Answer: Option E. All the statements are correct.
Explanation:
Thyroid gland is an endocrine gland that is found in front of the neck, it consists of two lobes connected by isthimus. It is in front of the neck lying against and around the front of larynx and trachea. Thyroid gland secretes three hormones which are thyroxine, triiodothyronine and calcitonin. Thyroxine and trio doth tribune are thyroid hormone. The parathyroid glands are tiny structures found in the thyroid gland. A thick connective tissue separates the parathyroid glands from thyroid tissues. The thyroid gland is controlled by thyroid stimulating hormone called thyrotropin which is released from the anterior pituitary gland which triggers the release of thyroid hormones.
Free palmitate is activated to its coenzyme A derivative (palmitoyl-CoA) in the cytosol before it can be oxidized in the mitochondrion. If palmitate and [14C]coenzyme A are added to a liver homogenate, palmitoyl-CoA isolated from the cytosolic fraction is radioactive, but the palmitoyl-CoA isolated from the mitochondrial fraction is not. Why is this?
Answer:
Explanation:
The fatty acyl group condensed with CoA in the cytosol is first transferred to carnitine, releasing CoA and then transported to the mitochondrion, where it is condensed with CoA again. CoA cytosolic and mitochondrial pools are therefore kept separate and no radioactive CoA enters the mitochondrion from the cytosolic pool.
Final answer:
The palmitoyl-CoA is radioactive in the cytosolic fraction due to the initial activation with [14C]CoA. In the mitochondrial matrix, the carnitine shuttle uses a fresh molecule of CoA, not the radioactive one, resulting in non-radioactive palmitoyl-CoA within the mitochondria.
Explanation:
The explanation for why palmitoyl-CoA isolated from the cytosolic fraction is radioactive but the one from the mitochondrial fraction is not lies in the fatty acid oxidation process and the role of carnitine palmitoyltransferase system. Initially, palmitate is activated to palmitoyl-CoA in the cytosol by acyl-CoA synthetase, which requires coenzyme A (CoA) and ATP. This activated form, which now contains [14C]CoA and is radioactive, is then transferred to the mitochondrion with the help of the carnitine palmitoyltransferase system.
The carnitine shuttle transports fatty acids into the mitochondrial matrix for β-oxidation. It involves the conversion of palmitoyl-CoA to acyl-carnitine by carnitine palmitoyltransferase I (CPTI) on the outer mitochondrial membrane, and the transport of acyl-carnitine across the inner mitochondrial membrane by carnitine-acylcarnitine translocase. Once in the matrix, carnitine palmitoyltransferase II (CPTII) converts acyl-carnitine back to palmitoyl-CoA, which enters the β-oxidation pathway. However, this pathway uses a fresh molecule of CoA that is already present in the mitochondrial matrix, and not the [14C]CoA originally attached to palmitate in the cytosol. Hence, the palmitoyl-CoA in the mitochondrial fraction is not radioactive.
Suppose Alia recently learned that she inherited a mutant RB1 allele from her mother, who had retinoblastoma. RB1 is a tumor suppressor gene that is related to retinoblastoma. Why would Alia be at higher risk for getting retinoblastoma at an earlier age than her sister, Francine, who inherited a normal RB1 allele from their mother?
Answer:
Tumour suppressor genes can readily convert into tumour causing genes if mutations arise in them.
As Alia has one of the alleles that is a mutant, she will have more chances of getting retinoblastoma if the other allele gets mutated. A person who has inherited two functioning alleles of a tumour suppressor gene will require both the alleles to get mutated for the development of the tumour, hence her sister, Francine, who inherited a normal RB1 allele will have lesser chances.
Answer:
A person who inherited two functioning alleles of a tumor suppressor gene needs both alleles to mutate for tumors to develop
Explanation:
How often do human skin cells divide? Why might that be? Compare this rate to how frequently human neurons divide. What do you notice?
Answer:
explained
Explanation:
The skin cells divide every day, as the epithelial layer is the outer protective layer, it needs to be replaced from time to time, which helps in healing wear and tear. Our skin cells divide once in each day.
In contrast, the nerve cells do not divide at all. That is their frequency of cell division is zero. They cannot divide because they lack centriole in them.
Human skin cells undergo division approximately every 27 days. This rapid turnover is necessary to replenish and repair the skin, which is exposed to constant wear and tear, environmental factors, and potential damage.
On the other hand, human neurons generally do not undergo division once they are fully developed. Unlike skin cells, neurons are post-mitotic, meaning they have exited the cell cycle and have limited or no ability to divide further.
The notable difference in the division rate between skin cells and neurons highlights the contrasting needs and functions of these cell types.
Thus, skin cells require frequent division to maintain the integrity and functionality of the skin, while neurons focus on specialized functions and long-term stability rather than continuous replication.
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Many renewable energy sources are intermittent. What does this mean, and why does it pose a challenge to the use of these sources for electrical power generation
Answer:
Explanation:
Many intermittent
Intermittent renewable energy sources which are intermittent are difficult and challenging which means that they have an disrupting effect on the conventional procedures used to plan the daily maintenance and operations of the electric grid. The power fluctuates over different and many time horizons which will force the operators to adjust it before time i.e an hour ahead to day ahead to adjust the operating procedures manually.
Describe the similarities and differences between voltaic and electrolytic cells. Be sure to address the following items in your description: spontaneous reduction or oxidation reactions and which electrode(s) they occur at charges on electrodes .
Answer:
Both cell are opposite to each other in function and make
Explanation:
a) A voltaic cell works opposite to that of electrolytic cell. It produces chemical energy from electrical energy while the later one produces electrical energy from chemical energy.
b) The chemical reactions in voltaic cell are spontaneous while chemical reactions in electrolytic cells are non –spontaneous.
c) The anode and cathode carry opposite charges in the two cells. In case of voltaic cell, anode carries negative charge while cathode carries positive charge.However, in case of electrolytic cell both cathode and anode bear opposite charges wrt voltaic cell
1. What are growth factors?
2. Suggest an intracellular regulatory molecule (from your text) that growth factors might stimulate to control cell division. Please describe how it works
3. In the following experiment, you will test the effect of two different growth factors on the growth of your three cell lines. Please describe an experimental design with appropriate negative controls Remember from Module 1 Interactive Lab: A negative control is one that does not have the substance for which you are testing. For example, if you were testing the effect of a fuel additive on your car's gas mileage, a negative control would be a measurement of the gas mileage without using the additive.
4. Do you think all cell lines will respond the same way to the growth factors? Construct a hypothesis for this experiment.
Growth factors stimulate cell division and may signal intracellular molecules such as MAPK to encourage this process. An experiment using growth factors with different cell lines and a control group can be designed to measure cell growth and proliferation. Different cell lines may respond differently to the same growth factors due to their unique characteristics.
Explanation:Growth factors are a group of proteins that stimulate cell division. They bind to receptors on the cell surface and activate signaling pathways, resulting in cell proliferation.
An example of an intracellular regulatory molecule that growth factors might stimulate is the Mitogen-Activated Protein Kinase (MAPK). When a growth factor binds to its receptor, it triggers a cascade of reactions, including the activation of MAPK. The activated MAPK then moves into the cell nucleus, where it stimulates the transcription of genes involved in cell division.
To conduct the experiment mentioned in your question, one method is to culture three separate cell lines, each with its unique growth conditions. Two groups of each cell line should be set up: one exposed to a growth factor, another serving as the negative control, not exposed to the growth factor. The growth and proliferation of cells can then be measured over a set period, comparing those exposed to the growth factor with those not.
It's reasonable to hypothesize that not all cell lines will respond identically to the growth factors. Each cell line has its own distinct characteristics and may have different receptors or intracellular signaling pathways that may be more or less responsive to a particular growth factor.
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Growth factors are ligands that promote cell growth and stimulate cell division. They bind to cell-surface receptors called RTKs, initiating signaling pathways that control cell division. An intracellular regulatory molecule that growth factors might stimulate is the MAP kinase protein, which triggers the expression of proteins involved in cell division. In an experimental design, groups with and without growth factors can be compared to determine their specific effects on cell growth. It is hypothesized that different cell lines will respond differently to growth factors based on their varying sensitivities and requirements.
Explanation:1. What are growth factors?
Growth factors are ligands that promote cell growth and stimulate cell division. These ligands bind to cell-surface receptors called receptor tyrosine kinases (RTKs), which initiate signaling pathways that control cell division.
One intracellular regulatory molecule that growth factors might stimulate is the MAP kinase protein. When activated by the RAS G-protein, the MAP kinase protein triggers the expression of proteins that interact with other cellular components to initiate cell division.
To test the effect of two different growth factors on the growth of three cell lines, you can have three groups: Group A (control) without any growth factors, Group B with growth factor 1, and Group C with growth factor 2. By comparing the growth of the three cell lines in each group, you can identify the specific effects of the growth factors on cell growth.
A possible hypothesis for this experiment is that different cell lines will respond differently to the growth factors. This is based on the understanding that different cell types have different sensitivities and requirements for growth factors, which may result in variations in their response to the stimuli.
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Which of the following takes place during translation?
a) DNA replication
b) The conversion of genetic information from DNA nucleotides into RNA nucleotides
c) The conversion of genetic information from the language of proteins to the language of enzymes
d) The conversion of genetic information from the language of nucleic acids to the language of proteins
Answer:
d) The conversion of genetic information from the language of nucleic acids to the language of proteins
Explanation:
Answer:
The answer is: d) The conversion of genetic information from the language of nucleic acids to the language of proteins.
Explanation:
This is what is known as the central dogma of biology. All genetic information is found in genes, which are segments of DNA that contain genetic information. The human genome contains approximately 30000 genes. However, only a small part is codifying.
For this information to happen, the first thing is that it must be copied (replication) and this process happens in the kernel. DNA is copied to messenger RNA (transcription). This information is then used for protein construction (translation), which occurs in the cytoplasm.
These three processes are known as the central dogma of biology, which is that information goes from DNA to RNA and from it to proteins.
Which of these statements is completely true about the WT situation above?
A. Gene Z is expressed in the presence of Stimulus A and blocked in the absence of A
B. Gene Z is blocked in the presence of Stimulus A and expressed in the absence of A
C. Gene Z is blocked in the presence of Stimulus A and blocked in the absence of A
D. Gene Z is expressed in the presence of A and expressed in the absence of A
E. None of the above
Answer:
The answer is Letter B
Explanation:
Gene Z is blocked in the presence of Stimulus A and expressed in the absence of A
A male Flagus fly with the Barkus phenotype is crossed with a female who has the wild-type phenotype. A second cross is performed in which the female Flagus fly has the Barkus phenotype and the male has the wild-type phenotype. This second cross is known as a:
Answer:
Reciprocal cross
Explanation:
a reciprocal cross is a breeding experiment designed to test the role of parental sex on a given inheritance pattern. In this type of test cross, male of one genotype and female of different genotype are crossed and the opposite is true as well. ( i.e the female of one genotype and male of different genotype are crossed.)
As we can see in the male Flagus fly with the Barkus phenotype is being crossed with a female with the wild-type phenotype.
In the second test, we can see that the opposite is true of our first statement where the female Flagus fly has the Barkus phenotype and the male has the wild-type phenotype after being crossed again thereby reversing the genotype.
Color-blindness is an X-linked recessive disorder. Mike is color-blind. His wife, Meg, is homozygous for normal color vision allele. Using Punnett squares, derive and compare the genotypic and phenotypic ratios for the offspring of this marriage.
a. What is Mikes's genotype?b. What is Meg's genotype?c. What is the genotypic ratio of offspring?d. What is the phenotypic ratio of offspring?e. If they have eight children, how many of them would you expect to be color-blind?
Mike's genotype is XcY, Meg's is XX. Using Punnett square, the genotypic ratio of their offspring is 1:1 and the phenotypic ratio is 100% normal vision. If they have eight children, none are expected to be color-blind.
Explanation:In terms of genetics, color blindness is an X-linked recessive disorder. Mike, being color-blind, must have the genotype XcY. His wife, Meg, has normal color vision and is homozygous dominant, so her genotype is XX.
When we use a Punnett Square to derive the genotypic ratio of their offspring, we see that all daughters will be heterozygous carriers (XcX) and all sons will have normal vision (XY). Thus, the genotypic ratio is 1:1 (XcX: XcY).
Regarding the phenotypic ratio, all offspring are expected to have normal vision because they will inherit at least one normal vision allele. Therefore, the phenotypic ratio is 100% normal vision.
If Mike and Meg have eight children, because female offspring will be carriers and male offspring will possess the normal vision allele, we would expect none of them to be color-blind.
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In the Cori cycle, when glucose is degraded by glycolysis to lactate in muscle, the lactate is excreted into the blood and returns to the liver. In the liver, lactate is converted back into glucose by gluconeogenesis. For each given enzyme, identify whether it is involved in the glycolysis pathyway, gluconeogenesis pathway, both pathways, or neither pathway.1. glyceraldehyde 3-phosphate dehydrogenase O glycolysis O gluconeogenesis O both O neither 2. glucose-6-phosphatase O glycolysis O gluconeogenesis both O neither 3. alcohol dehydrogenase O glycolysis O gluconeogenesis both neither 4. phosphoenolpyruvate carboxykinase O glycolysis O gluconeogenesis O both 5. phosphofructokinase-1 O glycolysis O gluconeogenesis both O neither 6. phosphoglycerate mutase O glycolysis gluconeogenesis O both neither 7. hexokinase O glycolysis O gluconeogenesis. O both neither 8. pyruvate dehydrogenase O glycolysis gluconeogenesis
Answer:
Explanation:
1. glyceraldehyde 3-phosphate dehydrogenase: O both
2. glucose-6-phosphatase: O gluconeogenesis
3. alcohol dehydrogenase: O neither
4. phosphoenolpyruvate carboxykinase: O gluconeogenesis
5. phosphofructokinase-1: O glycolysis
6. phosphoglycerate mutase: O both
7. hexokinase: O glycolysis
8. pyruvate dehydrogenase: O neither
In a normal cell,______ of the errors that occur during DNA replication are ____. A. 10%......repaired by the exonuclease activity of DNA polymerase B. 99%.....repaired by the exonuclease activity of DNA polymerase C. 99%.....repaired by the mismatch repair pathway D. 99%......not repaired prior to mitosis
Answer:
The correct option is B. In a normal cell, 99% of the errors that occur during DNA replication are repaired by the exonuclease activity of DNA polymerase
Explanation:
DNA proofreading can be described as a method by which in which the DNA polymerases check and repair any errors they might have made during the replication of the DNA. The exonuclease activity of the DNA interprets whenever a wrong nucleotide has been placed and corrects it.
Without the method of proofreading, the DNA would undergo many errors during the process of replication which will lead to many faulty proteins being formed.
About 99% of the errors that occur during DNA replication are repaired by the mismatch repair pathway, which helps in recognizing misincorporated bases, excising them, and correcting the sequence to maintain genetic stability.
Explanation:In a normal cell, the correct answer is that 99% of the errors that occur during DNA replication are repaired by the mismatch repair pathway. DNA replication is an incredibly accurate process with DNA polymerase adding nucleotides to the growing DNA strand, and also proofreading each new addition for errors. When errors do slip past the proofreading activity of DNA polymerase, the mismatch repair system corrects most of these errors by recognizing the misincorporated base, excising it, and resynthesizing the correct sequence. This mechanism helps maintain the genetic stability by reducing the occurrence of mutations that can lead to conditions like cancer. In the absence of proper mismatch repair, these errors can persist and lead to more permanent genetic damage.
A protein contains an ER signal sequence at amino acid positions 7 to 15. At amino acids 25 to 40 the protein also contains a mitochondrial signal sequence. First, describe in one sentence each, what the sequence would look like (i.e. amino acid makeup) for ER and mitochondrial localization. Then, answer where does this protein traffic to (can be one word)? Why? Can it traffic to the other organelle at a later time (can be one word)? Why or why not?
A protein with both an ER and a mitochondrial signal sequence at the N-terminus will be trafficked first to the ER due to the co-translational recognition of the ER signal sequence. Once the protein begins translocation into the ER, it cannot be trafficked to the mitochondria later.
An Endoplasmic Reticulum (ER) signal sequence is typically located at the N-terminus of the protein and is characterized by a hydrophobic or positively charged region that directs the protein to the ER membrane. Mitochondrial signal sequences are also usually found at the N-terminus and include a stretch of hydrophobic and positively charged amino acids that target the protein to the mitochondrial matrix.
Given that both signal sequences are at the N-terminus of the protein, the ER signal sequence will direct the protein to the ER first as these sequences are recognized co-translationally as the protein emerges from the ribosome. Proteins that have an ER signal sequence are generally targeted to the ER for secretion or for insertion into the ER membrane. As for whether the protein can traffic to the mitochondria at a later time, the answer is no; once a protein has been targeted to the ER and begins its translocation, it cannot be redirected to another organelle.
Suppose that RNA polymerase was transcribing a eukaryotic gene with several introns. In what order would the RNA polymerase encounter the following elements in the DNA sequence of the gene?A-stop codonB-translation initiation codonC-TATA box elementD-3' UTRE-5' UTRF-splice acceptor site
RNA polymerase, during the transcription of a eukaryotic gene with introns, would encounter DNA elements in the order of TATA box, 5' UTR, translation initiation codon, splice acceptor site, stop codon, and 3' UTR. Transcription occurs in the 5' to 3' direction, beginning with the promoter region and moving downstream.
If RNA polymerase were transcribing a eukaryotic gene with several introns, the order in which it would encounter the elements in the DNA sequence of the gene would be as follows:
C-TATA box element - This is a component of the promoter region where transcription factors assemble and RNA polymerase II binds to start transcription.E-5' UTR (Untranslated Region) - The segment of mRNA that is transcribed first but is not translated into protein.B-translation initiation codon - This is typically the AUG codon, where the ribosome will begin the translation process.F-splice acceptor site - The location within the intron where splicing will occur during mRNA processing.A-stop codon - Signals the end of the protein-coding sequence within the mRNA.D-3' UTR (Untranslated Region) - The segment of mRNA that follows the coding sequence and is involved in regulation and stability of mRNA.It's important to note that RNA polymerase II transcribes the gene from the 5' to 3' direction, starting with the promoter region and moving downstream through the gene.
Securin is a cytoplasmic protein that binds to separase, an enzyme that degrades cohesin. When separase is bound to securin, separase is inactive. When separase is released, it immediately becomes active. Securin and separase remain attached to each other as long as securin is phosphorylated. Based on the role of securin, it must become dephosphorylated just prior to
metaphase.
anaphase.
prometaphase.
prophase.
telophase.
cytokinesis.
Answer:
Anaphase.
Explanation:
Two main process of cell division are mitosis and meiosis. The regulation of the cell cycle is maintained by the regulation of cyclins, cdks and important protein factors.
Securin and separase are important proteins that allow the transition of one phase of the cell cycle to the another phase. The securin helps in the separation of the chromosome and sister chromatid that occurs in anaphase. Hence, dephosphorylation must occur prior to anaphase.
Thus, the correct answer is option (2).
In mountain Koalas, brown coat color is recessive to gray coat color. A gray female rabbit gives birth to four gray-coated and three brown-coated baby rabbits. What can be deduced about the genotype of the baby rabbits’ father?
Answer:
gg
Explanation:
It is given that allele for brown colored coat is recessive to gray colored coat. Let the allele for brown color coat be represented by “g” and gray color coat be represented by “G”
Since brown color is a recessive trait. It will be present only in offspring having genotype “gg”. This means that genotype of both the parents will have “g” allele.
Since it is given that female is gray colored thus genotype of female would be “Gg”
So the father would have genotype “gg”
Which of the following is NOT part of genetic "linkage mapping"?
A. Calculating recombination frequencies using a testcross
B. Assuming that recombination frequency is proportional to the distance
between two linked genes
C. Finding and studying alternative (e.g., mutant) phenotypes of a character
D. Using information on DNA sequences
E. Studying characters two at a time
Answer:
The correct answer is option D.
Explanation:
Linkage mapping is the mapping process in which the genes present on the chromosomes are mapped on the base of their linkage. The linkage mapping helps in calculating how frequent recombination occurs using testcross.
Linkage mapping does not utilize the information that is on DNA sequences, however, it helps in assuming the distance of two linked genes is proportional of the recombination frequency. By the recombination, frequency mutations can be found and study.
Thus, the correct answer is option D.
Interferons are cytokines produced by host cells in response to intracellular infection. There are two types of interferons, type I and type II. Which of the following is a difference between these two types? Choose one:
A. Type I is part of innate, nonspecific immunity, while type II is part of adaptive, specific immunity.
B. Type I binds to extracellular pathogens, while type II binds to intracellular pathogens.
C. Type I includes interferon-alpha and interferon-beta, while type II includes interferon-omega.
D. Type I interferes with viral replication, while type II activates white blood cells.
Answer:
A. Type I is part of innate, nonspecific immunity, while type II is part of adaptive, specific immunity.
Explanation:
Type I interferons: Are produced early on during infection and are responsible for activation of the innate immune response, e.g Natural Killer cells.
Type II interferons: Are produced as part of the innate immune response and act as a link between innate immune response and activation of the adaptive immune response.
Interferon type I is part of innate immunity, while type II is part of adaptive immunity.
Explanation:The correct answer is option A: Type I interferons are part of the innate, nonspecific immunity, while type II interferons are part of the adaptive, specific immunity. Type I interferons, including interferon-alpha and interferon-beta, are produced by most cells in response to viral infection and have antiviral effects. Type II interferon, also known as interferon-gamma, is mainly produced by activated T cells and natural killer cells to activate white blood cells and enhance immune response.
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_______ results from the interaction of a number of gene loci specifying a given trait. Rather than stepwise inheritance patterns, phenotypic distributions are more curve-like.
Answer:continuous variation
Explanation:
Variation is the differences between individuals within a species. Variation could be caused by genetic inheritance or by environment factor(s). There are two types of variation which are; continuous variation and discontinuous variation.
''CONTINUOUS VARIATION results from the interaction of a number of gene loci specifying a given trait. Rather than stepwise inheritance patterns, phenotypic distributions are more curve-like.'' Examples of continuous variation are; skin colour, air colour, height and weight.
The law of independent assortment allows for genetic recombination. The following equation can be used to determine the total number of possible genotype combinations for any particular number of genes: 2g = Number of possible genotype combinations (where g is the number of genes)1 gene: 21 = 2 genotypes2 genes: 22 = 4 genotypes3 genes: 23 = 8 genotypesa) Consider the following genotype: YySsTt. How many different gamete combinations can be produced?
Answer:
Gametes that will be produced are:
YST, YSt, YsT, Yst, yST, ySt, ysT, yst
Explanation:
From YySsTt, it is obvious that as there 3 type of genes.
The number of gametes = [tex]2^{3}[/tex] = 8
5 genes = [tex]2^{5}[/tex] = 32 genotypes
10 genes = [tex]2^{10}[/tex] = 1024 genotypes
20 genes = [tex]2^{20}[/tex] = 1048576 genotypes
In a long bone, the osteons are:
A. lined up in the same direction as the diaphysis of the bone
B. lined up perpendicular to the long axis of the bone, in the direction of perforating canals.
C. arranged in an irregular pattern
D. are separated by medullary spaces
E. are lacking in the diaphysis of the bone
Answer:
B. lined up perpendicular to the long axis of the bone, in the direction of perforating canals.
Explanation:
Osteons are formations of concentric bone layers (lamellae) that tend to run parallel to the long axis of a bone. They sorround the Haversian canal, a canal that contains blood vessels that supply the osteocytes and are the structural unit of compact bone.
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Growth factors and cytokines both lead to tyrosine phosphorylation through receptors, but they do so through different mechanisms. What is the key difference between the receptors for these two classes of ligands in terms of their tyrosine kinase activity?
Answer:
The growth factor receptors have a kinase domain while the Cytokines receptors do not contain a kinase domain as part of their structure.
Explanation:
The two are signaling molecules that control cell activities in some manners, such paracrine, endocrine and autocrine manners.
The receptor kinase domain can be specific for substrate sites in which phosphorylation occurs.
Certain species of frogs in the genus Phyllobates have a powerful defensive adaptation−their skin can secrete a milky fluid that contains an extremely toxic compound called batrachotoxin (BTX). These frogs, which are found in Colombia, are known as poison dart frogs because some indigenous Colombian hunters coat the tips of their blowgun darts with the frogs' skin secretions. An animal hit by one of these darts dies quickly. What is the mechanism of action of BTX?
Answer:
Batrachotoxin is one of the most potent biotoxins, more toxic than strychnine (250x approx.), causing ventricular fibrillation and neuromuscular toxicity. Its action mechanism is producing a conformational change that allows a permanent state of the channel opened that increases the resting sodium permeability. This generates action potentials in excitable membranes of nerve and muscle; causing the blocking of neuromuscular transmission and evoking muscular contracture. Death results from respiratory paralysis. In addition, BTX also causes arrhythmias, ventricular tachycardia, and fibrillation.
Explanation:
Batrachotixin (BTX, PubChem CID: 6324647), obtained from frogs of the genus Phyllobates of the Choco rain forest of Colombia, activate permanently the voltage-gated sodium channels. Previous work demonstrated that a phenylalanine residue approximately halfway through pore-lining transmembrane segment IVS6 is a critical determinant of channel sensitivity to BTX. In addition, Li 2002 demonstrated an electrostatic ligand-receptor interaction at this site, possibly involving a charged tertiary amine on BTX.
- Cataldi, M. (2016). Batrachotoxin.
- Li, H. L., Hadid, D., & Ragsdale, D. S. (2002). The batrachotoxin receptor on the voltage-gated sodium channel is guarded by the channel activation gate. Molecular pharmacology, 61(4), 905-912.
- National Center for Biotechnology Information. PubChem Database. Batrachotoxin, CID=6324647, https://pubchem.ncbi.nlm.nih.gov/compound/Batrachotoxin (accessed on Dec. 4, 2019).
___________ is a measure of the degree to which the phenotypic variation of a given trait is due to genetic factors.
g Deficiencies in glucose 6‑phosphate dehydrogenase frequently present as anemia. Glucose 6‑phosphate dehydrogenase catalyzes in the pentose phosphate pathway. A deficiency in glucose 6‑phosphate dehydrogenase causes a dearth of in all cells, but this is particularly problematic in red blood cells because they lack mitochondria. Its primary role in red blood cells is to maintain levels of Red blood cells with low levels of this compound are more susceptible to
Answer:
Its primary role in red blood cells is to maintain levels of red blood cells with low levels of this compound are more suceptible to oxidative damage.
Explanation:
Pentose phosphte pathway helps in the formation of reducing equivalent named NADPH.
The so formed NADPH is used to maintain a normal cellular level of glutathione which act as antioxidant.
As glutathione is not mantained in its normal level, it ability to reduce to adverse effect of hydrogen peroxide is cut down.
As a result cell is more suceptible to oxidative damage.
Evaluate one strategy that manufactures can use to reduce the human costs of extracting minerals for use in cell phones. In a paragraph, explain how the strategy can help resolve the problem citing evidence to support your claim.
Answer:
Manufacturers could promote the recycling of cell phones.
Explanation:
Mining for the manufacture of cell phones and other electronic devices has caused at least six million refugees in the Democratic Republic of Congo in the last two decades. The high demand for minerals important for mobile technology causes humanitarian and environmental conflicts in the Democratic Republic of Congo. This is a serious issue that should be discussed among telephone companies to establish measures that will lower the human costs of depleting these ores.
One way to alleviate these problems would be through the recycling of disused electronic devices. Telephone companies could set up telephone collection points that are not in use and thus recover the ores needed to make new devices.
Minerals such as coltan, tungsten and cassiterite are widely used components in the manufacture of cell phones and electronic devices, many of which could be recovered and reused, but there is little awareness about cell phone recycling around the world.
The sodium-potassium ion exchange pump_____________.
A) must reestablish ion concentrations after each action potential.
B) transports sodium ions into the cell during depolarization.
C) transports potassium ions out of the cell during repolarization.
D) moves sodium and potassium opposite to the direction of their electrochemical gradients.
E) depends on a hydrogen gradient for energy.
Answer: option D
The sodium potassium ion exchange move sodium and potassium in opposite direction in electrochemical gradients.
Explanation:
The sodium potassium pump is found in many animals plasma membranes and its moves the sodium and potassium ion in opposite direction across the plasma membrane with the hydrolysis of ATP(adenosine triphosphate) to supply the needed energy. It is an active transport process.
Final answer:
The sodium-potassium pump moves sodium ions out and potassium ions into the cell against their concentration gradients, using ATP for energy, to maintain the correct ionic balances and the neuron's resting membrane potential.(Option D)
Explanation:
The sodium-potassium ion exchange pump plays a critical role in maintaining the electrochemical gradients necessary for the proper functioning of neurons. This pump actively transports three sodium ions out of the cell and two potassium ions into the cell, against their respective concentration gradients.
It uses energy in the form of ATP to carry out this process, which is essential for keeping the ionic concentrations at proper levels inside and outside the cell. By moving ions against their electrochemical gradients, the pump maintains the negative charge inside the cell, which is crucial for the transmission of nerve impulses.
Blood takes about 1.45 s 1.45 s to pass through a 2.00 mm 2.00 mm long capillary. If the diameter of the capillary is 5.00 μm 5.00 μm and the pressure drop is 2.35 kPa , 2.35 kPa, calculate the viscosity of blood. Assume laminar flow.
Answer:
266.21 Pa.s
Explanation:
By Hagen–Poiseuille equation,
ΔP = 128 μ LQ/ πd2
Where;
Δp is the pressure difference between the two ends,
L is the length,
μ is the dynamic viscosity,
Q is the volumetric flow rate,
d = diameter
t = is the time
Δp = 2.35 KPa = 2.350Pa
L= 2.00 mm = 0.002m
μ = ??? (unknown)
Q = V × A
= (L/t) A
= Lπ[tex]d^{2}[/tex]/ 4t ( since A = [tex]d^{2}[/tex])
d= 5.00 μm = 5.00 × 10 -6
t = 1.45s
∴ 2350 = (128 μ L/ π[tex]d^{2}[/tex] ) × (L π[tex]d^{2}[/tex] / 4t)
Q(volumetric flow rate) = 32μ[tex]L^{2}[/tex] / [tex]d^{2}[/tex]t
From ΔP = 128 μ LQ/ π[tex]d^{2}[/tex]
μ = ΔP π[tex]d^{2}[/tex] / LQ
= (2350 [tex]d^{2}[/tex] / t) / 32L2
= [2350 × (5.00 × 10 -6 ) ( 1.45) / 32( 0.002)2]
= [(0.034075/ (32 × 0.000004)]
= 0.034075/ 0.000128
= 266.21 Pa.s
In MendAliens, normal head top is dominant to Bart head top, in which the top of the head is jagged. Suppose a MendAlien homozygous for normal head shape is crossed with one homozygous for Bart head shape (see picture above). What are the genotypes of the two parents? What are the genotypes in the F1 generation? What are the phenotypes present in the F1 generation?
Answer:
The genotypes of the two parents: normal head shape (HH) and Bart head shape (hh)
The genotypes in the F1 generation= Hh
The phenotypes present in the F1 generation= All normal head shape
Explanation:
Let's assume that the allele "H" is responsible for the normal head shape while the allele "h" gives Bart head shape. According to the given information, both the parents are homozygous. The genotype of the parent with normal head shape would be "HH" while that of the one with Bart head shape would be "hh". Since the normal head shape is dominant, all the F1 hybrid progeny would exhibit "normal head shape".
The genotypes of the two MendAlien parents are homozygous dominant (HH) and homozygous recessive (hh), respectively. The F1 generation will all have heterozygous genotypes (Hh) and exhibit the dominant phenotype of a normal head shape.
The MendAlien scenario described involves a monohybrid cross, which is a basic genetic cross considering one trait at a time. We are told that a homozygous dominant individual (with a normal head shape) is crossed with a homozygous recessive individual (with a Bart head shape).
The genotype of the homozygous dominant parent would be HH (where 'H' represents the allele for a normal head shape) and the genotype of the homozygous recessive parent would be hh (where 'h' represents the allele for a Bart head shape).
When these individuals are crossed, the F1 generation will all be heterozygous (Hh), receiving one dominant allele for a normal head shape from the dominant parent and one recessive allele for a Bart head shape from the other parent. As a result, all F1 offspring will display the phenotype of the dominant trait - normal head shape.
Dr. Smith's parents have normal hearing. However, Dr. Smith has an inherited form of deafness. Deafness is a recessive trait that is associated with the abnormal allele d. The normal allele at this locus, associated with normal hearing, is D. Dr. Smith's parents could have which of the following genotypes?A) DD and ddB) dd and ddC) Dd and DdD) Dd and DD
Answer:
The correct option is C) Dd and Dd
Explanation:
To explain the answer, lets make a punnet square with a cross between Dd and Dd.
D d
D DD Dd
d Dd dd
As deafness is a recessive trait, this means that both the alleles in a person should be same and recessive for that trait to occur in the phenotype. A cross between heterozygous carriers show that there will be a 25% probability that the offsprings produced will get the recessive alleles and hence they will be deaf. 50% of the offsprings will have the chance to be carriers like their parents. And there willbe a 25% chance that they will have normal homozygous alleles.