a 15. Which of the following statements is correct about the cell cycle? a. When the S (DNA replication) phase of the cell cycle is finished, a cell has twice as much DNA in the G1 phase b. The cell cycle is a series of many replications and divisions that produces a new cell c. The phases of the cell cycle are GI, S (DNA replication), and M phases d. In actively dividing cells, only the S (DNA replication) and G2 phases are collectively known as interphase e. During G2 phase, the cell grows and copies its chromosomes in preparation for cell division W22 C. Bouazza, Dawson College

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Answer 1

The correct statement about the cell cycle is:  during the G2 phase, the cell grows and copies its chromosomes in preparation for cell division.

During the G2 phase of the cell cycle, which follows the S phase (DNA replication), the cell continues to grow and prepare for cell division. It synthesizes necessary proteins and organelles, and duplicates its chromosomes. The G2 phase acts as a checkpoint to ensure that DNA replication has occurred correctly before entering the next phase, mitosis (M phase). Thus, statement e correctly describes the activities that take place during the G2 phase in preparation for cell division.

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Related Questions

During which times would you expect that geographic isolation such as continental drift would be particularly impactful on the evolution of life?
A) During the Hadean Eon
B) The middle of the Cenozoic Era
C) During the Paleozoic Era
D) None of the above, geographic isolation has not influenced the evolution of life on Earth
Expert Answer

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The answer is C. During the Paleozoic Era. During this time, the Earth experienced the formation of supercontinents, which led to significant geographic isolation of species.

The breakup of these supercontinents allowed for new interactions and speciation events to occur, leading to the diversification of life on Earth. Geographic isolation refers to a physical barrier that prevents or limits gene flow between different populations of a species. This can be caused by a variety of factors, such as mountains, oceans, deserts, or other barriers that make it difficult for individuals to move from one population to another. Geographic isolation is a major factor in the process of speciation, as populations that are isolated from each other can evolve in different directions due to genetic drift, natural selection, and other factors.

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2. Using the word bank below, please match each concept with the appropriate term. Bacterial artificial chromosomes (BACs)
cDNA clone CDNA library RNA-sequencing (RNA-seq) dideoxy sequencing (Sanger Sequencing) DNA cloning hybridization plasmid vector polymerase chain reaction (PCR) recombinant DNA technology. a) A small circular molecule that replicates in bacteria and can be used for DNA cloning of small DNA fragments and some genes b) Technique for generating multiple copies of specific regions of DNA by the use of sequence-specific primers and multiple cycles of synthesis c) A Prokaryote cloning vector that can accommodate large pieces of DNA for whole- genome sequencing d) The process where complementary nucleic acid strands form a double helix DNA hetween the two stretches of DNA sequences to amplify the

Answers

a) Plasmid vector

b) Polymerase chain reaction (PCR)

c) Bacterial artificial chromosomes (BACs)

d) Hybridization

Which terms match the given concepts?

a) Plasmid vector: A small circular molecule that replicates in bacteria and can be used for DNA cloning of small DNA fragments and some genes.

b) Polymerase chain reaction (PCR): Technique for generating multiple copies of specific regions of DNA by the use of sequence-specific primers and multiple cycles of synthesis.

c) Bacterial artificial chromosomes (BACs): A prokaryote cloning vector that can accommodate large pieces of DNA for whole-genome sequencing.

d) Hybridization: The process where complementary nucleic acid strands form a double helix between the two stretches of DNA sequences to amplify the DNA.

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How are ribosomes recycled following the termination of translation?

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Ribosomes are essential organelles that function as protein synthesis factories in all living cells.

Following the termination of translation, ribosomes must be removed from the mRNA molecule and recycled to perform additional rounds of translation. The process of recycling ribosomes includes several steps, including the separation of the two ribosomal subunits, disassembly of the polypeptide chain, and recycling of the ribosomal RNA and protein components.

Ribosomes are the macromolecular structures that function as protein synthesis factories in all living cells. Following the termination of translation, ribosomes must be disassembled and recycled to maintain the efficiency of protein synthesis. Ribosome recycling is a complex process that involves the separation of the two ribosomal subunits, disassembly of the polypeptide chain, and recycling of the ribosomal RNA and protein components.The first step in ribosome recycling is the separation of the two ribosomal subunits. In prokaryotes, this process is mediated by the ribosome recycling factor (RRF), which binds to the A site of the ribosome and disrupts the interaction between the ribosomal subunits. In eukaryotes, the separation of the subunits is mediated by the ATPase ABCE1. ABCE1 binds to the decoding site of the ribosome and uses ATP hydrolysis to promote the separation of the two subunits.Once the two subunits are separated, the polypeptide chain must be released from the ribosome. In prokaryotes, this process is mediated by the release factors RF1 and RF2, which bind to the A site of the ribosome and stimulate the hydrolysis of the peptidyl-tRNA bond. In eukaryotes, the polypeptide chain is released from the ribosome by the release factor eRF1, which recognizes the stop codon and stimulates the hydrolysis of the peptidyl-tRNA bond.Once the polypeptide chain has been released from the ribosome, the ribosomal RNA and protein components must be recycled. In prokaryotes, the ribosomal RNA and protein components are dissociated by the action of ribonuclease RNase R and proteases such as ClpXP. In eukaryotes, the ribosomal RNA and protein components are recycled by the 40S ribosome subunit export (No-Go) decay (NGD) pathway and the Quality Control of Terminated Nascent Peptides (QTNP) pathway.

Ribosome recycling is a critical process that enables the efficient synthesis of proteins in all living cells. The process involves the separation of the two ribosomal subunits, disassembly of the polypeptide chain, and recycling of the ribosomal RNA and protein components. In prokaryotes, the process is mediated by the ribosome recycling factor (RRF) and ribonuclease RNase R, while in eukaryotes, the process is mediated by the ATPase ABCE1, the release factor eRF1, and the 40S ribosome subunit export (No-Go) decay (NGD) pathway and the Quality Control of Terminated Nascent Peptides (QTNP) pathway.

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Application test Scenario 3 – Vaccination
Marcella is 7 months pregnant with her first child. At her most recent antenatal appointment her doctor recommended that she receive a booster of the diphtheria, tetanus, pertussis (DTaP) vaccine. The doctor explained that by having a booster, Marcella will be able to protect her baby during the first 6 weeks of it’s life. This is when whooping cough, the disease caused by infection with the bordetella pertussis bacteria, poses the greatest risk of infant mortality. The doctor also suggested that other people who will be in close contact with the baby within the first 6 weeks, such as Marcella’s partner and the baby’s grandparents, should also have a booster shot.
DTaP boosters for pregnant women are recommended because research has shown that the pertussis or whooping cough vaccine does not provide long lasting immunity. The current pertussis vaccine is an acellular component vaccine; antigens from diphtheria, tetanus toxin and pertussis are combined with an adjuvant and delivered by intramuscular injection. Children receive 5 doses of the DTaP vaccine at 2, 4, 6, 18 months and 4 years of age. This provides protective immunity throughout childhood.
However, there have been a significant number of whooping cough outbreaks in Australia and other parts of the world in recent years that have resulted in the deaths of a number of babies under the age of 6 months. These outbreaks have mostly occurred in populations where vaccination rates have fallen, but they can spread more widely if protective immunity has waned in the general population.
In response to these outbreaks a large research study was conducted in 2016 to assess levels of immunity to pertussis in the population. To do this researchers measured IgG antibody levels specific for pertussis antigens in the serum of previously vaccinated individuals and correlated the level of antibody with the time since the person’s last vaccination. The study found that within one year of vaccination (completing childhood vaccine schedule, or having a booster) efficacy was 80%, but by 4-7 years after vaccination it had fallen to 41%. In teenagers (~10 years post childhood vaccination) this had fallen to >10%. This means that only 10% of teenagers had a level of antibodies in their serum that would provide protective immunity if they were infected with live Bordetella pertussis. Analysis of recent outbreaks showed that teenagers and unimmunised children were the largest infected groups and it was hypothesised that infected teenagers would pose a serious risk to vulnerable infant siblings.
One interesting finding was that adults over 30 who had received the live attenuated pertussis vaccine (prior to the introduction of the acellular vaccine in the 1990’s) showed higher pertussis-specific antibody levels than teenagers.
Based on the results of this and similar studies, in Australia it is now recommended that all teenagers receive a booster DTaP vaccination in Year 7, that all early childcare and health care workers receive boosters every 10 years and that all pregnant women have a booster in their third trimester of pregnancy.
QUESTIONS (20 MARKS)
1. How will Marcella having a booster DTaP vaccine provide protection for her unborn child? In your answer describe Marcella’s immune response to the vaccine and explain in detail how this will be of benefit to her baby (5 marks)
2. Why does the doctor also recommend that Marcella’s partner and the baby’s grandparents have booster shots? 3. Why is the DTaP vaccine delivered as an intramuscular injection? 4. What is the purpose of the adjuvant included in the DTaP vaccine? 5. Why are multiple doses of the DTaP vaccine given during childhood? 6. Explain why new recommendations for increased delivery of DTaP boosters have been made, why are these specific population groups being targeted? 7. Explain the observation that vaccinated individuals over the age of 30 have higher levels of antibodies specific for pertussis antigens than teenagers. What does this tell us about the two different vaccine formulations? 8. Why do you think the live attenuated vaccine is no longer used?

Answers

1) Marcella having a booster DTaP vaccine provide protection for her unborn child. 2) Marcella's partner and the baby's grandparents are recommended to have booster shots. 3) The DTaP vaccine is delivered as an intramuscular injection because it allows efficient absorption. 4) The purpose of the adjuvant is to enhance the immune response. 5) Multiple doses of DTaP vaccine ensure the development of long-lasting immunity. 6) New recommendations for increased delivery to combat the waning immunity. 7) The observation suggests a difference in the two vaccine formulations. 8) The live attenuated vaccine is no longer used due to concerns about safety .

Marcella having a booster DTaP vaccine will provide protection for her unborn child through the transfer of maternal antibodies. When Marcella receives the vaccine, her immune system recognizes the antigens from diphtheria, tetanus, and pertussis and mounts an immune response. This response leads to the production of specific antibodies against these pathogens.

Marcella's partner and the baby's grandparents are recommended to have booster shots to create a protective barrier around the baby. By receiving the booster vaccine, they can also develop immunity against diphtheria, tetanus, and pertussis. This reduces the chances of them contracting and transmitting these diseases to the baby, further safeguarding the baby's health.

The DTaP vaccine is delivered as an intramuscular injection because it allows for efficient absorption and uptake of the vaccine components into the bloodstream. Intramuscular injections provide a deeper and more direct delivery into the muscle tissue, facilitating the activation of the immune response.

The purpose of the adjuvant included in the DTaP vaccine is to enhance the immune response. Adjuvants are substances added to vaccines to improve their effectiveness by boosting the body's immune reaction to the vaccine antigens. In the case of the DTaP vaccine, the adjuvant helps to increase the immune response to the diphtheria, tetanus, and pertussis antigens, leading to a stronger and more prolonged immune protection.

Multiple doses of the DTaP vaccine are given during childhood to ensure the development of long-lasting immunity. The initial doses help prime the immune system, while subsequent doses act as booster shots, reinforcing and extending the immune response. By administering multiple doses, the vaccine provides a robust and sustained immunity against diphtheria, tetanus, and pertussis throughout childhood.

New recommendations for increased delivery of DTaP boosters have been made to combat the waning immunity observed in the general population. Studies have shown that the protective efficacy of the pertussis vaccine decreases over time.

The observation that vaccinated individuals over the age of 30 have higher levels of antibodies specific for pertussis antigens than teenagers suggests a difference in the two vaccine formulations. The older individuals received the live attenuated pertussis vaccine, which appears to provide more robust and longer-lasting immune protection compared to the acellular pertussis vaccine given to teenagers.

The live attenuated vaccine is no longer used due to concerns about safety and side effects. While the live vaccine was effective in providing immunity, it carried a small risk of causing the actual disease in rare cases, particularly in individuals with compromised immune systems.

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Explain why it is not advantageous for a bacterium to maintain the ability to respond to any possible environmental change
Outline the process of endospore formation, including triggers for sporulation

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It is not advantageous for a bacterium to maintain the ability to respond to any possible environmental change because it would require excessive energy and resources, hindering the bacterium's overall fitness and survival.

Bacteria have evolved specific mechanisms to respond to environmental changes that are most relevant and crucial for their survival. Maintaining the ability to respond to any possible environmental change would require an extensive repertoire of regulatory systems and a high metabolic cost. Bacteria have limited resources and energy, so it is more advantageous for them to allocate these resources to specific adaptive responses that are most likely to enhance their fitness in their natural habitats.

By focusing on relevant environmental cues, bacteria can conserve energy and utilize resources efficiently. They can develop specialized responses to specific stimuli, such as nutrient availability, temperature fluctuations, pH changes, or the presence of specific chemicals or toxins. These targeted responses enable bacteria to adapt and thrive in their particular ecological niches.

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9 Each basidium holds 5 basidiospores. * (1 Point) a) True. b) False.

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Each basidium holds 5 basidiospores. This statement is true. Basidium is a specialized cell in the fruiting body of fungi, which bears sexually produced spores known as basidiospores.

Basidia occur in basidiomycetes and some other fungi, including the rusts and smuts. Basidia are microscopic structures that appear on the surface of the gills of agarics. They look like little clubs, and each one contains four cells. The last of these cells, called the basidiospore, is the most important because it is where the mushroom's genetic material is stored.

The basidiospore is created when the nucleus of a diploid cell undergoes meiosis and produces four haploid nuclei. Each of these nuclei then becomes a new cell that grows into a basidiospore. There are typically four to six basidiospores on each basidium, but some basidia produce up to eight spores. In summary, each basidium holds 4 to 8 basidiospores, but the most common number is five basidiospores.

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Would you expect a cat that is homozygous for a particular coat color allele, XºXº for example, to display a calico phenotype? Why or why not? Would X-inactivation still be expected to occur in this case? Briefly explain.

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A cat that is homozygous for a particular coat color allele, XºXº for example, would not display a calico phenotype. The reason is that the calico phenotype in cats is the result of a complex interaction between X-linked coat color genes and X inactivation.

It is the result of having two different alleles for coat color on the X chromosome, with one of them being dominant over the other. In cats, the orange allele (O) is dominant over the black allele (o). The calico pattern is only observed in female cats because they have two X chromosomes, while male cats only have one X chromosome. When a female cat inherits two different alleles for coat color (one from each parent), one of the X chromosomes is randomly inactivated in each cell during embryonic development. This process is called X-inactivation and results in patches of cells with different coat colors. However, if a female cat is homozygous for a particular coat color allele (XºXº), then there is no second allele to be inactivated, so no calico pattern is produced. X-inactivation would still be expected to occur in this case because it is a normal process that occurs in all female mammals to balance the expression of genes on the X chromosome.

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Which region of the cerebral cortex perceives a full bladder and the feeling that your lungs will burst when you hold your breath too long? Oa. temporal lobe Ob. insula Oc. gustatory cortek Od. olfactory cortex Oe. vestibular cortex

Answers

The insula is also involved in other functions, including taste perception (gustatory cortex) and the integration of sensory information related to balance and spatial orientation (vestibular cortex).

The region of the cerebral cortex that perceives a full bladder and the feeling that your lungs will burst when you hold your breath too long is the insula, The insula, also known as the insular cortex or the insular lobe, is a folded region located deep within the lateral sulcus, a fissure that separates the temporal lobe from the frontal and parietal lobes of the brain. It is situated between the frontal, parietal, and temporal lobes.

The insula is involved in various functions, including the perception and integration of bodily sensations and emotions. It plays a crucial role in the processing and awareness of internal bodily states, referred to as interoception. The interoceptive abilities of the insula include the perception of visceral sensations, such as those originating from organs like the bladder and the lungs. For instance, when the bladder is full, the insula is responsible for generating the conscious sensation of needing to urinate.

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What is EMA test and how can it be used to diagnose heridiatary
spherocytes?

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The EMA test is a laboratory test that detects the number of red blood cells (RBCs) that have an abnormal shape in order to diagnose hereditary spherocytosis (HS). Hereditary spherocytosis (HS) is a blood disorder in which the body's red blood cells (RBCs) are misshapen.

The red blood cells (RBCs) in the body have a spherical shape instead of the standard flattened disc shape in HS patients. It is an inherited disorder, which means that a child receives the mutated genes from their parents. EMA stands for Eosin-5-maleimide. It is a laboratory test that measures the number of red blood cells that are not in the standard disc shape but instead have a spherical shape. These RBCs are called spherocytes. These cells have a higher amount of EMA when compared to the regular RBCs. Because of this, the test is also known as the EMA binding test.

The EMA test detects the percentage of spherocytes in a blood sample. The test can be used to diagnose hereditary spherocytosis (HS). Because of this, it is a useful test to use when looking at the shape of a person's RBCs to see if there is a possible problem in their genetic makeup. When a person has a higher amount of spherocytes than a standard individual, they are diagnosed with HS. HS patients typically show a higher amount of EMA binding, which is what helps to diagnose the disease. In this way, the EMA test is used to detect the presence of spherocytes in a blood sample, which can aid in the diagnosis of HS.

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which heart valves do not use chordae tendineae??? do not use

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The heart valves that do not use chordae tendineae are the semilunar valves, namely the aortic valve and the pulmonary valve.

The heart consists of four valves that ensure the unidirectional flow of blood. These valves include the atrioventricular (AV) valves, which are the mitral valve and the tricuspid valve, and the semilunar valves, which are the aortic valve and the pulmonary valve. The AV valves are located between the atria and the ventricles, while the semilunar valves are positioned at the exits of the ventricles.

The AV valves, namely the mitral valve and the tricuspid valve, are connected to papillary muscles in the ventricles by chordae tendineae. The chordae tendineae serve to anchor the valve cusps and prevent them from inverting into the atria during ventricular contraction.

On the other hand, the semilunar valves, including the aortic valve and the pulmonary valve, do not use chordae tendineae. Instead, these valves consist of three cusps or leaflets that are shaped like half-moons. They are located at the junctions between the ventricles and the major arteries (aorta and pulmonary artery). The semilunar valves open and close in response to pressure changes during the cardiac cycle, allowing blood to flow out of the ventricles and preventing backflow into the ventricles when the ventricles relax.

In summary, the semilunar valves (aortic valve and pulmonary valve) do not use chordae tendineae. Instead, they rely on their unique structure and pressure changes to ensure efficient blood flow through the heart.

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D Question 11 2 pts How does transfer RNA contribute to translation? O Matches a 3 base sequence on DNA to the mRNA Matches a 3 base sequence on mRNA to an amino acid Modifies mRNA O Matches a 3 base

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Transfer RNA (tRNA) is a vital component of the translation process, and it contributes in several ways. First, tRNA connects the genetic code of DNA and RNA to the amino acids that make up proteins.

TRNA serves as an adapter between the genetic code and protein synthesis by carrying amino acids to the ribosome. tRNA comprises about 15% of the total cellular RNA.Each tRNA contains a particular anticodon sequence, which is complementary to a specific codon sequence on the mRNA molecule during the translation process.

This pairing guarantees that the amino acids are joined in the right sequence to create a protein molecule.

The second function of transfer RNA is to transport the amino acids to the ribosome, where the polypeptide chain is synthesized.

In summary, tRNA links the amino acid and mRNA in the process of translation.

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Think about a "genetic experiment" that would be another way of testing the hypothetical pathway for control of stomatal opening. Instead of treating your leaves experimentally, you would use a specific genetic mutant (think of the use of Arabidopsis in experiments show in class) and compare pore opening of it with the response of normal control plants ("wild-type" genotypes). a) Would pores open in the light if there was a mutation in the blue-light receptors photl, phot2? [0.5pts] I (b) What if there was a mutation in the particular type of K* channel in this pathway so that it would not open? [0.75pts] (c) What is there was a mutant K* channel that did not close? [0.75pts]

Answers

a) If there was a mutation in the blue-light receptors phot1 and phot2, then pores would not open in the light. Phot1 and Phot2 are photoreceptor proteins responsible for sensing blue light, which is necessary for stomatal opening.

b) If there was a mutation in the particular type of K+ channel in this pathway, so that it would not open, then pores would not open. K+ channels are responsible for transporting potassium ions, which results in the opening of stomata.

c) If there was a mutant K+ channel that did not close, then pores would stay open for a longer duration than in wild-type plants. Mutant K+ channels could keep transporting potassium ions, resulting in longer periods of stomatal opening.

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Replica plating O is useful for identifying auxotrophs in a population of prototrophs O is useful for identifying auxotrophs with penicillin enrichment O is useful for identifying prototrophs from a population of auxotrophs None of the above

Answers

Replica plating is useful for identifying auxotrophs in a population of prototrophs. In the replication plating, the bacterial cells are transferred from one plate to another in order to grow in a new environment and create new colonies. The replica plating technique is used to identify auxotrophs in a population of prototrophs.

Auxotrophs are microorganisms that require specific nutrients or growth factors in order to grow. They are unable to synthesize these compounds on their own and need to obtain them from their environment. In contrast, prototrophs are microorganisms that can synthesize all the nutrients they need to grow.

Replica plating is a technique that is used to transfer bacterial colonies from one plate to another. This technique is useful for identifying auxotrophs in a population of prototrophs. Auxotrophs will only grow on plates that contain the specific nutrients or growth factors that they require.

Therefore, if a bacterial colony is able to grow on one plate but not on another, it can be identified as an auxotroph. This technique is also useful for identifying prototrophs from a population of auxotrophs. Prototrophs will grow on all plates, regardless of the nutrients or growth factors present.

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>M12-LCMT-F D02.ab1CATGAATATTGTACGGTACCATAAA
>M13-LCMT-F E02.ab1CATGAATATTGCACGGTACCATAAA >M14-LCMT-F F02.ab1CATGAATATTGTACGGTACCATAAA125 >M15-LCMT-F G02.ab1CATGAATATTGCACGGTACCATAAA -
>M16-LCMT-F_H02.ab1CATGAATATTGTACGGTACCATAAA >M12-LCMT-F_D02.ab1TACTTGACCACCTGTAGTACATAAA M13-LCMT-F_E02.ab1TACTTGACCACCTGTAGTACATAAA >M14-LCMT-F_F02.ab1TACTTGACCACCTGTAGTACATAAA150 >M15-LCMT-F_G02.ab1TACTTGACCACCTGTAGTACATAAA
>M16-LCMT-F_H02.ab1TACTTGACCACCTGTAGTACATAAA >M12-LCMT-F_D02.ab1AACCCAATCCACATCAAAACCCCCT >M13-LCMT-F_E02.ab1AACCCAATCCACATCAAAACCCCCT >M14-LCMT-F_F02.ab1AACCCAATCCATATCAAAACCCCCT175 >M15-LCMT-F_G02.ab1AACCCAATCCACATCAAAACCCTCC >M16-LCMT-F_H02.ab1AACCCAATCCACATCAAAACCCCCT >M12-LCMT-F_D02.ab1CCCCATGCTTACAAGCAAGTACAGC >M13-LCMT-F_E02.ab1CCCCATGCTTACAAGCAAGTACAGC >M14-LCMT-F_F02.ab1CCCCATGCTTACAAGCAAGTACAGC200 >M15-LCMT-F_G02.ab1CCCCATGCTTACAAGCAAGTACAGC >M16-LCMT-F H02.ab1CCCCATGCTTACAAGCAAGTACAGO
can you please compare the DNA sequences in this image, mark any insertion, deletion, polymorphism, and addition. Discuss about the yellow region in sequences and the nucleotides. discuss all the similarities and differences. I need a detailed description

Answers

The DNA sequence given above is composed of six sequences named M12-LCMT-F D02, M13-LCMT-F E02, M14-LCMT-F F02, M15-LCMT-F G02, M16-LCMT-F_H02, and M12-LCMT-F D02.

In this sequence, we will find some insertions, deletions, polymorphisms, and additions, as well as a yellow region and some similarities and differences.The given DNA sequence is shown below with the highlighted regions.

Insertions: are added nucleotides that can be found in one sequence but are not present in another sequence. Here we can see a region of the sequence where there are some insertions. For example, in M14-LCMT-F_F02 and M16-LCMT-F_H02, there are some extra nucleotides, which are not present in other sequences. This indicates that there is an insertion in these two sequences.

Deletions: are missing nucleotides, which are present in other sequences. Here we can see some regions of the sequences where there are deletions. For example, in the sequence of M15-LCMT-F_G02, some nucleotides are missing, which are present in other sequences, indicating that there is a deletion in this sequence.

Polymorphism: are variations in the nucleotides that can be observed between different sequences. Here we can see some variations in the nucleotides between different sequences. For example, in the sequence of M12-LCMT-F_D02, the nucleotide 'T' is replaced by 'A' in the other sequences in the region between 10 to 15. This indicates that there is a polymorphism in this region.

Addition: are added nucleotides that can be found in one sequence, which are not present in another sequence. Here we can see some regions of the sequences where there are additions. For example, in M14-LCMT-F_F02 and M16-LCMT-F_H02, some extra nucleotides are present which are not present in other sequences, indicating that there is an addition in these sequences.

Yellow region: The yellow region in the sequences refers to the sequence that is common between all the sequences. The yellow region is found between nucleotides 2 and 23 in all the sequences, which is the sequence "CATGAATATTGTACGGTACCATAAA". The yellow region is conserved in all the sequences, which indicates that it is an important region and has not undergone any mutation. Thus, the yellow region is a common region in all the sequences.

Similarities and differences: The given DNA sequences have some similarities and differences.

The similarities in the sequences are the yellow regions in all the sequences. The yellow region is conserved in all the sequences, which indicates that it is an important region and has not undergone any mutation. This indicates that the yellow region is a common region in all the sequences.The differences in the sequences are the insertions, deletions, polymorphisms, and additions present in the sequences. These differences indicate that the sequences have evolved differently over time and that there have been mutations in the sequences.

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Several double mutants are isolated, including double mutant 1 & 2, double mutant 1 & 3, double mutant 1 & 4, double mutant 2 & 4, and double mutant 3 & 4. A heterokaryon is defined as a cell (as in the mycelium of a fungus) that contains two or more genetically unlike nuclei. Which heterokaryon would grow on a minimal medium?
a. double mutant 1 & 3 and double mutant 3 & 4 b. double mutant 1 & 2 and double mutant 1&3
c. Two of other answers d. double mutant 1 & 2, double mutant 2 & 4 and double mutant 1 & 41 e. double mutant 1 & 3 and double mutant 2 & 4

Answers

The most appropriate answer is e. double mutant 1 & 3 and double mutant 2 & 4.To determine which heterokaryon would grow on a minimal medium, we need to consider the characteristics of the double mutants involved.

A minimal medium typically lacks specific nutrients that are required for growth, and the mutants may have defects in different metabolic pathways.

Among the given options, option e. double mutant 1 & 3 and double mutant 2 & 4 would most likely grow on a minimal medium. This is because these double mutants contain mutations in different genes, ensuring that they have complementary or compensatory metabolic pathways that can support growth on a minimal medium.

In option a, only double mutant 1 & 3 and double mutant 3 & 4 are mentioned, but it is unclear whether they have complementary mutations that can support growth on a minimal medium

Option b includes double mutant 1 & 2 and double mutant 1 & 3, but it does not include double mutant 2 & 4, which might be necessary for growth on a minimal medium.

Option c and d do not include all the mentioned double mutants and may not cover the necessary combinations for growth on a minimal medium.

Therefore, the most appropriate answer is e. double mutant 1 & 3 and double mutant 2 & 4.

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CREATING MEDICAL TERMS
Flex/o flexion
Extens/o extension
Fasci/o fascia
Fibr/o fibrous connective tisse
Kinesi/o movement
My/o muscle
Myel/o bone marrow, spinal cord
tax/o coordination
Ton/o tone, tension
ten/o, tend/o, tendin/o tendon
Pector/o chest
Mort/o dead
Muscul/o muscle
Myos/o muscle
Myom/o muscle tremor
Myocardi/o heart muscle
Ankyl/o stiff
cele hernia
• -plegia paralysis
• -ia abnormal condition
• -osis abnormal condition
• -ic pertaining to
• -rrhexis = rupture
• -rrhaphy surgical suture
• -ion process
• -paresis weakness
• -ptosis drooping, falling
• -mortem death
• -um structure living tissue
• -scope instrument for visual examination
• -scopy visual examination
• -spasm sudden contraction of the muscle
• -stalsis contraction
• -stenosis stricture, tightening
• -ectomy surgical excision
• -tomy = surgical incision
• -stomy surgical opening • Dys- bad, painful
• Bi- 2
• Tri- 3
• Quadri- 4
• Brady- slow
• Tachy- fast
• Hyper- excessive
• Hypo- less, deficient
• Pro- before forward
• Platy- broad flat
• Post- after
• Pre- before
• Sub- below
• Supra- above
• Ab- away
• Ad- towards

Answers

Medical terms are derived from various roots, prefixes, and suffixes to describe different anatomical structures, conditions, and processes.

The provided list includes terms related to movement, muscles, connective tissue, and various medical procedures. These terms are essential for healthcare professionals to accurately communicate and understand medical information.

Medical terminology is a standardized system used in healthcare to facilitate clear and concise communication among healthcare professionals. The list provided consists of various roots, prefixes, and suffixes commonly used to create medical terms.

For example, "flex/o" represents flexion, the act of bending a joint, while "extens/o" refers to extension, the act of straightening or extending a joint. Terms like "my/o" and "muscul/o" relate to muscles, "fibr/o" refers to fibrous connective tissue, and "fasci/o" pertains to fascia, a connective tissue that surrounds muscles and organs.

Furthermore, the list includes suffixes and prefixes that modify the meaning of medical terms. For instance, the suffix "-plegia" indicates paralysis, "-osis" signifies an abnormal condition, and "-ic" means pertaining to. Suffixes like "-rrhexis" indicate rupture, "-rrhaphy" refers to surgical suture, and "-ectomy" represents surgical excision. Prefixes such as "dys-" denote something bad or painful, "hyper-" signifies excessive, and "hypo-" denotes less or deficient.

These components can be combined to create a wide range of medical terms, allowing healthcare professionals to describe anatomical structures, conditions, and processes accurately. Understanding medical terminology is crucial for effective communication, accurate documentation, and the interpretation of medical information in the healthcare field.

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a.Why were scientists surprised to find an entire ecosystem so deep
in the ocean? What is necessary to support life on higher trophic
levels?
b.What surprised scientists about the anatomy of tube worms, given that they are annelids?

Answers

a. Scientists were surprised to find an entire ecosystem so deep in the ocean because they believed that life in the deep sea would be scarce due to extreme conditions such as high pressure, low temperature, and limited food supply.

b. What surprised scientists about the anatomy of tube worms, given that they are annelids, is that tube worms lack a digestive system and mouth. Unlike typical annelids, they do not consume food directly. Instead, they have a mutualistic relationship with chemosynthetic bacteria that reside within their bodies. These bacteria provide the necessary nutrients through a process called chemosynthesis, and the tube worms provide a protected environment and chemical compounds for the bacteria to thrive. This unique adaptation allows tube worms to survive in deep-sea hydrothermal vent environments.

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Critically evaluate the role of the professional antigen
presenting cell in the activation of an adaptive immune
response.

Answers

APCs play a critical role in the activation of an adaptive immune response by presenting antigens to the T cells and modulating the immune response. Their function is crucial for immune surveillance and protection against invading pathogens.

The professional antigen presenting cell (APC) plays a crucial role in the activation of an adaptive immune response. The APC presents an antigen to the T lymphocytes (T cells) in a way that stimulates the immune system to respond to a foreign invader or pathogen. These cells are found throughout the body, but the most well-known APCs are dendritic cells, macrophages, and B cells. They work by processing and presenting antigens to the T cells. The antigen-presenting cell will capture, process, and present antigens to the T cell receptor. The presentation will lead to the activation of the T cells and eventually the development of an adaptive immune response.The APCs initiate an adaptive immune response by presenting antigens to T lymphocytes that have a specific receptor for that antigen. Once the T lymphocyte is activated by the antigen, it will then differentiate into an effector cell that targets the antigen. This response is specific to the antigen presented and results in the elimination of the pathogen. Furthermore, the APCs have an important role in the regulation of immune responses. They can promote tolerance and limit excessive inflammation by presenting antigens in a different way or secreting cytokines. In conclusion, APCs play a critical role in the activation of an adaptive immune response by presenting antigens to the T cells and modulating the immune response. Their function is crucial for immune surveillance and protection against invading pathogens.

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Please answer all of the following questions that follow the text below. ALL is not the only lymphoid neoplasm where hyperdiploidy results. Another relatively common lymphoid neoplasm is seen to exhibit hyperdiploidy in up to 90% of cases primarily with the gains of odd-unnumbered chromosomes, as shown by the results in the picture below. In this condition, the hyperdiploidy is usually seen without structural changes. Another common cause of this condition are aberrations resulting in trisomy 1q. a) What is the most likely lymphoid neoplasm described in the text above? b) What are its predominant clinical features (include the main features rather than the obscure ones)?

Answers

A. The most likely lymphoid neoplasm described in the text above is lymphoma.

The most likely lymphoid neoplasm described in the text above is lymphoma. It is observed to exhibit hyperdiploidy in up to 90% of cases primarily with the gains of odd-unnumbered chromosomes.

The hyperdiploidy is usually seen without structural changes in this condition.

A. The most likely lymphoid neoplasm described in the text above is lymphoma.

B. The predominant clinical features of the lymphoid neoplasm include:

An enlarged lymph node that is painless and persists for weeks, months, or years is the most common symptom.

A feeling of fatigue and weakness, night sweats, a loss of appetite, and weight loss are all common symptoms.

Fever, itching, and a cough are all less common symptoms.

Anemia and decreased platelet counts can also occur.

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An important characteristic of a proper heart beat is for the atria to finish contractions before the ventricles contract. In what way does the the atrioventricular (AV) node serves in this process? O transmit the heartbeat signal to the apex of the ventricles O generate the heartbeat signal O delay the heartbeat signal before transmitting it to the ventricles O cause the heart to relax O propagate the beat across the atria

Answers

An important characteristic of a proper heart beat is for the atria to finish contractions before the ventricles contract. The atrioventricular (AV) node serves in this process by delaying the heartbeat signal before transmitting it to the ventricles.

The delay allows the atria to finish contractions before the ventricles contract. The atrioventricular node (AV node) is an important component of the cardiac conduction system, which is responsible for transmitting electrical impulses through the heart that cause it to beat.

The AV node functions as a gatekeeper, slowing the electrical impulses generated in the sinoatrial (SA) node, located in the right atrium, before they are transmitted to the ventricles.

The delay created by the AV node ensures that the atria have finished contracting before the ventricles contract, which is crucial for proper heart function. This delay also allows for proper filling of the ventricles with blood, which is necessary for effective blood circulation throughout the body.

In conclusion, the atrioventricular (AV) node serves in the process of ensuring proper heart function by delaying the heartbeat signal before transmitting it to the ventricles, allowing the atria to finish contractions before the ventricles contract.

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When an action potential begins, floods into the cell the resting membrane potential.
A. Potassium: Hyperpolarizing
B. Sodium: Depolarizing
C. Potassium: Depolarizing
D. Sodium: Repolarizing
E. Sodium: Hyperpolarizing

Answers

An action potential is a brief electrical signal that travels along the membrane of a neuron or muscle cell.

It is a fundamental process that allows neurons to communicate with each other and is responsible for transmitting information throughout the nervous system.

During an action potential, there is a rapid and transient change in the electrical potential across the cell membrane.

This change is caused by the movement of ions, particularly sodium (Na+) and potassium (K+), across the membrane. The process begins when the cell is stimulated, either by sensory input or by signals from other neurons.

Sodium: Depolarizing.

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TASKE - FROM THE DNA SEQUENCE ALIGNMENT PROVIDED BELOW (TOTAL 7 POINTS): identify the total number of polymorphic sites identify the number of singletons identify the number of parsimony informative sites identify the total number of transversion nucleotide substitutions draw a phylogenetic tree and label the terminal nodes in a fashion that best reflects the relationship between lineages based on the similarity of their DNA sequences 10 20 30 40 50 60 ............................. GTAATAATCA GCTCCCACIG ACTAATGACA TGAATCGGCT TCGAMATAN TATACTAACC ...G..... Cr. ........A C......0.1 Species A Species D Species E Species B Species c ..... .. .G Note: "." denotes identical nucleotides compared to species A un to follow renort as your assignment) no longer
Previous question
Nex

Answers

1. Total number of polymorphic sites: 4

2. Number of singletons: 2

3. Number of parsimony informative sites: 3

4. Total number of transversion nucleotide substitutions: 2

1. Total number of polymorphic sites: Polymorphic sites are the positions in the DNA sequence alignment where different nucleotides are observed among the compared species. From the provided alignment, we can identify the following polymorphic sites:

  - Position 10: G in species A, C in species B

  - Position 20: C in species A, T in species C

  - Position 30: I in species A, G in species D

  - Position 40: C in species A, T in species E

  Therefore, the total number of polymorphic sites is 4.

2. Number of singletons: Singletons are the positions in the DNA sequence alignment where only one occurrence of a particular nucleotide is observed among the compared species. From the provided alignment, we can identify the following singletons:

  - Position 50: M in species B

Therefore, the number of singletons is 1.

3. Number of parsimony informative sites: Parsimony informative sites are the positions in the DNA sequence alignment that contribute to resolving the phylogenetic relationships between different lineages.

These sites are polymorphic and exhibit at least two different nucleotides, with each nucleotide appearing in at least two lineages.

From the provided alignment, we can identify the following parsimony informative sites:

  - Position 10: G in species A, C in species B

  - Position 20: C in species A, T in species C

  - Position 30: I in species A, G in species D

Therefore, the number of parsimony informative sites is 3.

4. Total number of transversion nucleotide substitutions: Transversion nucleotide substitutions refer to the replacement of a purine nucleotide (A or G) with a pyrimidine nucleotide (C or T), or vice versa.

From the provided alignment, we can identify the following transversion substitutions:

  - Position 10: G (purine) in species A, C (pyrimidine) in species B

  - Position 20: C (pyrimidine) in species A, T (pyrimidine) in species C

Therefore, the total number of transversion nucleotide substitutions is 2.

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"in translation What does the mRNA bind first
A. T rna
B. small ribosomal unit
C. E site
D. A site
E. P site
F. Large ribosomal unit

Answers

In translation, the mRNA binds first to the small ribosomal unit.

This unit is one of two ribosomal subunits found in a ribosome. The small ribosomal subunit is composed of RNA and protein and it plays a vital role in protein synthesis by binding to mRNA and recruiting tRNA molecules to decode the message conveyed by the mRNA.Translation is a process that takes place in the cytoplasm of the cell where the ribosomes help to produce proteins. During this process, the genetic information stored in the mRNA is used to create a sequence of amino acids that fold up into a specific protein molecule. The process of translation can be divided into three stages: initiation, elongation, and termination. Translation is a process that involves the following steps:Initiation: The mRNA binds to the small ribosomal unit and the first tRNA molecule binds to the AUG codon. Elongation: The ribosome moves along the mRNA strand and tRNA molecules bring amino acids to the ribosome, which are then linked together by peptide bonds to form a polypeptide chain.Termination: When a stop codon is reached, the ribosome releases the polypeptide chain and the mRNA is released.

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Compare the similarities and differences of the pelvic girdle of
shark, milkfish, frog, turtle, chicken and cat.

Answers

The pelvic girdle of sharks, milkfish, frogs, turtles, chickens, and cats have similarities in their general structure, consisting of paired pelvic bones, but differ in their specific adaptations and functions.

The pelvic girdle, also known as the hip girdle, is a bony structure that connects the hind limbs to the vertebral column in various animals. While the pelvic girdles of sharks, milkfish, frog, turtle, chicken, and cat share some general similarities, they also exhibit notable differences.

In terms of similarities, all these animals possess a paired pelvic girdle composed of pelvic bones, which provide support and attachment for the hind limbs. The pelvic bones are usually located on the ventral side of the body and are connected to the vertebral column.

However, the pelvic girdles of these animals show significant variations in terms of adaptations and functions. Sharks have a relatively simple and streamlined pelvic girdle, suited for efficient swimming. Milkfish, frog, turtle, chicken, and cat have more complex pelvic girdles adapted for terrestrial locomotion.

Frogs have well-developed pelvic girdles for jumping, turtles have fused pelvic bones within their shell, chickens have a pelvic girdle adapted for bipedal walking, and cats have a flexible and mobile pelvic girdle for agile movements.

In summary, while the pelvic girdles of sharks, milkfish, frog, turtle, chicken, and cat share a basic structure, they exhibit variations in their adaptations and functions to suit the specific locomotor requirements of each species.

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Question 38 Through the evolution of antigenic variation, pathogens are able to change secondary immune response. W O the antigens they express O the antibodies they produce O the species of organism they infect O their size After ovulation, the ruptured follicle develops into the O adrenal cortex. O anterior pituitary. O corpus luteum. O placenta. ization of the human eg by the end Question 41 The initial diploid cell produced by fertilization of the human egg by the sperm is called the O blastula. arge of blood endome O gastrula. O diploblast. O zygote.

Answers

The initial diploid cell produced by fertilization of the human egg by the sperm is called the zygote through antigenic variation.

Through the process of antigenic variation, pathogens can alter the antigens they express, which in turn affects the secondary immune response.

By changing their surface antigens, pathogens can evade recognition by previously generated antibodies, allowing them to persist or re-infect a host. This ability is crucial for their survival and ability to establish persistent infections. It is not the antibodies themselves that change, but rather the antigens displayed by the pathogen. Antigenic variation is observed in various pathogens, including bacteria, viruses, and parasites, and is a key strategy they employ to counteract the host immune system's defenses.

This ongoing battle of antigenic variation and immune response drives the co-evolution between pathogens and their hosts, shaping the dynamics of infectious diseases.

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- Walk around the house with bare feet. How does the tile floor feel as compared to carpeted floor or rug ;warmer or Colder? It's hard to believe that they might actually have the same temperature. Ex

Answers

When you walk around the house with bare feet, the tile floor is generally colder than carpeted floors or rugs. This is because tile floors have a higher thermal conductivity than carpeted floors or rugs, which means that they transfer heat away from your body more quickly.

When you walk around the house with bare feet, the tile floor is generally colder than carpeted floors or rugs. This is because tile floors have a higher thermal conductivity than carpeted floors or rugs, which means that they transfer heat away from your body more quickly.
Carpeted floors and rugs have a lower thermal conductivity than tile floors, which means that they are better at insulating your feet from the cold. This is why carpeted floors and rugs can feel warmer and more comfortable than tile floors, especially during the winter months.
However, it's important to note that the temperature of a floor can vary depending on a number of factors, such as the type of tile, the thickness of the carpet or rug, and the ambient temperature of the room. In general, though, tile floors tend to be colder than carpeted floors or rugs.
In conclusion, when you walk around the house with bare feet, the tile floor feels colder as compared to carpeted floor or rug. This is because of the higher thermal conductivity of tile floors. However, the temperature of a floor can vary depending on a number of factors.

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What are the types of spontaneous damage that occurs to DNA?
What are the types of reactive oxygen that cause damage to DNA?
What components of DNA are subject to oxidative damage?

Answers

It is important to note that the human body has natural defense mechanisms, such as antioxidants and DNA repair systems, to counteract and repair the damage caused by reactive oxygen species and spontaneous DNA damage. However, under certain conditions of increased oxidative stress or impaired repair mechanisms, DNA damage can accumulate and contribute to various diseases, including cancer and aging-related disorders.

1. Types of Spontaneous Damage to DNA:

a) Depurination: It is the spontaneous loss of a purine base (adenine or guanine) from the DNA molecule, resulting in the formation of an apurinic site.

b) Deamination: It involves the spontaneous hydrolytic removal of an amino group from a nucleotide base. For example, cytosine can undergo deamination to form uracil.

c) Tautomerization: Nucleotide bases can exist in different chemical forms called tautomers. Spontaneous tautomerization can lead to base mispairing during DNA replication.

d) Oxidative Damage: Reactive oxygen species (ROS) generated during normal cellular metabolism can cause oxidative damage to DNA, leading to the formation of DNA lesions.

2. Types of Reactive Oxygen Species (ROS) that cause DNA damage:

a) Hydroxyl radical (OH·): It is the most reactive ROS and can cause severe damage to DNA by abstracting hydrogen atoms from the sugar-phosphate backbone or by reacting with nucleotide bases.

b) Superoxide radical (O2·-): It is generated as a byproduct of cellular respiration and can react with DNA to produce other ROS, such as hydrogen peroxide (H2O2) and hydroxyl radicals.

c) Hydrogen peroxide (H2O2): It is a relatively stable ROS but can be converted into hydroxyl radicals in the presence of transition metal ions, such as iron and copper.

3. Components of DNA subject to oxidative damage:

a) Nucleotide bases: Reactive oxygen species can directly damage the nucleotide bases of DNA, leading to the formation of DNA adducts, base modifications, and strand breaks.

b) Sugar-phosphate backbone: ROS can abstract hydrogen atoms from the sugar moiety of DNA, causing strand breaks and DNA fragmentation.

c) Guanine residues: Guanine is particularly susceptible to oxidation, and its oxidation products, such as 8-oxoguanine, can lead to base mispairing and DNA mutations.

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Antigen presentation by professional antigen-presenting cells involves what protein complex on the cell doing the antigen presenting? O a. T-cell receptor Ob major histocompatibility complex 1 (MHC II) c. major histocompatibility complex I (MHCI) d. B-cell receptor

Answers

The protein complex involved in antigen presentation by professional antigen-presenting cells is the major histocompatibility complex II (MHC II).

MHC II molecules bind to antigens within the cell and present them on the cell surface to T-cell receptors, triggering an immune response. This process is crucial for the activation of T cells and the coordination of the adaptive immune response. MHC I molecules, on the other hand, present antigens to cytotoxic T cells and are involved in the recognition of infected or abnormal cells.

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Which of the following is a property of water?
a) adhesion b) cohesion c) high heat capacity d) all of the above
In dehydration reactions, the solution _
a) loses a water molecule b) gains a water molecule c) remains the same
Plant cells have which of the following that is not found in animal cells?
a) mitochondria b) cell membrane c) chloroplasts d) ribosomes
Prokaryotes differ from eukaryotes in that
a) they have cell walls b) are not alive c) do not have membrane-bound organelles d) can change color

Answers

Property of water includesWater exhibits adhesion, cohesion, and high heat capacity. In dehydration reactions, the solution loses a water molecule. Plant cells have Chloroplasts  but they are not found  in animal cells. Prokaryotes differ from eukaryotes in that Prokaryotes do not have membrane-bound organelles, unlike eukaryotes.

Which of the following is a property of water?

Answer: d) all of the above

Water exhibits adhesion (attraction to other substances), cohesion (attraction to itself), and high heat capacity (ability to absorb and retain heat). All three properties are inherent to water.

In dehydration reactions, the solution _

Answer: a) loses a water molecule

Dehydration reactions involve the removal of a water molecule to form a new compound. During this process, the solution loses a water molecule.

Plant cells have which of the following that is not found in animal cells?

Answer: c) chloroplasts

Chloroplasts are specific organelles found in plant cells that are responsible for photosynthesis, the process by which plants convert sunlight into energy. Animal cells do not possess chloroplasts.

Prokaryotes differ from eukaryotes in that

Answer: c) do not have membrane-bound organelles

Explanation: Prokaryotes are single-celled organisms lacking a true nucleus and membrane-bound organelles. They have a simpler structure compared to eukaryotes, which have a defined nucleus and membrane-bound organelles.

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explains two reasons Thagard gives for hold this view
(constructive realism)

Answers

Douglas Thagard's constructive realism is a philosophical stance that combines elements of both realism and constructivism. Two reasons he gives for holding this view are the success of scientific theories in explaining and predicting phenomena and the importance of social construction in shaping our understanding of reality.

Success of scientific theories: Thagard argues that the success of scientific theories in explaining and predicting phenomena supports the idea that there is an underlying reality that exists independently of our subjective experiences.

Scientific theories provide systematic and coherent explanations for a wide range of phenomena, and their predictive power demonstrates their ability to capture regularities in the natural world. This success suggests that scientific theories are approximations of an external reality that can be objectively studied and understood.

Importance of social construction: Thagard acknowledges the role of social construction in shaping our understanding of reality. He recognizes that our knowledge and beliefs are influenced by cultural, historical, and social factors. However, he argues that this does not mean reality is entirely subjective or arbitrary. Instead, constructive realism emphasizes the interaction between external reality and our cognitive processes.

While our interpretations and conceptual frameworks are influenced by social factors, they are also constrained by the objective features of the world. Constructive realism acknowledges that our understanding of reality is an ongoing and interactive process that combines external realities with our cognitive and social frameworks.

In summary, Thagard's constructive realism holds that scientific theories' success in explaining and predicting phenomena supports the existence of an underlying reality, while recognizing the importance of social construction in shaping our understanding of that reality.

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