the exons to form the mature transcript.C-The spliceosome removes the Shine-Delgarno sequence from the 5' end of the mRNA.D-The spliceosome adds a stretch of about 170 adenine nucleotides to the 3' end of the mRNA.E-The spliceosome adds a 7-methylguanosine residue to the 5' end of the mRNA.
Answer:
B-The spliceosome removes the introns and joins the exons to form the mature transcript.
Pre-mRNA splicing is the precise removal of introns by the spliceosome from pre-mRNA to produce mature messages (mRNA). The spliceosome is critical for cell function, and faulty pre-mRNA splicing causes disease.Thus, option B is correct.
What are the main characteristics of spliceosome?The incredibly intricate macromolecular device in charge of pre-mRNA splicing is known as the spliceosome. It comes together in a highly dynamic manner from five U-rich short nuclear RNAs (snRNAs) and more than 200 proteins.
A newly created precursor messenger RNA (pre-mRNA) transcript is converted into a mature messenger RNA through the molecular biology process of RNA splicing (mRNA).
Exons are rejoined when the introns (RNA's non-coding regions) have all been removed (coding regions).
For a cell to operate properly, the spliceosome is necessary, and improper pre-mRNA splicing results in illness.
Therefore, The spliceosome removes the introns and joins the exons to form the mature transcript.
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Why can an exogenous protein, protein that is added by experimentalists that has the same amino acid sequence as endogenous protein, be resistant to RNA interference while the endogenous protein is susceptible to RNA interference?
(A) The genetic code is degenerate, meaning that there is more than one codon for a single amino acid.
(B) The exogenous protein consists of introns rendering the exogenous protein resistant.
(C) The endogenous protein contains introns which is why its susceptible to RNA interference.
(D) The genetic code is degenerate, meaning that there is only one codon for a single amino acid.
Answer:
(A) The genetic code is degenerate, meaning that there is more than one codon for a single amino acid.
Explanation:
RNA interference occurs when small single-stranded RNA molecules inhibit the expression of the mRNA having a sequence complementary to them. The sequence of mRNA is read in the form of triplets during the process of translation. The base triplets make the genetic code and specify amino acids to be added to the protein. One genetic code specifies a particular amino acid but some amino acids have more than one genetic code, that is, the genetic code is degenerate.
Therefore, an exogenous protein with the same amino acid sequence as that of the endogenous protein may be resistant to the RNA interference as its mRNA has alternative genetic codes for the same amino acid. For example, if mRNA with "GCU" code is susceptible to RNA interference, the mRNA with GCA may be resistant to it. Though both specify the amino acid alanine.
The photographs show comb jellies and an anglerfish in a benthic ecosystem.
Which feature of these organisms' bodies allows them to withstand the
extreme pressure of this environment?
O
A. Thick skin
O
B. A lack of air pockets
O
C. The presence of chemosynthetic bacteria
Answer:
B. A lack of air pockets
Explanation:
Answer: A lack of air pockets
Explanation:
The organism in the bethic zone can survive in such high pressure area. This is because the organism there have no air pockets that provides them the ability to be balanced in the water.
The human beings cannot survive under such pressure because we have air pockets inside the body which makes the body unstable and the pressure will be 1100 times more under such depth.
The fishes do not have air pockets that makes them able to withstand such pressure.
Wedges change the _______________ of the applied force.
magnitude
direction
The answer is direction.
I hope this helps!!!
Starch is a polymer of glucose molecules in plants with a role analogous to that of glycogen in animals. Starch synthesis requires ADP-glucose generated by ADP-glucose pyrophosphorylase. The biochemical mechanism of ADP-glucose pyrophosphorylase catalysis is similar to that of UDP-glucose pyrophosphorylase. What is the driving force for the ADP-glucose pyrophosphorylase reaction? hydrolysis of ADP-glucose Ohydrolysis of pyrophosphate O production of pyrophosphate O production of ADP-glucose Suppose a researcher introduces a mutation into the glucosidase domain of the mammalian glycogen debranching enzyme. The mutation inhibits the activity of the glucosidase but does not affect the other functions of the enzyme. The researcher then introduces the mutated enzyme into mammalian cells that do not express wild type glycogen debranching enzyme. Predict the effect of the mutation on glycogen metabolism. O linear glycogen chains with minimal branching O glycogen molecules with branches containing five or more glucose residues O glycogen molecules with branches containing a single glucose residue O glycogen molecules with branches containing four glucose residues
Answer:
"Ohydrolysis what do you mean"
O mean??
Which of the following is not a function associated with the cytoskeleton of bacteria? Plays a role in cell division. Plays are role in maintaining cell shape. Plays a role in segregating the bacterial chromosomes during cell division. Plays a role in the division of the cytoplasm into distinct compartments.
Answer:
Segregating the bacerial chromosomes during cell division is not a function associated with the cytoskeleton of bacteria
Explanation:
A cytoskeleto is in the cytoplasm of cells including archaea and bacteria.Cytoskeleton is responsible for the cell division, 'cell morphogenesis' is the process responsible for generation of the complex shapes of adults from the cells, DNA partitioning and cell motility. The structure, behaviour and function of the cytoskeleton varies upon every organism and the type of cell.Which of the following types of glial cells are the most abundant and versatile, and aid in making exchanges between capillaries and neurons? a. microglia b. ependymal cells c. oligodendrocytes d. astrocytes
Answer:
The correct option is d. Astrocytes are the most abundant and versatile glial cells, and aid in making exchanges between capillaries and neurons.
Explanation:
Glial cells have the function of protecting neurons and keeping them together, that is, they act as support cells. Among them we find:
Astrocytes are the most abundant glial cells, star-shaped and located in the central nervous system (in the brain and spinal cord). Its function is to maintain an appropriate chemical environment for neurons to exchange information, helping the exchange between neurons and capillaries. In addition, they wrap the capillaries forming part of the blood-brain barrier.
The other options are also glial cells but with different functions:
Microglia: They are extremely small glial cells of the central nervous system, their function is to elicit neuronal wastes and defend or protect the brain from external microorganisms .
Ependymal cells: These are ciliated glial cells of the central nervous system that form the epidendum and cover the ventricles of the brain and the central duct of the spinal cord.
Oligodendrocytes: This type of glial cell involves some neuronal axons, provides support for neuron prolongations and produces myelin.
Ependymal cells are the most abundant and versatile type of glial cells in the CNS. They produce cerebrospinal fluid, which acts as a circulatory fluid in the brain and spinal cord. Ependymal cells help cushion and protect neurons.
Explanation:Ependymal cells are the most abundant and versatile type of glial cells in the central nervous system (CNS). They aid in making exchanges between capillaries and neurons by producing cerebrospinal fluid (CSF). CSF acts as a circulatory fluid that performs some functions of blood in the brain and spinal cord. Ependymal cells filter the blood to produce CSF, which helps cushion and protect the neurons.
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1. On the planet Susru, there are three types of bears; those who like honey-nut cheerios, those who like multi-grain, and those who like plain cheerios. The phenotype is determined in an epistatic way by two loci:
a. HNNT, with alleles H (dominant) and h (recessive), and
b. MLTGRN, with alleles M (dominant) and m (recessive)
2. In a cross of a HHMM bear and an hhmm bear, and the Fls like honey-nut. A cross of two F1 bears produces the following sums two-locus genotype counts:
a. All F2 bears with at least one H allele: 801
b. All F2 bears with at least one M allele but no H alleles: 200.
3. What number of F2 bears with the hhmm genotype would produce an F2 data set that is consistent with a dominant mode of inheritance at the 1% level of significance (Hint: Think Chi-square)?
A. 55.
B. 37.
C. 118.
D. 90
E. 15
Answer:
A. 55.
Explanation:
The three types of bears are:
1. Those who like honey-nut cheerios (let that be H)
2. Those who like multigrain (let that be M)
3. Those who like plain-cheerios (let that be P)
We were told that the phenotype is determined in an Epistatis way by two loci: i.e Epistatis Dominance Mode with:
a) HNNT : with alleles Hh
b) MLTGRN : with alleles Mm
In a cross of HHNT and MLTGRN, all F(1) like honey-nut. i.e (HhMm)
Furthemore, the process continuous with a cross of two F1 bears i.e Interbreeding of F(1) bears produced bears with:
Bears who like honey-nut cheerios, Multi-grain and Plain cheerios in F(2) in the ratio of 12:3:1. This can be explained as follows:
Dominant Epistatis for types of bears in planet Susru.
(the table can be found in the attached file below)
From the table, Let:
H= honey-nut Cheerios
M= multi-grain
P= plain
From above table, (H) is dominant to (h) and epistatic to allels (M) and (m). Hence, it will mask the expression of M/m alleles. Therefore in F(2),
• bears with H-M (9/16) and H-mm (3/16) will produce bears who like honey-nut cheerios;
• bears with hhM- (3/16) will produce bears who like multigrain and,
• those with hhmm (1/16) genotype will produce bears who like Plain.
Thus the normal dihybrid ratio 9:3:3:1 is modified to 12:3:1 ratio in F2 Generation.
Now that the Epistatis Dominance Mode of Inheritance ratio is 12:3:1
12/16= 0.75
=75%
75% = 801 (i.e Phenotype: Honey-nut and Genotype: HHMM,HhMM,HHMm,HhMm,Hhmm)
3/16= 0.1875
=18.75%
18.75% = 200 (i.e Phenotype: Multigrain and Genotype : hhMM,hhMm).
If 75% is 801, then 100% will be;
(100*801)/75=x
x= 1068
100% = 1068
This implies that, the number of F2 bears with the hhmm genotype (plain) that would produce an F2 data set will be:
(1/16) * 1068 = 67( Expected) but observed will be 55.
(The Chi-Square table can be found in the attached file below)
During glycogenolysis, glycogen is broken down and converted to glucose-6-phosphate, which can enter glycolysis or be used by the liver to raise blood glucose levels. Complete the following sentences describing glycogen breakdown. Move the appropriate term to each blank. Some terms will be used more than once, and two terms will not be used at all.
1. The enzyme ________ removes terminal glucose residues from glycogen by cleaving ________ linkages.
2. Enzyme activity stops when the enzyme reaches a point _______ glucose residues from a branch point, which is a(n) ________ linkage.
3. The ________ of the debranching enzymes moves three glucose residues to another branch, connecting them by a(n) ________ linkage.
4. The ________ activity of the debranching enzyme removes the glucose at its ________ linkage.
5. The enzyme ________ continues removing terminal glucose residues
Answers:
Explanation:
1. The enzyme _Glycogen phosphorylase_ removes terminal glucose residues from glycogen by cleaving _α-[1→4 ]-glycosidic bond_linkages.
2. Enzyme activity stops when the enzyme reaches a point _Four_ glucose residues from a branch point, which is a(n) _α-[1→6]-glycosidic bond_ linkage.
3. The _glucosyltransferase_ of the debranching enzymes moves three glucose residues to another branch, connecting them by a(n) _α-[1→6]-glycosidic bond_ linkage.
4. The _glucosidase_ activity of the debranching enzyme removes the glucose at its _α-[1→6]-glycosidic bond_ linkage.
5. The enzyme _glucosidase_ continues removing terminal glucose residues.
Scientists observe that cells are found in every part of every organism. Which
does this evidence support?*
O
Cells are able to get rid of waste.
O
Cells are the basic unit of life.
О
Cells are able to take in energy. O
Cells are able to reproduce.
The triglycerides in animals tend to be solids (i.e., fats) at room temperature whereas the triglycerides in plants tend to be liquids (i.e., oils) at room temperature. Based on this fact, what can you conclude about the characteristics of the fatty acids in animal triglycerides compared to the fatty acids in plant triglycerides
Answer:
The animal triglycerides have a higher proportion of saturated fatty acids than the plant triglycerides.
Explanation:
The triglycerides have three fatty acids attached to the single glycerol molecule by ester linkage. The fatty acids may be saturated or unsaturated. The triglycerides with saturated fatty acids such as the animal triglycerides exhibit a relatively tight packing of these fatty acids and make them solid at room temperature.
On the other hand, the triglycerides with unsaturated fatty acids such as the plant triglycerides are liquid at room temperature since the presence of double bonds in the hydrocarbon chains leads to weaker interactions and does not allow the tight packing of the fatty acids.
In 1861, yellow fever struck the Mississippi Valley, hitting Memphis, TN, most forcefully. Half of the citizens fled and one-quarter of the remaining population died. Historical records indicate that the epidemic ceased only after winter frosts arrived. Why did the frost stop the epidemic? Could yellow fever epidemics reappear in the U.S.?
Yellow fever is an illness produced by a zoonotic virus called Flavivirus. The vector of this virus is the mosquito called Aedes aegypti. Is known that the life of insects is affected by the environment temperature afecting their metabolism or ability to move. The lower temperature limit for Aedes aegypti is around 10 °C, a temperature below which mosquitoes become torpid and unable to move. During the winter frost in the winter frost the temperatures were less than 3,33 °C, so the mosquito Aedes aegypti couldn't survive at this temperatures.
Final answer:
Winter frosts stop the yellow fever epidemic by killing the mosquito vector population. While yellow fever has been largely eradicated in the U.S., there is a risk of reemergence if infected individuals come into contact with the Aedes aegypti mosquito.
Explanation:
The frost stopped the yellow fever epidemic because it killed the mosquito vector population. Yellow fever is transmitted to humans by mosquito vectors, such as Aedes aegypti. When winter frosts arrive and temperatures drop, mosquitoes become less active and their population decreases, leading to a decline in the spread of the virus.
As for the reemergence of yellow fever epidemics in the U.S., while the disease has been largely eradicated in North America, there is still a risk of new outbreaks if infected individuals come into contact with Aedes aegypti mosquitoes. With increasing globalization and travel, it is possible for the virus to be reintroduced to North America, especially in densely populated urban areas.
Brown fat cells produce a protein called thermogenin in their mitochondrial inner membrane. Thermogenin is a channel for facilitated transport of protons across the membrane. What will occur in the brown fat cells when they produce thermogenin?
Answer:
Produce Heat
Explanation:
Thermogenin, also known as the uncoupling protein is produced by brown fat cells. It causes leads to the uncoupling of oxidative phosphorylation by allowing the transport of protons across the cell membrane and the energy is dissipated as heat.