ANSWER:
Based on her family history, she has only a 50% chance of inheriting a mutated BLM allele and the associated risk of colon cancer.
Option: (D)
EXPLANATION:
Bloom syndrome is a heredity disorder which causes reduced height and develops skin rashes when exposed to sunlight. This may lead to formation of skin cancer.Haploinsufficiency is a condition in which one allele is inactive and the existing allele is not sufficient to produce the required gene product for the normal function. BLM is a gene required for the making of a protein called RecQ helicase.There is no cure for the bloom syndrome and children with this disease should be given proper nourishment.Bloom syndrome occurs due to the 'mutation of the BLM gene' located in the 15th chromosome.Any help would be appreciated! Will give brainliest!!
Answer:
hey sebastian!
Explanation:
well the 4 is d) and 5 is a)
5. a)Crossing over occurs during prophase I of meiosis before tetrads are aligned along the equator in metaphase I. By meiosis II, only sister chromatids remain and homologous chromosomes have been moved to separate cells.
A 47-year old man presents with severe pain in his right great toe. He is unable to bear weight on his right foot and there was no apparent trauma. He reports the pain started last night and continued to increase in severity. Based on his physical exam and laboratory findings he is diagnosed with gout. Gout can be caused by several different mechanisms and blood samples are drawn and urinalysis is performed. The results from this individual are listed below.Patient Plasma Urate(mg/100ml) Urinary Uric Acid(mg/24hr)Normal <7.0 413±78Patient 10.5 215 What is the best explanation for the presentation of gout in this individual and please explain why the other choices are not consistent with the data or what additional information you may need to decide the root cause?1. overproduction of purines2. decreased salvage of purines3. decreased urinary excretion of uric acidAfter further discussion with the patient, he tells you that he recently switched from diet soda to regular soda because he read that the ‘fake sugar’ diet soda was bad for him; he tells you he drinks between 4 and 5 L of soda a day. (NOTE: The primary sugar in most sodas is fructose or high fructose corn syrup.)What is the potential impact of this dietary modification on the presentation of gout?
Answer:
the best explanation is 1. overproduction of purines
Explanation:
Purines are nitrogenous base that are broken down to uric acid which is excreted through the kidney as a constituent of urine. Different conditions that lead to an impaired removal of plasma uric acid cause gout. Soda drinks and many soft drinks contain low levels of purines and high level of fructose. Fructose is the only carbohydrate known to increase uric acid levels by increase the rate of degradration and synthesis of purine
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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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.
The concentration of agarose gel affects the resolution of DNA separation. For a standard agarose gel electrophoresis, a 0.8% gives good separation or resolution of large 5–10kb DNA fragments, while 2% gel gives good resolution for small 0.2-1kb fragments. 1% gels is often used for a standard electrophoresis. If you have very similar sized DNA molecules that are running too close together on an agarose gel, what solution would you apply to resolve this issue?
Answer:
Use a higher % agarose gel.
Explanation:
Agarose gels have a porous matrix. The higher the concentration of agarose, the smaller the pores, so larger DNA molecules will have more difficulty moving through the gel and they will run slower than small DNA molecules.
The higher % agarose gel has thus a better resolving power (the measurable interval between two entities -the DNA bands- is smaller). For that reason, a 2% agarose gel will allow you to differentiate better between two bands of close molecular weight, if you let the DNA fragments run long enough.
The immune system Choose one:
A. is nonspecific in its responses to invaders.
B. is composed entirely of cells, rather than tissues and organs.
C. kills all nonself cells.
D. is a system that differentiates self from nonself to neutralize harmful invaders.
Answer:
The correct answer is D. is a system that differentiates self from nonself to neutralize harmful invaders.
Explanation:
A defense system that protects the host body from pathogen infection is called the immune system. This immune system is made up of many different barriers which helps in preventing illness from pathogen attack.
Physical and chemical barriers provide the first line of defense to the body for example skin, mucous membrane, lysozyme in saliva provide first line of defence. Second line of defense is provided by non-specific immune cells like mast cells, macrophages, dendritic cells, etc.
The third line of defense is provided by pathogen-specific immune cells like T lymphocytes and B lymphocytes. These cells have the ability to differentiate self and non-self cell and kills the non-self cell that come from outside the body.
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.
What is the difference between heterozygous and homozygous individuals? What is the difference between heterozygous and homozygous individuals? Heterozygotes carry two copies of a gene while homozygotes only carry one. Homozygotes have one chromosome while heterozygotes have two similar chromosomes. All of the gametes from a homozygote carry the same version of the gene while those of a heterozygote will differ. The homozygote will express the dominant trait and the heterozygote will express the recessive trait.
Answer:
All of the gametes from a homozygote carry the same version of the gene while those of a heterozygote will differ.
Explanation:
A heterozygous individual carries both dominant and recessive alleles of a gene while a homozygous individual carries two copies of either dominant or recessive alleles of a gene. For example, the genotype TT and tt have two copies of dominant and recessive alleles respectively and are therefore homozygous genotypes. On the other hand, the genotype "Tt" is a heterozygous genotype.
An individual with a homozygous genotype would produce all the gametes having the same allele of the gene. The homozygous genotype "TT" would produce all the gametes with one copy of the "T" allele. An individual with a heterozygous genotype makes two types of gametes. The genotype "Tt" would produce 50% gametes having a "T" allele and rest 50% having a "t" allele. Segregation of alleles during meiosis produces different types of gametes in a heterozygous Individual.
Heterozygous individuals have two different gene copies and express the recessive trait. Homozygous individuals have two identical gene copies and express the dominant trait.
Explanation:The difference between heterozygous and homozygous individuals lies in their gene composition and expression. Heterozygotes have two different copies of a gene, while homozygotes have two identical copies. Heterozygotes have two similar chromosomes, while homozygotes have one chromosome carrying the same version of the gene.
All the gametes from a homozygote carry the same version of the gene, while those from a heterozygote will differ. In terms of trait expression, homozygotes will express the dominant trait, while heterozygotes will express the recessive trait.
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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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Wedges change the _______________ of the applied force.
magnitude
direction
The answer is direction.
I hope this helps!!!
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
The westward migration of Assateague Island might be halted or even reversed if all of the groins, jetties, and seawalls around Ocean City were removed. How might removal of all of these structures affect the risk of environmental damage to properties in Ocean City?
In 1933 after the hurricane ocean city inlet are formed. In order to make the inlet steady and to prevent sand flow in north direction but the natural dune is hindered. When these structures are removed, it causes drastic soil erosion and submergence of the land.
Explanation:Along the coasts of Maryland and Virginia, Assateague Island stretches for 37 miles. The islands in these part are barrier consisting of ocean currents and storms which reforms the land form. Sea level changing and movement of offshore sediments are affected.
The sand from dunes and upper beaches are removed in severe winter which reduces the width of beach. But in case of summer gentler wave action restores the shoreline.The erosion of sand takes place due to severe storms which is carried away by flood waters and re-deposited in marshes moving along westward and bring closer to mainland.
Answer:
This island is characterized for being an island that acts as a natural barrier and the erosion on it is constant, it is also extensively modified due to the fluctuation of the sea. In this way, the structures are built to prevent constant erosion and modifications caused by the sea. If these structures are considered to be removed, fluctuation by the sea will further erode this barrier island and water will enter, causing a flood to the oceanic island.
Explanation:
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.
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??
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.
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.
Your dog has puppies. A pair of homologous chromosomes inside a puppy's cell
A. must be genetically identical
B. must have the same alleles in the same locations
C. were both inherited from the same parent
D. will fight with each other to determine which one will be dominant
E. will have the same genes at the same locations
Answer:
E. will have the same genes at the same locations
Explanation:
Homologous chromosomes are the pairs of chromosomes. The members of a homologous pair are genetically and morphologically similar to each other. One chromosome of a homologous pair is inherited from the father while the other one comes from the mother.
Genes have two or more alleles. The alleles of a gene occupy the corresponding position on the homologous chromosomes. These specific positions of alleles of a gene are called loci. Therefore, a particular locus is occupied by alleles of the same gene on two homologous chromosomes.
Answer:
The correct answer is E. will have the same genes at the same locations.
Explanation:
Homologous chromosomes are the pair chromosomes one from each parent that have similarities in length, size, gene position, and centromere position. Homologous chromosomes have the same genes at the same location on the chromosome.
One gene can be found in multiple forms called alleles but a homologous chromosome can have a maximum of two alleles of a gene and these alleles can be the same(homozygous) or different(heterozygous) and can be dominant or recessive.
So a pair of homologous chromosomes inside a puppy's cell will have the same gene at the same location but can have different alleles of that gene.
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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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.
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.
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.
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.
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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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.
The first step in the catabolism of amino acids is the removal of the nitrogen as ammonia, forming a keto acid that can enter one of the carbon catabolic pathways. The alpha-keto acid pyruvate can be formed from the amino acids alanine, cysteine, glycine, serine and threonine. Consider the route for alanine catabolism. Anine reacts with to produce pyruvate and. This reaction is catalyzed by aspartate aminotransferase. alanine aminotransferase. alanine dehydrogenase. The substrate for the first step can be regenerated by reacting with NAD^+ This reaction is catalyzed by glutamate dehydrogenase. alpha-ketoglutarate dehydrogenase. alanine dehydrogenase. aspartate dehydrogenase. The coenzyme/prosthetic group required in the first reaction is thiamine pyrophosphate. biotin. pyridoxal phosphate. lipoic acid.
Coenzymes are organic compounds that many enzymes require for catalytic activity. They are frequently vitamins or vitamin derivatives.
The coenzyme prosthetic group required in the first reaction is thiamine pyrophosphate.
What is thiamine pyrophosphate?Thiamine pyrophosphate is a vitamin B1 derivative produced by the enzyme thiamine diphosphokinase.
Thiamine pyrophosphate is a cofactor found in all living systems that catalyzes a variety of biochemical reactions.
Thus, we can conclude that option A is correct.
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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).
Indicate whether or not the following antibody combinations could be used to successfully perform your assay. If a pair can’t be used successfully, indicate why not. (6pts)
• Rabbit anti-HA polyclonal antibody as antibody (II) and Goat anti-rabbit polyclonal antibody conjugated to HRP as antibody (III) Circle one: yes no
• Mouse anti-HA monoclonal antibody as antibody (II) and Goat anti-rabbit polyclonal antibody conjugated to HRP as antibody (III) Circle one: yes no
• Mouse anti-HA monoclonal antibody as antibody (II) and Goat anti-mouse polyclonal antibody conjugated to HRP as antibody (III) Circle one: yes no
Answer:
YesNo;YesExplanation:
Yes: The assay would be successful since antibody (III) is raised against rabbit therefore it would recognize and bind to the epitopes on antibody (II) because it is also a rabbit antibody. No: The assay would unsuccessful since antibody (III) is raised against rabbit and would not bind to the epitopes on antibody (II) which is mouse anti-HA.Yes: The assay would be successful and the signal would be amplified since the polyclonal mouse antibody (III) would bind all the epitopes of the monoclonal mouse antibody (II).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
Should any of Marjorie's family members whom she comes into contact with be considered at risk of pertussis? If so, who?A) Yes, the five-year-old cousin should be considered at risk.B) Yes, the aunt should be considered at risk.C) Yes, the six-week-old cousin should be considered at risk.D) No, no one should be considered at risk.
Answer:
C) Yes, the six-week-old cousin should be considered at risk.
Explanation:
young are at higher risk of pertussis as compared to older children and adult because young infants are not fully immunized. At the young age of 6 week, the cousin may not have started the DPT vaccination. so babies at this age gets easily affected when they are in contact with the infected people.
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)