Explain the importance of the following in prokaryotic and/or eukaryotic DNA replication, as described in the BCH3703 course material: 4.1 topoisomerase (5) 4.2 metal ions (5) 4.3 telomeres

Answers

Answer 1

4.1 Topoisomerase is important for relieving DNA tension during replication

4.2 metal ions act as cofactors for replication enzymes, and 4.3 telomeres protect chromosome ends and prevent genomic instability.

4.1 Topoisomerase:

Topoisomerase is important in both prokaryotic and eukaryotic DNA replication. It is an enzyme responsible for relieving the strain or tension that builds up ahead of the replication fork during DNA unwinding. It achieves this by cutting and rejoining the DNA strands, allowing them to rotate and unwind.

Topoisomerase plays a crucial role in preventing DNA damage, maintaining DNA integrity, and facilitating the smooth progression of DNA replication.

4.2 Metal Ions:

Metal ions, such as magnesium (Mg2+) and manganese (Mn2+), are essential cofactors in both prokaryotic and eukaryotic DNA replication. They are required by several enzymes involved in DNA replication, including DNA polymerases and DNA ligases.

Metal ions stabilize the structure of these enzymes, promote their catalytic activity, and facilitate the proper binding of nucleotides during DNA synthesis. They are also involved in the coordination of nucleotide triphosphates (NTPs) and the correct positioning of the DNA template. Overall, metal ions are crucial for the efficient and accurate replication of DNA.

4.3 Telomeres:

Telomeres are specific DNA sequences located at the ends of eukaryotic chromosomes. They play a vital role in maintaining genomic stability during DNA replication.

Telomeres function as protective caps, preventing the loss of essential genetic information during each round of DNA replication. Due to the nature of DNA replication, the lagging strand is unable to be fully replicated at the very end, resulting in the gradual shortening of the telomeres with each replication cycle.

Telomeres provide a buffer zone and prevent the erosion of critical genetic material. They also facilitate the replication of the very ends of chromosomes through the action of the enzyme telomerase, which helps to extend the telomeric DNA.

Proper regulation and maintenance of telomeres are crucial for preserving chromosomal integrity and preventing genomic instability.

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Related Questions

Name the building block that makes up 40% of the plasma
membrane. (one word)

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The building block that makes up 40% of the plasma membrane is phospholipids.

The plasma membrane is composed primarily of a bilayer of phospholipids. Phospholipids are a type of lipid molecule that consists of a hydrophilic (water-loving) head and two hydrophobic (water-repelling) tails. The hydrophilic heads face the aqueous environment both inside and outside the cell, while the hydrophobic tails are sandwiched between them, forming the interior of the membrane.

These phospholipids arrange themselves in a bilayer structure, with the hydrophilic heads oriented towards the aqueous surroundings and the hydrophobic tails facing inward. This arrangement creates a stable barrier that separates the cell's internal contents from the external environment, controlling the movement of substances in and out of the cell.

Due to their abundance and fundamental role in forming the plasma membrane, phospholipids make up a significant portion of it, accounting for approximately 40% of its composition. Other components of the plasma membrane include proteins, cholesterol, and various types of lipids, but phospholipids are the primary building blocks responsible for its structural integrity and selective permeability.

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1. What does the last tRNA bring in? Explain.
2. What is the DNA Complement and DNA Template of the mRNA
codons 5 ’ A U G C G U A A A U G G A G G G U A G A A U U C A A G U
A A ?

Answers

1. The last tRNA brings in the amino acid corresponding to the last codon of the mRNA sequence during protein synthesis.

The tRNA molecule carries the specific amino acid that is complementary to the mRNA codon. The ribosome, which facilitates protein synthesis, recognizes the codon on the mRNA and matches it with the appropriate tRNA carrying the corresponding amino acid. This process ensures that the correct amino acid is added to the growing polypeptide chain, following the genetic code.

2. To determine the DNA complement and DNA template of the mRNA sequence, we need to use the rules of complementary base pairing. The base pairs in DNA are adenine (A) with thymine (T) and cytosine (C) with guanine (G).

Given the mRNA codon sequence: 5' A U G C G U A A A U G G A G G G U A G A A U U C A A G U

The DNA complement sequence is obtained by replacing each base in the mRNA with its complementary base in DNA:

DNA Complement: 5' T A C G C A T T T A C C T C C C A T C T T A A G T

The DNA template strand is the reverse complement of the mRNA sequence, as it serves as the template for mRNA synthesis during transcription:

DNA Template: 3' A C G C U A A A U G G A G G G U A G A A U U C A A G

In the DNA template, the bases are read in the opposite direction (from 3' to 5') compared to the mRNA sequence. The DNA template strand is complementary to the mRNA sequence, allowing RNA polymerase to synthesize a complementary mRNA molecule during transcription.

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Question 24 1.82 pts Which of the following combinations is potentially harmful? O An Rh+ mother that has an Rh- fetus An Rh- mother that has an Rh- fetus O An Rh- mother that has an Rh+ fetus An Rh+

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The combination that is potentially harmful is an Rh- mother with an Rh+ fetus. During pregnancy, there is a potential for incompatibility between the Rh factor of the mother and fetus.

The Rh factor refers to a specific antigen present on the surface of red blood cells. An Rh+ fetus inherits the Rh antigen from an Rh+ father, while an Rh- mother does not have the Rh antigen.

If an Rh- mother carries an Rh+ fetus, there is a risk of Rh incompatibility. This can occur if fetal blood enters the maternal bloodstream during pregnancy or childbirth. The mother's immune system recognizes the Rh antigen as foreign and produces antibodies against it. Subsequent pregnancies with Rh+ fetuses can lead to an immune response where the maternal antibodies attack the fetal red blood cells, causing a condition known as hemolytic disease of the newborn (HDN) or erythroblastosis fetalis. HDN can result in severe anemia, jaundice, and other complications in the fetus or newborn.

To prevent harm, Rh- mothers who are at risk of Rh incompatibility are typically given Rh immune globulin (RhIg) during pregnancy to prevent the formation of antibodies against the Rh antigen.

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When blood pressure increases, Multiple Choice O O O baroreceptors detect the change in the carotid arteries. the cardioregulatory center decreases parasympathetic stimulation heart rate and stroke vo

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When blood pressure increases, baroreceptors detect the change in the carotid arteries, and the cardioregulatory center decreases parasympathetic stimulation, resulting in an increase in heart rate and stroke volume.

Baroreceptors are specialized sensory receptors located in the carotid arteries and aortic arch that detect changes in blood pressure. When blood pressure increases, these baroreceptors are activated and send signals to the cardioregulatory center in the brain.

The cardioregulatory center, which is part of the autonomic nervous system, responds to the increased blood pressure by decreasing parasympathetic stimulation and increasing sympathetic stimulation. This leads to a decrease in vagal tone (parasympathetic activity) and an increase in sympathetic activity.

The decrease in parasympathetic stimulation results in a decrease in the release of acetylcholine, which normally slows down the heart rate. As a result, the heart rate increases.

Additionally, the increase in sympathetic activity leads to the release of norepinephrine, which increases the force of contraction of the heart muscle, resulting in an increased stroke volume.

Overall, these responses work together to help normalize blood pressure by increasing cardiac output and maintaining adequate perfusion to the body's tissues.

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<The complete question is>

When blood pressure increases, Multiple Choice Option 1. baroreceptors detect the change in the carotid arteries. 2.the cardioregulatory center decreases 3.  parasympathetic stimulation heart rate and stroke volume increase, 4.norepinephrine secretion increase

The heterozygous jiggle beetles represents pleiotropy. O polygenic. O incomplete dominance. codominance. complete domiance. Question 40 What can be concluded about the green allele and hot pink allele. O The green allele is recessive and the hot pink allele is dominant. O The green allele and pink allele are recessive. O The green allele is dominant and the hot pink allele is recessive. O The green allele and pink allele are dominant.

Answers

The green allele is recessive, and the hot pink allele is dominant in the case of the heterozygous jiggle beetles.

Based on the information provided, we can conclude that the green allele is recessive, and the hot pink allele is dominant. Pleiotropy refers to a single gene having multiple effects on an organism, which is not evident from the given context. Polygenic inheritance involves multiple genes contributing to a trait, which is also not mentioned in the scenario. Incomplete dominance occurs when neither allele is completely dominant over the other, resulting in an intermediate phenotype in heterozygotes. Codominance occurs when both alleles are expressed equally in the phenotype of heterozygotes. Complete dominance occurs when one allele is completely dominant over the other, resulting in the expression of only one allele in the phenotype of heterozygotes.

Since the scenario states that the beetles are heterozygous, meaning they carry two different alleles, we can deduce that the hot pink allele must be dominant because it is expressed in the phenotype. The green allele, on the other hand, is recessive because it remains unexpressed in the presence of the dominant hot pink allele. Therefore, the correct conclusion is that the green allele is recessive, and the hot pink allele is dominant.

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1. Describe three differences between prokaryotic and
eukaryotic cells.
2. Discuss the major differences between a plant cell and an
animal cell.

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Prokaryotic and eukaryotic cells have fundamental differences that separate them in terms of structure, function, and overall complexity. Here are three differences between prokaryotic and eukaryotic cells Prokaryotic cells do not have a nucleus, while eukaryotic cells have a nucleus.

Eukaryotic cells have membrane-bound organelles, whereas prokaryotic cells do not. Eukaryotic cells are more complex than prokaryotic cells. A plant cell and an animal cell are similar in that they are both eukaryotic cells and have many similarities in terms of structure and function. However, there are some significant differences between the two. Here are some major differences between a plant cell and an animal cell Plant cells have cell walls, while animal cells do not.

Plant cells contain chloroplasts, which are responsible for photosynthesis, while animal cells do not have chloroplasts. Plant cells have large central vacuoles, while animal cells have small vacuoles or none at all. Plant cells have a more regular shape, while animal cells can take on various shapes. Plant cells store energy as starch, while animal cells store energy as glycogen.

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a) Compare and contrast the basal states of glucocorticoid and retinoid X receptors and their activation mechanisms by their cognate steroid hormones which lead to gene transcription. (20 marks)

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Glucocorticoid Receptor (GR) and Retinoid X Receptor (RXR) are both nuclear receptors that function as transcription factors.

Here is a comparison and contrast of their basal states and activation mechanisms:

Basal State:

Glucocorticoid Receptor (GR): In the absence of its ligand (e.g., cortisol), the GR resides in the cytoplasm as part of a multiprotein complex.

Retinoid X Receptor (RXR): RXR can exist in both the cytoplasm and the nucleus.

Activation Mechanisms:

Glucocorticoid Receptor (GR): Upon binding of cortisol (the cognate hormone), the GR undergoes a conformational change, leading to dissociation from HSPs.

Retinoid X Receptor (RXR): RXR can be activated by its cognate ligand, 9-cis retinoic acid (9-cis RA), or through heterodimerization with other nuclear receptors.

Gene Transcription:

Glucocorticoid Receptor (GR): Activation of the GR by cortisol leads to the recruitment of coactivators to the GREs on target genes.

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Which of the following will most likely happen to a population when the size of the population far overshoots their carrying capacity? (such as the deer on St. Matthew's island) O the population will exhibit exponential growth O the population crashes. the birth rate increases and the death rate decreases. the growth rate remains unchanged.

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The most likely outcome to occur when the size of a population far overshoots their carrying capacity is that the population will crash. A population crash refers to a rapid decrease in the size of a population that results from the inability of the environment to support the population's carrying capacity.

The carrying capacity is the maximum number of individuals in a species that can be supported by a given habitat without causing any negative impacts on the environment.There are a number of factors that can contribute to a population crash, such as disease, predation, environmental degradation, and resource depletion. When a population overshoots their carrying capacity, competition for resources increases, which can lead to reduced food availability, malnutrition, and starvation. The death rate increases, and the birth rate decreases as a result of the scarcity of resources.

So, when the size of a population far overshoots their carrying capacity, the most likely outcome is that the population will experience a crash. This is due to the increased competition for resources, which leads to a decrease in the birth rate and an increase in the death rate.

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What group of floral products compensated for the loss of cut flower production during the 1970 s and 1980 s when imports replaced most of domestic production? orchard plants vegetable plants All of these foliage plants A greenhouse covering material subject to a large radiant heat loss is polyethylene glass wood polyacrylic Every greenhouse should have an emergency electrical generator. The generator should ideally have a capacity of 1 kilowatt for every square foot of greenhouse floor. 1000 2,000 3000 4000 Water droplet size in a fog cooling system is 2 microns 40 Microns 11 microns 10 microns

Answers

The group of floral products that compensated for the loss of cut flower production during the 1970s and 1980s when imports replaced most of domestic production includes foliage plants.

Polyethylene is a cost-effective material that provides good insulation and durability, making it suitable for greenhouse applications. Glass is a transparent option that allows ample light transmission while offering longevity and weather resistance. Polyacrylic, another material used for greenhouse coverings, combines good light transmission with resistance to weathering and aging.

Having a properly sized generator ensures that essential equipment, such as heating, cooling, and ventilation systems, can continue to function during power outages or emergencies. The 1 kilowatt per square foot guideline helps provide adequate power supply to meet the energy demands of the greenhouse and maintain optimal growing conditions. In a fog cooling system, the water droplet size typically ranges around 10 microns.

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3) Staphylococcus aureus infections are know to cause: A. impetego B. Scalded skin syndrome C. Endocarditis D. All of these 4) Prions cause: A. Kuru B. Scrapie C.,boxine spongiform encephalopathy. D. All of the above 5) A sexually transmitted disease that is recurrent because of viral latency is. A chancroid B. Herpes C. Syphilis D gonorrhea E. PID 6) Pathogenicity of tetanus. Is due to: A. Exotoxin B. Endotoxin C.invasive action of the organism D. Collagenase activity 7) Disorders in w high pseudomonas aeruginosa has been implicated are: A. Infections in cystic. Fibrosis patients B. Conjunctivitis C. Burn sequelae Otis. Media E. Al of these D. 8) German measles is also termed: A. bubeloa B. Ribella C.variola D. Varicella 9) Which of the following are true: A. Attenuated whole agent vaccines can offer life long immunity B. Subunit vaccines produce the most side effects C. Toxin vaccines do not require boosters D. All of the above

Answers

1. Staphylococcus aureus infections can cause impetigo, scalded skin syndrome, and endocarditis.

2. Prions are responsible for causing diseases such as kuru, scrapie, and bovine spongiform encephalopathy (BSE).

3. Herpes is a sexually transmitted disease that can recur due to viral latency.

4. The pathogenicity of tetanus is due to the production of an exotoxin by the bacteria.

5. Pseudomonas aeruginosa has been implicated in various disorders, including infections in cystic fibrosis patients, conjunctivitis, burn sequelae, and otitis media.

6. German measles is also known as rubella.

7. Attenuated whole agent vaccines can provide long-lasting immunity, subunit vaccines may produce side effects, and toxin vaccines may require boosters.

1. Staphylococcus aureus is a bacterium known to cause multiple infections. It can lead to impetigo, a superficial skin infection, as well as scalded skin syndrome, a more severe blistering condition. Additionally, it can cause endocarditis, an infection of the inner lining of the heart.

2. Prions are abnormal proteins that can cause neurodegenerative diseases. Kuru is a prion disease transmitted through cannibalistic rituals, scrapie affects sheep and goats, and bovine spongiform encephalopathy (BSE), also known as "mad cow disease," affects cattle.

3. Herpes is a sexually transmitted infection caused by the herpes simplex virus (HSV). The virus establishes latency in nerve cells, leading to recurrent outbreaks of symptoms such as painful sores or blisters.

4. Tetanus is caused by the bacterium Clostridium tetani. Its pathogenicity is primarily due to the production of a neurotoxin called tetanospasmin, which affects the nervous system and leads to muscle stiffness and spasms.

5. Pseudomonas aeruginosa is a bacterium associated with various infections. It can cause infections in cystic fibrosis patients, including respiratory infections. It is also known to cause conjunctivitis (pink eye), otitis media (middle ear infection), and can contribute to burn wound infections and their sequelae.

6. German measles is another term for rubella, a viral infection characterized by a rash and fever. It is caused by the rubella virus and can cause complications, especially if contracted during pregnancy.

7. Attenuated whole agent vaccines, such as the measles, mumps, and rubella (MMR) vaccine, can provide long-lasting immunity. Subunit vaccines, which contain only specific parts of the pathogen, may produce fewer side effects compared to whole agent vaccines.

Toxin vaccines, such as the diphtheria and tetanus vaccines, may require periodic boosters to maintain immunity against the toxins produced by the pathogens.

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Which of the following is not a dietary recommendation? a. Consume 0 grams of trans fats.
b. Consume 48 grams of dietary fiber. c. Consume no more than 50 grams of sugar, and preferably less than 36 grams. d. Consume no more than 80 grams of protein, and preferably less than 50 grams.
e. Consume no more than 2300 mg (2.3 grams) of sodium, and preferably less than 1500 mg.

Answers

Option (d) "Consume no more than 80 grams of protein, and preferably less than 50 grams" is not a dietary recommendation.

Option (d) is not a dietary recommendation because it suggests limiting protein intake to no more than 80 grams, preferably less than 50 grams. However, protein requirements can vary based on factors such as age, sex, body weight, activity level, and overall health. The appropriate amount of protein intake for an individual depends on their specific needs and goals, such as muscle building, weight management, or medical conditions. There is no universally recommended limit on protein intake, and it is generally advised to consume an adequate amount of protein to support overall health.

On the other hand, options (a), (b), (c), and (e) are dietary recommendations commonly advised for maintaining a healthy diet. These recommendations focus on avoiding trans fats, consuming an adequate amount of dietary fiber, limiting sugar intake, and controlling sodium intake for optimal health.

In summary, option (d) "Consume no more than 80 grams of protein, and preferably less than 50 grams" is not a general dietary recommendation, as protein requirements vary among individuals.

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What is the function of Troponin C, Troponin I and Troponin T? How do they each cause muscle contraction? Include detail

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Troponin C, Troponin I, and Troponin T are three subunits of the troponin complex found in muscle cells. They play crucial roles in regulating muscle contraction, specifically in skeletal and cardiac muscles.

Troponin C (TnC): Troponin C is a calcium-binding protein that is essential for muscle contraction. It binds to calcium ions (Ca2+) when the concentration of Ca2+ increases in the cytoplasm of muscle cells, triggering a series of events that lead to muscle contraction.

Troponin I (TnI): Troponin I is another subunit of the troponin complex that inhibits the interaction between actin and myosin, two key proteins involved in muscle contraction. Troponin I prevents muscle contraction in the absence of calcium ions. When calcium ions bind to troponin C, it causes a conformational change in troponin I, relieving its inhibitory effect on actin.

Troponin T (TnT): Troponin T is the third subunit of the troponin complex and plays a structural role in muscle contraction. Troponin T binds to tropomyosin, another protein that is associated with the actin filament. When troponin C binds to calcium ions, it induces a conformational change in troponin T, which in turn shifts the position of tropomyosin.

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To correct sickle-cell anemia via gene therapy using a viral vector, the cells that would need to be collected from a sickle cell patient are called:
a. embryonic stem cells.
b. mesenchymal stem cells.
c. totipotent stem cells.
d. hematopoietic stem cells.
e. neural stem cells.

Answers

To correct sickle-cell anemia via gene therapy using a viral vector, the cells that would need to be collected from a sickle cell patient are hematopoietic stem cells. The correct option is d.

Hematopoietic stem cells are the cells responsible for generating the various types of blood cells, including red blood cells. In sickle-cell anemia, there is a mutation in the gene that codes for hemoglobin, resulting in the production of abnormal hemoglobin molecules that cause the characteristic sickle-shaped red blood cells.

To correct this mutation, gene therapy can be performed by introducing a functional copy of the gene into the patient's cells. Hematopoietic stem cells are an ideal target for gene therapy in sickle-cell anemia because they are the precursor cells that give rise to red blood cells.

By collecting hematopoietic stem cells from the patient, modifying them with the functional gene using a viral vector (such as a modified virus), and then reintroducing these genetically modified cells back into the patient's body, it is possible to restore normal hemoglobin production and alleviate the symptoms of sickle-cell anemia.

Therefore, the correct answer is d.

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Signal transduction- yeast genetics
in one sentence, what does alpha factor in the WT 'a' cell do?
(In terms of cell cycle/budding and FUS1 transcription)

Answers

In terms of cell cycle/budding and FUS1 transcription, the alpha factor in the WT 'a' cell induces the pheromone response pathway, leading to cell cycle arrest and activation of transcription factors that initiate FUS1 transcription.

In Saccharomyces cerevisiae, alpha factor is a peptide pheromone that activates a cell signaling pathway that controls mating and cell cycle progression. Alpha factor activates the G protein-coupled receptor, Ste2p, initiating a cascade of signal transduction events that result in the activation of the mitogen-activated protein kinase (MAPK) pathway. The pheromone response pathway results in cell cycle arrest and activation of transcription factors that initiate the transcription of mating-specific genes, including the FUS1 gene.

FUS1 encodes a protein involved in cell fusion and mating. The pheromone response pathway is a model system for studying signal transduction in yeast genetics, as many of the signaling proteins and pathways are conserved in higher eukaryotes.

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You are interested in developing CRISPR mutation alleles of human gene CCR5. You first look up the gene sequence on public database GenBank. Based on the sort of mutant alleles you want to create you decide to design 3 guide RNA target sites within the first 1000bp of the gene (shown below).
Each target site should be 20 bp long and it must have a protospacer adjacent motif (PAM), which has the form NGG, immediately downstream (3’) of the target site. N means any base. The DNA sequence below shows the coding strand only, in the 5’--> 3’ direction.
1 cttcagatag attatatctg gagtgaagaa tcctgccacc tatgtatctg gcatagtgtg 61 agtcctcata aatgcttact ggtttgaagg gcaacaaaat agtgaacaga gtgaaaatcc 121 ccactaagat cctgggtcca gaaaaagatg ggaaacctgt ttagctcacc cgtgagccca 181 tagttaaaac tctttagaca acaggttgtt tccgtttaca gagaacaata atattgggtg 241 gtgagcatct gtgtgggggt tggggtggga taggggatac ggggagagtg gagaaaaagg 301 ggacacaggg ttaatgtgaa gtccaggatc cccctctaca tttaaagttg gtttaagttg 361 gctttaatta atagcaactc ttaagataat cagaattttc ttaacctttt agccttactg 421 ttgaaaagcc ctgtgatctt gtacaaatca tttgcttctt ggatagtaat ttcttttact 481 aaaatgtggg cttttgacta gatgaatgta aatgttcttc tagctctgat atcctttatt 541 ctttatattt tctaacagat tctgtgtagt gggatgagca gagaacaaaa acaaaataat 601 ccagtgagaa aagcccgtaa ataaaccttc agaccagaga tctattctct agcttatttt 661 aagctcaact taaaaagaag aactgttctc tgattctttt cgccttcaat acacttaatg 721 atttaactcc accctccttc aaaagaaaca gcatttccta cttttatact gtctatatga 781 ttgatttgca cagctcatct ggccagaaga gctgagacat ccgttcccct acaagaaact 841 ctccccggta agtaacctct cagctgcttg gcctgttagt tagcttctga gatgagtaaa 901 agactttaca ggaaacccat agaagacatt tggcaaacac caagtgctca tacaattatc 961 ttaaaatata atctttaaga taaggaaagg gtcacagttt ggaatgagtt tcagacggtt 1021 ataacatcaa agatacaaaa catgattgtg agtgaaagac tttaaaggga gcaatagtat
Come up with 3 guide RNA target sites

Answers

Three guide RNA target sites within the first 1000 base pairs of the CCR5 gene, each 20 bp long with a PAM (NGG) immediately downstream: Target Site 1: 61-80 bp (AGTCCTCATAAATGCTTACT), Target Site 2: 101-120 bp (CCACCTAAGATCCTGGGTCC), Target Site 3: 181-200 bp (TAGTTAAAACTCTTTAGACA).

What are three guide RNA target sites within the first 1000 base pairs of the CCR5 gene, each 20 bp long with a protospacer adjacent motif (PAM) in the form of NGG immediately downstream?

Based on the given DNA sequence, we need to design three guide RNA target sites within the first 1000 base pairs (bp) of the CCR5 gene. Each target site should be 20 bp long and have a protospacer adjacent motif (PAM) in the form of NGG immediately downstream of the target site.

Here are three possible guide RNA target sites:

Target Site 1: 61-80 bp

   Target sequence: AGTCCTCATAAATGCTTACT

   PAM sequence: GGT

Target Site 2: 101-120 bp

   Target sequence: CCACCTAAGATCCTGGGTCC

   PAM sequence: AGA

Target Site 3: 181-200 bp

   Target sequence: TAGTTAAAACTCTTTAGACA

   PAM sequence: AAA

For Target Site 1, we selected the sequence starting from position 61 and ending at position 80. The target sequence is AGTCCTCATAAATGCTTACT, and the PAM sequence is GGT.

For Target Site 2, we chose the sequence starting from position 101 and ending at position 120. The target sequence is CCACCTAAGATCCTGGGTCC, and the PAM sequence is AGA.

For Target Site 3, we selected the sequence starting from position 181 and ending at position 200. The target sequence is TAGTTAAAACTCTTTAGACA, and the PAM sequence is AAA.

These guide RNA target sites can be used for CRISPR-Cas9 gene editing experiments to introduce specific mutations in the CCR5 gene.

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Header ol Footer Text BLOOD COMPOSITION QUESTIONS 1. Fill in the blank for the following statements about blood composition a. The blood consists of 55% of plasma and 45% of formed elements. b. Normal

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Blood composition: The blood consists of 55% of plasma and 45% of formed elements.

The blood consists of 55% plasma and 45% formed elements. Plasma is a complex mixture of water, proteins, nutrients, electrolytes, nitrogenous wastes, hormones, and gases. Plasma is mainly water containing many dissolved solutes including proteins such as antibodies, albumin, fibrinogen, and globulin. Formed elements refer to red blood cells, white blood cells, and platelets. Red blood cells, or erythrocytes, are the most abundant formed element. They contain hemoglobin and transport respiratory gases. White blood cells, or leukocytes, are less abundant than red blood cells but have important defensive roles. Platelets are cell fragments that play a key role in blood clotting. Normal blood pH is 7.35 to 7.45. The body works to maintain this narrow pH range as it is essential for proper physiological functioning.

Blood is a complex and vital fluid that contains a variety of components. Blood consists of plasma, which is 55% of the total volume, and formed elements, which are 45% of the total volume. Formed elements include red blood cells, white blood cells, and platelets. Red blood cells transport respiratory gases and are the most abundant formed element, while platelets are involved in blood clotting. Normal blood pH is a narrow range between 7.35 and 7.45, which is essential for proper physiological functioning.

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Which description describes a reflex arc, specifically, that of the Patellar tendon. If, (+)= activation of (-)= inhibition of O Both A & C OA) Stimulus-> (+)Sensory neuron-> (+)Interneuron-> (+)Motor neuron OB) Stimulus-> (+)Sensory neuron-> (+)Interneuron-> (+)Motor neuron OC) Stimulus-> (+)Sensory neuron-> Both (1) & (2) where (1) (+)Interneuron-> (-)Motor neuron (2)-(+) Motor neuron D

Answers

The correct description that describes the reflex arc of the Patellar tendon is option C, Stimulus -> (+) Sensory neuron -> Both (1) and (2), where (1) represents the activation of an interneuron and (2) represents the activation of a motor neuron.

In this reflex arc, a sensory neuron is activated in response to a stimulus, in this case, the stretching of the patellar tendon. Both an interneuron and a motor neuron receive sensory information from the sensory neuron. The motor neuron can then be activated or inhibited by the interneuron. A coordinated response to the stimulus is made possible by this modulation.

When the Patellar tendon is stretched beyond what is normal, the interneuron may inhibit the motor neuron, preventing overexertion of the muscles and acting as a safeguard. On the other hand, if the stretch is within a normal range the motor neuron may be activated by the interneuron causing the quadriceps muscle to contract as needed and the leg to extend.

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The correct sequence of layers in the wall of the alimentary canal, from internal to external, is a.mucosa, muscularis, serosa, submucosa. b.submucosa, mucosa, serosa, muscularis. c.mucosa, submucosa, muscularis, serosa. d.serosa, muscularis, mucosa, submucosa.

Answers

The correct sequence of layers in the wall of the alimentary canal, from internal to external, is mucosa, submucosa, muscularis, serosa.

The correct option is C.

Mucosa, submucosa, muscularis, serosa.What is the alimentary canal?The alimentary canal is a muscular tube that begins at the mouth and extends through the pharynx, esophagus, stomach, small intestine, and large intestine to the anus. It is composed of four distinct layers of tissues that function together to perform digestion and absorption of nutrients from food.

These layers are referred to as mucosa, submucosa, muscularis, and serosa.The four layers of the alimentary canal are:Mucosa: The mucosa is the innermost layer of the alimentary canal. It is made up of three layers of tissues: the epithelium, the lamina propria, and the muscularis mucosae. It produces mucus, enzymes, and hormones that aid in digestion.Submucosa: The submucosa is the second layer of the alimentary canal. It is composed of connective tissues that contain blood vessels, nerves, and lymphatics. It also contains glands that produce mucus, enzymes, and hormones.

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Which of the following will most likely disrupt the Hardy-Weinberg equilibrium that xists for a population of small rodents ving in a habitat with ample resources? a. The rodents reproduce frequently and have large litters, so the population size is increasing. b. Mate selection is completely random within the population of rodents. c. The population continues to remain isolated from other populations of the rodent. d. The coding region of a gene is altered in sperm produced by a particular male that mates with several of the female rodents, which produce many progeny as a result.

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The option that is most likely to disrupt the Hardy-Weinberg equilibrium in a population of small rodents living in a habitat with ample resources is: The coding region of a gene is altered in sperm produced by a particular male that mates with several of the female rodents, which produce many progeny as a result. So, option D is accurate.

The Hardy-Weinberg equilibrium describes the genetic equilibrium that occurs in an ideal, non-evolving population. It is based on several assumptions, including random mating, no genetic drift, no gene flow, no mutation, and no selection.

In this scenario, if the coding region of a gene is altered in the sperm produced by a male and is passed on to a large number of progeny, it introduces a genetic change into the population. This alteration can disrupt the equilibrium by changing the allele frequencies. As the altered gene spreads through the population, it can result in a departure from the expected genotype frequencies predicted by the Hardy-Weinberg equilibrium.

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True or False: A piece of silver can be cut indefinitely into pieces and still retain all of the properties of silver Al Truc. All particles, including subatomic particles that make up the element, possess the proporties of the element. B) True. Atoms are the smallest units of matter, are indivisible, and possess the properties of their element. C) False. Once the pieces are smaller than an atom of silver, the pieces no longer retain the properties of silver D) False. Silver atoms are too small to possess the properties of silver E) False. As a piece of silver is cut into smaller pieces, the atoms begin to take on the properties of smaller elements on

Answers

The statement "False. Once the pieces are smaller than an atom of silver, the pieces no longer retain the properties of silver" is the correct answer to this question.

Elements are made up of atoms that are identical in nature, including their physical and chemical properties. This is valid for silver as well. A silver atom can be cut into several pieces and still maintain its silver properties.

However, once the pieces are reduced to less than one silver atom, they lose their chemical properties as they no longer have the silver properties.

Once the pieces are smaller than an atom of silver, the pieces no longer retain the properties of silver.

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Three genotypes in a very large population have, on average, the following values of survival and fecundity, regardless of their relative frequencies: Genotype A1A1 A1A2 A2A2 Survival to adulthood (viability) 0.80 0.90 0.50 Number of offspring 3.0 4.0 8.0 Absolute fitness 2.4 3.6 4.0 Which of the following best describes what will happen at this locus in the long run? There will be a stable polymorphism because the heterozygote has a higher survival rate than either homozygote. Nothing will happen because the differences among genotypes in survival and fecundity cancel each other out. Allele A2 will be fixed eventually. One allele will be fixed but we cannot predict which one. Allele Al will be fixed eventually.

Answers

The population under observation has three genotypes: A1A1, A1A2, and A2A2. These genotypes have survival rates of 0.80, 0.90, and 0.50, and fecundity rates of 3.0, 4.0, and 8.0, respectively.

The absolute fitness of these genotypes is 2.4, 3.6, and 4.0, respectively. Which of the following statements best describes what will happen to the locus in the long run? Allele A2 will eventually become fixed is the correct option. This is due to the fact that allele A2 has the highest fitness of the three alleles, with a fitness of 4.0, and will thus outcompete the other two alleles in the population over time. Eventually, A2 will become the only allele present in the population because it is more effective at reproducing and surviving than A1. Over time, A2 will increase in frequency while A1 will decrease, and ultimately, A2 will become fixed in the population because it will be the only allele remaining.

Therefore, allele A2 will be fixed eventually. The statement "There will be a stable polymorphism because the heterozygote has a higher survival rate than either homozygote" is incorrect.

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You have isolated a microbe from the soil and sequenced its genome. Please discuss how you could use the sequence information to identify the organism and establish if it is a prokaryotic or eukaryotic microorganisms

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To identify the organism and establish whether it is a prokaryotic or eukaryotic microorganism after isolating a microbe from the soil and sequencing its genome, the following steps could be taken: Assemble the genome sequencing reads into a contiguous sequence (contig).

Contigs are produced by sequencing the DNA multiple times and assembling the resulting DNA sequences together. During this process, overlapping regions are identified and used to construct a single continuous DNA sequence.Step 2: Using a genome annotation software, a genome annotation is made. The annotation process identifies genes and noncoding sequences, predicts gene function, and assigns them to functional classes. Gene identification can help determine whether the organism is prokaryotic or eukaryotic.

Comparison of the genome sequence with sequences of known organisms in a database. The comparison of genome sequences is commonly used to identify microbes, as sequence similarity is an indicator of evolutionary relatedness. In the case of eukaryotes, a comparison of gene sequences can also be used to identify and classify organisms.Another way of establishing whether an organism is prokaryotic or eukaryotic is by looking at the organization of the genome. Prokaryotic genomes are generally simpler in their organization, with no nucleus or organelles, and they have a circular chromosome. Eukaryotic genomes, on the other hand, are usually larger and more complex, with multiple chromosomes, a nucleus, and various organelles such as mitochondria, chloroplasts, and endoplasmic reticulum.

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Statement 1: Dendritic cells are phagocytes with professional antigen-presenting properties Statement 2: Neutrophils circulate as part of the blood and act as surveillance to detect presence of pathogens O Statement 1 is true Statement 2 is false. O Statement 2 is true. Statement 1 is false. O Both statements are true. O Both statements are false points Statement 1: Fever is a sign of pathogen infection. Statement 2: Vasodilation is a type of immune response that can cause redness and swelling at the infection site. O Statement 1 is true. Statement 2 is false, O Statement 2 is true. Statement 1 is false. O Both statements are true. O Both statements are false Which of the following describes passive immunity? O vaccination for polio O allowing oneself to become infected with chicken pox O catching a common cold O antibodies transferred to the fetus from the mother across the placenta If Peter is allergic to peanuts and Paul is not, what is the precise molecular difference in Peter's bloodstream responsible for this? O Peter's blood has mast cells and basophils carrying IgEs that match an antigen on peanuts. Peter's blood has mast cells and basophils carrying IgGs that match an antigen on peanuts. O Peter's blood has mast cells and basophils carrying IgMs that match an antigen on peanuts O Peter's blood has mast cells and basophils carrying IgAs that match an antigen on peanuts Sive Answer 1 points Statement 1: The cell-mediated immune response is brought about by T cells Statement 2: In humoral immunity, some B cells become memory cells which are long-lived cells that can recognize an antigen that once already infected the body O Statement 1 is true. Statement 2 is false. Statement 2 is true. Statement 1 is false O Both statements are true Both statements are false.

Answers

Dendritic cells are phagocytes with professional antigen-presenting properties. Neutrophils circulate as part of the blood and act as surveillance to detect presence of pathogens.

The correct answer is that statement 1 is true and statement 2 is false. Fever is a sign of pathogen infection. Vasodilation is a type of immune response that can cause redness and swelling at the infection site. The correct answer is that both statements are true.

Passive immunity is antibodies transferred to the fetus from the mother across the placenta.The precise molecular difference in Peter's bloodstream responsible for this is Peter's blood has mast cells and basophils carrying IgEs that match an antigen on peanuts.

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What is a shared derived characteristic for the stramenopiles?
What is a shared primitive characteristic for this group? How do
these differ from autopomorphies and synapomorpies?

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The shared derived characteristic for the stramenopiles is the presence of two flagella. The presence of chlorophyll c, on the other hand, is a shared primitive characteristic of the stramenopiles.

A shared derived characteristic for the stramenopiles is the presence of two flagella.

One of the flagella has a smooth surface, while the other has fine, hair-like projections known as "straw-like" or "hairy" flagella. This unique flagellar arrangement is a distinguishing feature of the stramenopiles.

A shared primitive characteristic for the stramenopiles is the presence of chlorophyll c.

This type of chlorophyll pigment is also found in other algal groups. Chlorophyll c is considered primitive because it is a common feature among various algal lineages and not specific to the stramenopiles.

Stramenophiles are a specific group of organisms that share common characteristics, including the presence of two flagella with distinct structures. Autapomorphies are uniquely derived characteristics specific to individual taxa, while synapomorphies are shared derived characteristics that indicate common ancestry between multiple taxa.

Therefore, the shared derived characteristic and shared primitive characteristic for the stramenopiles is the presence of two flagella and chlorophyll c respectively.

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Fatty acid breakdown generates a large amount of acetyl CoA. What will be the effect of fatty acid breakdown on the activity of the Pyruvate Dehydrogenase (PDH) Complex?
a. The activity of the PDH complex would remain the same b. The activity of the PDH complex would decrease c. The activity of the PDH complex would increase

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Pyruvate dehydrogenase (PDH) complex is a cluster of multienzyme that facilitates the conversion of pyruvate into acetyl-CoA.

Acetyl-CoA is a critical energy-generating molecule that helps provide energy to the human body. The PDH complex is regulated via negative feedback inhibition, which helps to control the rate of metabolism of pyruvate. Negative feedback inhibition happens when high energy levels in the body act as an inhibitor to metabolic pathways, leading to a reduction in enzyme activity.

Acetyl-CoA is a compound that is produced by a range of metabolic pathways, including fatty acid breakdown. When there is an increase in acetyl-CoA, the body will increase the activity of the Pyruvate Dehydrogenase (PDH) Complex. It's because Acetyl-CoA also serves as a key regulator of PDH activity.Acetyl-CoA regulates the activity of the PDH complex by inhibiting its activity. When there is an increase in acetyl-CoA, the PDH complex will be inhibited, which will help to control the rate of metabolism of pyruvate. Thus, we can say that the activity of the PDH complex would increase when there is an increase in acetyl-CoA.

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At values of Y (fractional saturation below 0,1 and
above 0,9, the slopes of Hill plots tend to a value of 1,
indicating an absence of cooperativity. True or False
SUPPORT YOUR ANSWER.

Answers

The statement "At values of Y (fractional saturation below 0,1 and above 0,9, the slopes of Hill plots tend to a value of 1, indicating an absence of cooperativity" is False.

The Hill plot is a linear plot with a log dose vs response curve. The Hill coefficient (nH) determines the degree of cooperativity in the Hill plot, which ranges from 0 to 1. If nH is equal to 1, there is no cooperativity; if nH is between 0 and 1, it indicates the degree of cooperativity.

The slope of the Hill plot at fractional saturation levels of less than 0.1 and greater than 0.9 is almost horizontal. At these limits, the concentration of binding sites on the protein is low.

The sites are almost entirely vacant at fractional saturation levels less than 0.1 and almost completely occupied at saturation levels greater than 0.9. As a result, the binding behavior at both ends is non-cooperative.

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Dehydration during exercise:
A. is dangerous if we lose 1-2% bodyweight
B. causes hyponatremia
C. may occur when runners are allowed to drink ad libidum
D. is a training technique to improve lactate t

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Dehydration during exercise can have various consequences, including danger if a person loses 1-2% of their body weight, the possibility of hyponatremia. It is not a recommended training technique to improve lactate threshold.

Dehydration during exercise can have significant impacts on the body and athletic performance. It is important to maintain proper hydration levels to ensure optimal functioning of the body's physiological processes.

A. Losing 1-2% of body weight through dehydration during exercise can be dangerous. Even a small percentage of dehydration can lead to decreased performance, increased heart rate, impaired thermoregulation, and reduced blood volume, which can affect cardiovascular function and increase the risk of heat-related illnesses.

B. Hyponatremia is a condition characterized by low sodium levels in the blood. While dehydration typically involves a loss of body fluids, hyponatremia occurs when excessive water intake dilutes the sodium concentration in the blood. This can be a potential risk during prolonged exercise if individuals consume large amounts of water without adequate electrolyte replenishment.

C. Allowing runners to drink ad libitum means they can drink freely as desired. In some cases, athletes may drink excessively during exercise, leading to hyponatremia or overhydration. Proper guidance and monitoring of fluid intake are important to prevent dehydration and hyponatremia.

D. Dehydration is not considered a training technique to improve lactate threshold. Lactate threshold training typically involves structured workouts designed to increase the body's ability to tolerate and clear lactate during intense exercise. Adequate hydration is important during training to support optimal performance and recovery.

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Suppose you have a couple who are both heterozygous for BOTH albinism and sickle cell anemia. Use A and a for the albinism alleles, and T and t for the sickle cell alleles. (Technically, the sickle alleles are codominant, but since we’re interested in the disease rather than sickle trait, we’ll use dominant/recessive notation.)
What are the genotypes for the couple described above? Their phenotypes? Keep in mind that a genotype must include two alleles per genetic locus! (Phenotype will be albino or not albino and sickle cell anemia or healthy.)

Answers

The genotypes of the couple described are AaTt for the male and AaTt for the female. Their phenotypes will depend on whether they express the recessive traits or not.

For the couple described, the male is heterozygous for both albinism (Aa) and sickle cell anemia (Tt), and the female is also heterozygous for both traits (AaTt). The genotype for each individual includes two alleles per genetic locus.

In terms of phenotypes, the presence of the dominant allele (A) for albinism means that neither the male nor the female will express the albino phenotype. Therefore, their phenotype will be non-albino.

For sickle cell anemia, the presence of the recessive allele (t) is necessary for the expression of the disease. Since both individuals are heterozygous for the sickle cell trait (Tt), they will not have sickle cell anemia. Instead, their phenotype for sickle cell will be healthy or unaffected.

To summarize, the genotypes of the couple are AaTt, and their phenotypes are non-albino and healthy for sickle cell anemia.

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Prokaryotic genomes can be said to be and as compared to eukaryotic ones. O gene dense; non-coding DNA poor gene poor, non-coding DNA rich gene poor; non-coding DNA poor O gene dense; non-coding DNA rich

Answers

Prokaryotic genomes can be said to be gene dense; non-coding DNA poor, as compared to eukaryotic ones. Prokaryotes have single, circular chromosomes which contain most of their genetic material, whereas eukaryotes have multiple linear chromosomes enclosed in a nucleus.

Prokaryotes are unicellular organisms that lack a true nucleus and membrane-bound organelles, while eukaryotes are organisms that have a true nucleus and membrane-bound organelles, like mitochondria, chloroplasts, and a Golgi apparatus. Eukaryotic DNA is wound around histones to form nucleosomes, which give the chromatin its structure and organization. Non-coding DNA accounts for the majority of the DNA in eukaryotes, while prokaryotes have a relatively small amount of non-coding DNA.Prokaryotic genomes are gene-rich because they have evolved to be very efficient. The high gene density is a result of the compact organization of prokaryotic genomes, which allows them to fit into a small cell. In comparison, eukaryotic genomes are much larger and more complex than prokaryotic ones. Eukaryotic DNA contains introns and exons, which can be alternatively spliced to produce a variety of protein isoforms. As a result, eukaryotic genomes are able to produce a greater diversity of proteins than prokaryotic ones.In conclusion, prokaryotic genomes are gene dense and non-coding DNA poor, while eukaryotic genomes are gene poor, non-coding DNA rich, and more complex.

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Explain the relationship between each of the following terms: (a) energy and work (b) potential energy and kinetic energy (c) free energy and spontaneous changes

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(a) Energy and work are related concepts in physics. Energy is a broad term that refers to the capacity of a system to do work or transfer heat.

It exists in different forms such as kinetic energy, potential energy, thermal energy, and more. Work, on the other hand, is a specific type of energy transfer that occurs when a force is applied to an object, causing it to move in the direction of the force. Work is the process of converting energy from one form to another or transferring it from one object to another. (b) Potential energy and kinetic energy are two forms of energy. Potential energy is the energy possessed by an object due to its position or condition. It is stored energy that can be converted into other forms, such as kinetic energy. Kinetic energy, on the other hand, is the energy possessed by an object due to its motion. It depends on the mass of the object and its velocity. When an object moves, its potential energy may be converted into kinetic energy, and vice versa. (c) Free energy and spontaneous changes are related to thermodynamics. Free energy (G) is a measure of the energy available in a system to do useful work. It takes into account both the enthalpy (H) and entropy (S) of the system through the equation: ΔG = ΔH - TΔS, where ΔG is the change in free energy, ΔH is the change in enthalpy, T is the temperature, and ΔS is the change in entropy. Spontaneous changes are processes that occur without the need for external intervention and tend to increase the disorder or entropy of a system. In thermodynamics, a spontaneous process occurs when the change in free energy (ΔG) is negative, indicating that the system's energy is decreasing and becoming more stable.

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