Match the role of the enzyme to their Gyrase DNA Ligase DNA polymerase Helicase [Choose ] The enzyme complex adds nucleotides in a leading a lagging fashion to generate new copies of DNA. The enzyme unwinds DNA to create a replication fork. The enzyme that forms a covalent bond in the phosphodiester backbone of DNA. ✓ The enzyme adds negative supercoils to the DNA to reduce strain on the DNA. The enzyme complex adds nu The enzyme that forms a cova The enzyme unwinds DNA to +

Answers

Answer 1

Matching the roles of enzymes to their respective functions:

- Gyrase: The enzyme adds negative supercoils to the DNA to reduce strain on the DNA.

- DNA Ligase: The enzyme that forms a covalent bond in the phosphodiester backbone of DNA.

- DNA polymerase: The enzyme complex adds nucleotides in a leading and lagging fashion to generate new copies of DNA.

- Helicase: The enzyme unwinds DNA to create a replication fork.

Gyrase is an enzyme that plays a crucial role in DNA replication and maintenance. It introduces negative supercoils into the DNA molecule, which helps to relieve the torsional strain that builds up during the unwinding of the double helix. By adding negative supercoils, gyrase prevents the DNA strands from becoming overly tangled and ensures the smooth progress of DNA replication and transcription.

DNA Ligase is an enzyme responsible for the formation of phosphodiester bonds in the DNA backbone. It plays a crucial role in DNA repair and replication by joining the Okazaki fragments on the lagging strand during DNA replication and sealing any nicks or gaps in the DNA molecule. DNA ligase effectively seals the breaks in the DNA backbone, allowing for the continuity and integrity of the DNA molecule.

DNA polymerase is a group of enzymes that are essential for DNA replication. They catalyze the addition of nucleotides to the growing DNA strand during DNA synthesis. DNA polymerases work in both the leading and lagging strands of DNA replication. The leading strand is synthesized continuously, while the lagging strand is synthesized in short fragments called Okazaki fragments. DNA polymerase plays a key role in accurate DNA replication, ensuring that the genetic information is faithfully copied.

Helicase is an enzyme that plays a central role in DNA replication by unwinding the DNA double helix. It uses energy from ATP hydrolysis to break the hydrogen bonds between the base pairs and separate the DNA strands, creating a replication fork. Helicase unwinds the DNA ahead of the replication fork, allowing access to the template strands and enabling the DNA polymerase to synthesize new complementary strands.

These enzymes work together during DNA replication to ensure the accurate duplication of genetic material. Gyrase and helicase prepare the DNA molecule for replication by unwinding and relieving strain, while DNA polymerase adds nucleotides to create new strands, and DNA ligase joins the fragments and seals any breaks in the DNA backbone. The coordinated actions of these enzymes ensure the faithful replication and transmission of genetic information during cell division and DNA repair processes.

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Related Questions

Cationic detergents are considered more effective because... Otheir positive charge is repelled by the negative charged surface of microbial cells O their positive charge is attracted to the negative charged surface of microbial cells O their negative charge is attracted to the negative charged surface of microbial cells their positive charge is attracted by the positive charged surface of microbial cells

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Cationic detergents are effective in fighting bacteria because their positively charged head is attracted to the negatively charged surface of microbial cells. When the detergent binds to the cell membrane, it disrupts the membrane's integrity and causes the cell contents to leak out.

Cationic detergents are considered more effective because their positive charge is attracted to the negative charged surface of microbial cells. An ionic detergent consists of a hydrophilic polar head, which has either a positive or negative charge, and a hydrophobic nonpolar tail, which is commonly a long alkyl chain.The most important feature of a cationic detergent is its positively charged head, which is why it's more effective against bacteria.

Cationic detergents, also known as cetylpyridinium chloride, benzalkonium chloride, and quaternary ammonium compounds, are effective against a variety of bacteria, including gram-positive and gram-negative bacteria. They act by disrupting the microbial cell membrane and causing the contents to leak  Cationic detergents are more effective because they are positively charged

Their positively charged head is attracted to the negative charge on the surface of microbial cells Cetylpyridinium chloride, benzalkonium chloride, and quaternary ammonium compounds are all examples of cationic detergents.Cationic detergents, such as these, cause bacterial cell membranes to rupture and leak out contents.

Cationic detergents are effective in fighting bacteria because their positively charged head is attracted to the negatively charged surface of microbial cells. When the detergent binds to the cell membrane, it disrupts the membrane's integrity and causes the cell contents to leak out.

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Final answer:

Cationic detergents like quaternary ammonium salts (quats) are effective because their positive charge is attracted to the negatively charged surface of microbial cells. This disrupts the bacterial membrane, killing the bacteria. They're frequently used in disinfectants for this reason.

Explanation:

Cationic detergents are considered more effective because their positive charge is attracted to the negatively charged surface of microbial cells. These detergents, such as quaternary ammonium salts (quats), contain a positively charged cation at one end attached to a long hydrophobic chain.

The cationic charge of quats confers their antimicrobial properties, which are diminished when neutralized. Due to this property, they can effectively disrupt the integrity of bacterial membranes, thereby effectively killing the bacterial cells.

These quats, including benzalkonium chlorides, are also found in a variety of household cleaners and disinfectants as they are stable, non-toxic, inexpensive, colorless, odorless, and tasteless.

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Explain the importance of lipid nanoparticle technology in RNA delivery system.

Answers

Lipid nanoparticle technology plays a crucial role in RNA delivery systems, enabling efficient and targeted delivery of RNA therapeutics.

Lipid nanoparticle technology is of paramount importance in the field of RNA delivery systems. These nanoparticles, composed of lipids, are designed to encapsulate and protect RNA molecules, ensuring their stability and preventing degradation. The main answer lies in their ability to facilitate efficient and targeted delivery of RNA therapeutics to specific cells or tissues in the body.

Lipid nanoparticles possess unique characteristics that make them ideal for RNA delivery. Firstly, their small size allows for easy penetration through biological barriers, such as cell membranes. This enables effective delivery of RNA molecules into the target cells, where they can exert their therapeutic effects. Additionally, the lipid-based structure of these nanoparticles enables them to interact with cell membranes, facilitating the internalization of the RNA cargo into the cells.

Moreover, lipid nanoparticles offer protection to the RNA molecules during circulation in the body. The lipid bilayer of the nanoparticles shields the RNA from enzymatic degradation and clearance by the immune system. This enhances the stability and half-life of the RNA therapeutics, increasing their efficacy and reducing the required dosage.

Furthermore, lipid nanoparticle technology allows for precise targeting of specific cells or tissues. By modifying the surface of the nanoparticles with ligands or antibodies that recognize cell-specific receptors, researchers can achieve selective delivery of RNA therapeutics to the desired cells. This targeted approach enhances the therapeutic efficiency and minimizes off-target effects, improving the safety profile of RNA-based therapies.

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Can
you help me to solve those questions?
Your male patient is in renal (kidney) failure. His recent blood tests indicated a hematocrit of 24%. (8 points) ■ Is this level of hematocrit normal or abnormal? Explain what information the hemato

Answers

A hematocrit level of 24% is considered abnormal or low. Hematocrit refers to the percentage of red blood cells (RBCs) in the total volume of blood.

Low hematocrit can indicate several conditions, and in the context of a patient with renal (kidney) failure, it can be attributed to several factors:

Anemia: Kidney failure can lead to decreased production of erythropoietin, a hormone responsible for stimulating red blood cell production in the bone marrow. Reduced erythropoietin levels can result in anemia, characterized by a low hematocrit level.

Blood loss: Patients with kidney failure may experience gastrointestinal bleeding or require frequent blood sampling for various tests. These factors can contribute to a decrease in hematocrit levels.

Fluid overload: Kidney failure can lead to fluid retention and an expansion of blood volume. Although the absolute number of red blood cells may be normal, the diluted blood volume can result in a lower hematocrit percentage.

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Compare and describe the differences and
similarities of artery muscle wall and large vein muscle
wall.

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Arteries have thicker muscle walls and more elastic fibers compared to large veins, allowing them to withstand higher blood pressure and maintain continuous blood flow, while veins have thinner muscle walls and valves to prevent backflow of blood.

Both artery and large vein muscle walls are composed of smooth muscle cells, elastic fibers, and collagen. Smooth muscle cells are responsible for the contraction and relaxation of the muscle wall, allowing for the regulation of blood flow. Elastic fibers provide elasticity to the walls, allowing them to stretch and recoil.

Arteries have thicker muscle walls compared to large veins. This thicker wall is necessary to withstand the higher pressure generated by the heart during systole (contraction phase). The increased muscle thickness and elasticity of arteries enable them to expand and recoil, maintaining continuous blood flow and preventing fluctuations in blood pressure.

In contrast, large veins have thinner muscle walls. While they still contain smooth muscle cells, the muscle layer is less prominent. Large veins are equipped with valves, which help to prevent the backflow of blood and ensure the unidirectional flow towards the heart.

The thinner muscle walls in veins allow them to accommodate larger volumes of blood and facilitate the return of blood to the heart against lower pressure.

In summary, both artery and large vein muscle walls contain smooth muscle cells, elastic fibers, and collagen, contributing to their contractile and elastic properties.

Arteries have thicker muscle walls and more elastic fibers, allowing them to withstand higher blood pressure and maintain continuous blood flow. Large veins have thinner muscle walls, but their structure is complemented by valves, facilitating the return of blood to the heart.

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D Question 6 1 pts People suffering from diarrhea often takes ORT therapy. What is the mechanism why ORT therapy works? OORT stimulates Na+, glucose and water absorption by the intestine, replacing fl

Answers

ORT or Oral Rehydration Therapy helps to replenish fluids and electrolytes in the body of people suffering from diarrhea.

This therapy is a simple, cost-effective, and efficacious way to prevent the deaths of millions of people each year. The mechanism by which ORT therapy works is that it stimulates the absorption of sodium (Na+), glucose, and water by the intestine, replacing the fluids that have been lost due to diarrhea.

The glucose present in the ORT solution is a source of energy that helps in the absorption of sodium and water into the bloodstream.

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Question 8.9 of 31 A FLAG QUESTION A species of butterfly is codominant for wing color. If a blue butterfly (D) mates with a yellow butterfly by what would their spring look like! Answers A-D А blue

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A species of butterfly is codominant for wing color. If a blue butterfly mates with a yellow butterfly, their offspring would be green. When two codominant alleles are inherited, both traits are seen in offspring.

The cross between blue (DD) and yellow (DD) butterfly would produce offspring with genotype Dd, resulting in green wings, which is the intermediate color between blue and yellow. The blending of both colors results in an entirely new color altogether that is green in this case.

The blending happens because neither allele is dominant. Codominance is the relationship between two different versions of a gene, where both alleles are expressed simultaneously. Codominance is different from incomplete dominance, which happens when two different alleles for the same trait combine and form an intermediate phenotype.

For example, a cross between a red (RR) and white (WW) flower produces pink (RW) flowers, which are a mix of both colors.In conclusion, when a blue butterfly (DD) mates with a yellow butterfly (DD), their offspring would have a green (Dd) phenotype.

The new color that is produced is the result of codominance, which is when both alleles are expressed simultaneously.

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1. Most vaccines are a collection of antigens delivered with an adjuvant. An adjuvant can..?
a. Improve the immune response to the vaccine.
b. Limit the growth of antigen-bearing microbes c. Inhibit antibody production.
d. Inhibit host B-cell division. e. Help degrade the vaccine.
2. True or False: If antibodies directed to the Rh factor on red blood cells are present, these antibodies can cause cell lysis similar lysis during mismatched blood transfusions that either anti-A or anti-B antibodies. 3. True or False: Patients suffering from Acquired Immunodeficiency Syndrome AIDS) after HIV infection die because of direct cytopathic effects of HIV on host cells.

Answers

1.They die from opportunistic infections, which occur because the immune system is unable to fight off infections due to the destruction of T helper cells.

2.False. Antibodies directed to the Rh factor on red blood cells, known as anti-Rh antibodies or anti-D antibodies, do not cause immediate cell lysis or hemolysis, similar to what happens during mismatched blood transfusions with anti-A or anti-B antibodies.

3.False. Patients suffering from Acquired Immunodeficiency Syndrome (AIDS) after HIV infection do not die primarily because of the direct cytopathic effects of HIV on host cells.

1. An adjuvant can improve the immune response to the vaccine. The antigen is a toxin or other foreign substance that induces an immune response in the body. An adjuvant is a component of a vaccine that enhances the body's immune response to an antigen. An adjuvant can be added to a vaccine to improve its effectiveness and to ensure that a person's immune system reacts to the vaccine in the desired way.

2. True. If antibodies directed to the Rh factor on red blood cells are present, these antibodies can cause cell lysis similar lysis during mismatched blood transfusions that either anti-A or anti-B antibodies.3. False. Patients suffering from Acquired Immunodeficiency Syndrome AIDS) after HIV infection do not die because of direct cytopathic effects of HIV on host cells. Instead, they die from opportunistic infections, which occur because the immune system is unable to fight off infections due to the destruction of T helper cells by HIV.

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In a large population of ragweed, genotype frequencies are in Hardy-Weinberg equilibrium with f(AA) = 0.04, f(Aa) = 0.32, f(aa) = 0.64. This locus is neutral with respect to fitness. Researchers sample 5 individuals from this population to establish a new population of ragweed in a national park. After several generations, the researchers return to the newly established population and find that the A allele has been lost. The most likely reason for this is: Non-random mating with respect to the A allele Drift caused by the sampling error in the founding population selected by the researchers Heterozygote advantage that decreased the homozygous individuals in the population New mutations that removed the A allele from the population Fluctuating selection pressure that vary over time or space

Answers

The most likely reason that the A allele has been lost in the new population of ragweed is due to drift caused by the sampling error in the founding population selected by the researchers.

A being passed on to the next generation should remain constant. However, when researchers sample 5 individuals from this population to establish a new population of ragweed in a national park, there is a chance that the frequency of the alleles will change due to sampling error.

The other options provided in the question, such as non-random mating, heterozygote advantage, new mutations, or fluctuating selection pressure, were not mentioned as factors in this scenario.

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Immune reconstitution inflammatory syndrome" (IRIS) occurs When the number of macrophages is normalized after antiretroviral therapy for HIV-AIDS Is caused by virus infection of a virus like HIV When

Answers

IRIS is an abnormal immunological response as the immune system heals and overreacts to past illnesses or microorganisms. After HIV-AIDS treatment, "immune reconstitution inflammatory syndrome" (IRIS) develops when macrophage numbers normalize.

It is not caused by HIV infection. HIV-positive people starting ART may develop IRIS. It causes an excessive inflammatory response to dormant microorganisms or opportunistic infections. HIV infection reduces immune cells, particularly macrophages. ART suppresses viral replication, restoring the immune system. Macrophages can normalize as the immune system recovers. This immunological recovery can cause a severe inflammatory response to pre-ART opportunistic illnesses or pathogens. Inflammation, tissue damage, and clinical decline can arise after immune system reconstitution.

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if the distance between the basil and the oregano is 16 in and the distance between the thyme and the oregano is 4 in, what is the distance between the basil and the thyme?

Answers

The distance between the basil and thyme is approximately 16.49 inches.

To find the distance between the basil and thyme, we can use the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides.

Let's assign variables to represent the distances between the plants:

Let x be the distance between the basil and the thyme.

Let y be the distance between the basil and the oregano.

Let z be the distance between the thyme and the oregano.

From the problem statement, we know that y = 16 in and z = 4 in.

Using the Pythagorean theorem, we can write:

x^2 = y^2 + z^2

x^2 = 16^2 + 4^2

x^2 = 256 + 16

x^2 = 272

Taking the square root of both sides, we get:

x = sqrt(272)

x ≈ 16.49 in

Therefore, the distance between the basil and thyme is approximately 16.49 inches.

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Design a messenger RNA transcript with the necessary prokaryotic
control sites that codes for the octapeptide
Lys-Pro-Ala-Gly-Thr-Glu-Asn-Ser.

Answers

A designed mRNA transcript for the octapeptide Lys-Pro-Ala-Gly-Thr-Glu-Asn-Ser require a promoter sequence, a Shine-Dalgarno sequence, a start codon, a coding region for the peptide, and a stop codon.

To design an mRNA transcript for the octapeptide Lys-Pro-Ala-Gly-Thr-Glu-Asn-Ser in a prokaryotic system, several key elements need to be included.

First, a promoter sequence is necessary to initiate transcription. The promoter sequence is recognized by RNA polymerase and helps to position it correctly on the DNA template.

Next, a Shine-Dalgarno sequence is required. This sequence, typically located upstream of the start codon, interacts with the ribosome and facilitates translation initiation.

Following the Shine-Dalgarno sequence, a start codon, such as AUG, is needed to indicate the beginning of the coding region for the octapeptide.

The coding region itself will consist of the corresponding nucleotide sequence for the octapeptide Lys-Pro-Ala-Gly-Thr-Glu-Asn-Ser. Each amino acid is encoded by a three-nucleotide codon.

Finally, a stop codon, such as UAA, UAG, or UGA, is required to signal the termination of translation.

By incorporating these elements into the mRNA transcript, the prokaryotic system will be able to transcribe and translate the genetic information to produce the desired octapeptide.

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Explain the potential consequences of mutations and how chromosomes determine the sex of a human individual. Determine autosomal and sex-linked modes of inheritance for single-gene disorders and explain what is meant by a carrier.

Answers

Mutations are a change in the genetic sequence, which could cause genetic disorders. The potential consequences of mutations can range from mild, such as producing an incorrect protein, to severe, such as completely preventing the protein from being produced or disrupting normal development or causing cancer.

The chromosomes determine the sex of a human individual because of the X and Y chromosomes. Females have two X chromosomes (XX), while males have one X and one Y chromosome (XY). If an egg cell is fertilized by a sperm cell that carries an X chromosome, the zygote will become a female. On the other hand, if an egg cell is fertilized by a sperm cell that carries a Y chromosome, the zygote will become a male.

Single-gene disorders could be inherited in two ways: autosomal and sex-linked. Autosomal inheritance occurs when the gene is located on one of the 22 pairs of autosomes. The mode of inheritance could be dominant or recessive. Sex-linked inheritance occurs when the gene is located on one of the sex chromosomes. For example, the hemophilia gene is located on the X chromosome and is recessive.

If a female carries one hemophilia gene on one of her X chromosomes, she is considered a carrier. On the other hand, if a male carries the gene on his X chromosome, he will develop hemophilia because there is no corresponding gene on the Y chromosome to mask the hemophilia gene's effects.

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mRNA degradation occurs in the cytoplasm
a- After exonucleolytic degradation 5–>3' as well as 3–>5'
b- By ribonucleoproteins
c- By endonucleolytic activity
d- By upf proteins
e- By deanilation

Answers

The correct option is B.

mRNA degradation occurs in the cytoplasm by ribonucleoproteins.

What is mRNA degradation?

Messenger RNA (mRNA) degradation is the method by which cells reduce the lifespan of mRNA molecules after they've served their purpose in the cell. The degradation of mRNA molecules begins with the removal of the 5′ cap structure, which is followed by the removal of the poly(A) tail by exonucleases in the 3′ to 5′ direction of the mRNA molecule. After the removal of the cap and tail, the mRNA molecule is broken down into smaller pieces by endonucleases or exonucleases.

This leads to the production of shorter RNA fragments that are then degraded into single nucleotides by RNases in the cytoplasm. The process of mRNA degradation involves a variety of proteins, including ribonucleoproteins, which are complexes of RNA and proteins.

Ribonucleoproteins are thought to be involved in all aspects of mRNA metabolism, from transcription and splicing to mRNA degradation. They bind to specific sequences in the mRNA molecule and help to regulate its stability and translation.MRNA degradation can occur through a variety of mechanisms, including exonucleolytic degradation 5–>3' as well as 3–>5', endonucleolytic activity, and upf proteins. However, ribonucleoproteins are the main proteins involved in mRNA degradation in the cytoplasm. Therefore, option B is correct.

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Select all that apply.
Isoelectric focusing:
always involves separation in two dimensions.
makes use of the fact that proteins have fairly unique pI's.
makes use of a gel with a pH gradient.
allows smaller molecules to migrate through pores in the gel more quickly than larger ones, all other things being equal.
utilizes an electric field to cause proteins to migrate towards the positive pole.

Answers

All the given options are best suited for Isoelectric focusing. Isoelectric focusing is a technique used for protein separation.

Isoelectric focusing involves two-dimensional separation, utilizes a gel with a pH gradient, and takes advantage of the unique isoelectric points (pI) of proteins. It allows smaller molecules to migrate faster through the gel pores, and an electric field is applied to guide proteins towards the positive pole.

Isoelectric focusing is a powerful method for separating proteins based on their isoelectric points (pI), which is the pH at which a protein carries no net charge. This technique does not always involve separation in two dimensions.

It can be performed in a single dimension, where proteins are separated according to their pI values only, or in two dimensions, combining isoelectric focusing with another separation method, such as SDS-PAGE (sodium dodecyl sulfate polyacrylamide gel electrophoresis), to achieve higher resolution.

The process of isoelectric focusing takes advantage of a gel with a pH gradient. The gel is prepared with a pH gradient that spans from acidic to basic regions.

When an electric field is applied, proteins migrate through the gel towards their respective isoelectric points, where their net charge is zero. This migration occurs because proteins move towards the pole (either positive or negative) that corresponds to their net charge.

In isoelectric focusing, smaller molecules tend to migrate through the pores in the gel more quickly than larger ones, assuming all other factors are equal. This is due to the differences in size and charge density between the molecules.

Smaller proteins can pass through the gel pores more easily, whereas larger proteins experience more hindrance and migrate at a slower rate.To guide the proteins during the separation process, an electric field is utilized. The electric field is applied across the gel, with one end being positive and the other negative.

This field induces movement of the charged proteins towards the pole that matches their net charge. By applying an electric field, the proteins are driven towards the positive pole, allowing for efficient separation based on their isoelectric points.

In summary, isoelectric focusing is a technique that utilizes a gel with a pH gradient and an electric field to separate proteins based on their isoelectric points.

While it can be performed in one or two dimensions, it is commonly used in combination with other techniques for higher resolution separations. The method takes advantage of the fact that proteins have distinct isoelectric points, and smaller proteins migrate more quickly through the gel pores than larger proteins, assuming other conditions are equal.

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use blood glucose as an example, explain how major organ systems
in the body work together to co ordinate how the glucose reaches to
the cells? in details please.

Answers

Blood glucose is an example of the way major organ systems in the body work together to coordinate how glucose reaches the cells. Glucose is a major source of energy for the body's cells, and the endocrine system works to regulate its levels in the bloodstream.

The pancreas, liver, and muscles are the primary organs involved in regulating glucose levels. The pancreas, for example, produces the hormones insulin and glucagon, which work together to maintain proper glucose levels. When glucose levels in the bloodstream are high, insulin is released by the pancreas. Insulin signals the liver and muscles to take up glucose, which helps to lower the concentration of glucose in the bloodstream. Conversely, when glucose levels are low, glucagon is released by the pancreas, which signals the liver to release stored glucose into the bloodstream to increase glucose concentration in the bloodstream.

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illustrate the classifications of cytological methods in
detail.

Answers

Cytological methods are techniques that are used in the laboratory for observing the cells of the living organism. The process involves the study of the cells under the microscope.



This is a type of light microscopy, which is used for observing the cells that are fixed to the slide. It is used to observe cells that are not stained, or cells that are stained with a basic dye such as hematoxylin. her specimens. Light microscopy can be used to observe living cells and tissues, and it can be used to detect cellular abnormalities. 2. Electron Microscopy: Electron microscopy is a technique that uses a beam of electrons to magnify the image of cells and other specimens.

This method is used to observe the cells that are living, and it helps to differentiate the cells that have a high refractive index. The cells that are living are differentiated from those that are dead. 
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An enzyme has KM of 5.5 mM and Vmax of 10 mM/min. If [S] is 10 mm, which will increase the velocity more: a 10-fold decrease in Km or a 10-fold increase in Vmax? Explain why with examples.

Answers

A 10-fold decrease in Km will increase the velocity more compared to a 10-fold increase in Vmax in this scenario because it allows the enzyme to achieve its maximum velocity at lower substrate concentrations, making the enzyme more efficient in catalyzing the reaction.

To determine which change, a 10-fold decrease in Km or a 10-fold increase in Vmax, will increase the velocity (V) of the enzyme more, we need to understand their effects on the enzyme kinetics.

Km is a measure of the substrate concentration at which the enzyme achieves half of its maximum velocity. A lower Km value indicates higher affinity between the enzyme and the substrate, meaning the enzyme can reach its maximum velocity at lower substrate concentrations. On the other hand, Vmax represents the maximum velocity that the enzyme can achieve at saturating substrate concentrations.

In this case, when [S] is 10 mM, it is equal to the Km value. If we decrease the Km by 10-fold (to 0.55 mM), it means the enzyme can achieve half of its maximum velocity at a lower substrate concentration. Therefore, a 10-fold decrease in Km will significantly increase the velocity because the enzyme will reach its maximum velocity even at lower substrate concentrations.

In contrast, a 10-fold increase in Vmax (to 100 mM/min) would not have as significant an effect on the velocity at the given substrate concentration. The enzyme can already reach its maximum velocity (10 mM/min) at the current substrate concentration (10 mM), so further increasing the Vmax will not have a substantial impact on the velocity.

Therefore, a 10-fold decrease in Km will increase the velocity more compared to a 10-fold increase in Vmax in this scenario because it allows the enzyme to achieve its maximum velocity at lower substrate concentrations, making the enzyme more efficient in catalyzing the reaction.

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8. Compare between the pace maker action potential and the cardiomyocytes action potential.

Answers

Pacemaker action potential is generated in the sinoatrial node of the heart. The pacemaker action potential is different from that of cardiomyocytes action potential due to its spontaneous and rhythmic nature.

The cells that are involved in the pacemaker action potential are more automatic and have less of a stable membrane potential. Cardiomyocyte action potential, on the other hand, is produced by the cardiac muscle cell that is located in the heart's muscular tissue.

The cardiomyocytes action potential is slow compared to that of the pacemaker action potential. The cardiomyocytes action potential is only triggered when the cells are stimulated, unlike the pacemaker action potential that is spontaneous and does not require stimulation to occur.
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1. Use a family tree to calculate the percentage of a hereditary defect in offspring (controlled by recessive allele) : a. Normal father (AA) and Carrier mother (Aa) b. Carrier father (Aω) and Carrier mother (Aω) c. Abuormal father (aa) and Carrier mother (Aa)

Answers

The family tree is used to calculate the percentage of a hereditary defect in offspring, which is controlled by the recessive allele. The following are the different scenarios:

a. Normal father (AA) and Carrier mother (Aa): When a normal father (AA) and a carrier mother (Aa) produce offspring, there is a 50% chance that the offspring will be carriers (Aa) and a 50% chance that the offspring will be normal (AA). The probability of the offspring having the hereditary defect is 0%.

b. Carrier father (Aω) and Carrier mother (Aω): When both parents are carriers (Aω), there is a 25% chance that the offspring will be normal (AA), a 50% chance that the offspring will be carriers (Aω), and a 25% chance that the offspring will have the hereditary defect (aa).

c. Abnormal father (aa) and Carrier mother (Aa): When an abnormal father (aa) and a carrier mother (Aa) produce offspring, there is a 50% chance that the offspring will be carriers (Aa) and a 50% chance that the offspring will have the hereditary defect (aa).

Therefore, the percentage of a hereditary defect in offspring in the above-mentioned scenarios is 0%, 25%, and 50%, respectively.

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True or False?
In osmosis, solutes move across a membrane from areas of lower water concentration to areas of higher water concentration.

Answers

The statement is False: In osmosis, solutes move across a membrane from areas of higher water concentration to areas of lower water concentration.

Osmosis is a special kind of diffusion that involves the movement of water molecules through a semi-permeable membrane (like the cell membrane) from an area of high concentration of water to an area of low concentration of water. It occurs in the absence of any external pressure.In reverse osmosis, however, pressure is applied to the high solute concentration side to cause water to flow from a region of high solute concentration to a region of low solute concentration.

It is used to purify water and to separate solutes from a solvent in industrial and laboratory settings.

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8) Which gland sits atop each kidney? A) adrenal B) thymus C) pituitary D) pancreas artery lies on the boundary between the cortex and medulla of the kidney. 9) The A) lobar B) arcuate C) interlobar D

Answers

The gland that sits at the top of each kidney is called adrenal gland (option A). The arcuate artery lies on the boundary between the cortex and medulla of the kidney (option B).

What is the adrenal gland?

The adrenal gland is a complex endocrine glands found above each kidney.

It is saddled with the responsibility of secreting steroid hormones namely; adrenaline and noradrenaline.

These hormones help regulate the following:

heart rateblood pressuremetabolism

Also, the arcuate arteries of the kidney are renal circulation vessels and can be found between the cortex and the medulla of the renal kidney.

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Consider the following segment of DNA, which is part of a linear chromosome: LEFT 5'....TGACTGACAGTC....3' 3'....ACTGACTGTCAG....5' RIGHT During RNA transcription, this double-strand molecule is separated into two single strands from the right to the left and the RNA polymerase is also moving from the right to the left of the segment. Please select all the peptide sequence(s) that could be produced from the mRNA transcribed from this segment of DNA. (Hint: you need to use the genetic codon table to translate the determined mRNA sequence into peptide. Please be reminded that there are more than one reading frames.) ...-Leu-Ser-Val-... ...-Leu-Thr-Val-... ...-Thr-Val-Ser-... ...-Met-Asp-Cys-Gln-... ...-Asp-Cys-Gln-Ser-...

Answers

Therefore, all of the provided peptide sequences could potentially be produced from the mRNA transcribed from this segment of DNA.

The peptide sequence(s) that could be produced from the mRNA transcribed from this segment of DNA are:

...-Leu-Ser-Val-...

...-Leu-Thr-Val-...

...-Thr-Val-Ser-...

...-Met-Asp-Cys-Gln-...

...-Asp-Cys-Gln-Ser-...

To determine the mRNA sequence, we need to transcribe the DNA sequence from the 3' to 5' direction. In this case, the RNA polymerase is moving from the right to the left of the segment.

The complementary RNA strand would be 5'....UGACUGACAGUC....3'.

Using the genetic codon table, we can translate this mRNA sequence into the corresponding peptide sequence:

Leu-Ser-Val

Leu-Thr-Val

Thr-Val-Ser

Met-Asp-Cys-Gln

Asp-Cys-Gln-Ser

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Briefly explain how Meselson and Stahl’s experiment was able to
determine the currently accepted model of DNA replication.

Answers

Meselson and Stahl's experiment provided evidence for the currently accepted model of DNA replication.

Meselson and Stahl conducted an experiment in 1958 to determine the mechanism of DNA replication. They used isotopes of nitrogen, N-14 (light) and N-15 (heavy), to label the DNA of bacteria. The bacteria were first grown in a medium containing heavy nitrogen (N-15) and then transferred to a medium with light nitrogen (N-14).

After allowing the bacteria to replicate their DNA once, they extracted DNA samples at different time intervals and analyzed them using density gradient centrifugation.

According to the currently accepted model of DNA replication, known as the semi-conservative replication model, the replicated DNA consists of one parental strand and one newly synthesized strand.

In the Meselson and Stahl experiment, they observed that after one round of replication, the DNA samples formed a hybrid band with intermediate density, indicating that the DNA replication was not conservative (entirely new or entirely parental strands), but rather semi-conservative.

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Evaluate the pulmonary pressures provided, and determine what portion of the respiratory pressure cycle is represented: Atmospheric pressure = 760 mmHg Intrapulmonary pressure= 763 mmHg Intrapleural p

Answers

According to the information we can infer that intrapulmonary pressure = 763 mmHg represents forced inspiration.

What represents the intrapulmonary pressure?

Intrapulmonary pressure refers to the pressure inside the lungs. During forced inspiration, the diaphragm and other respiratory muscles contract more forcefully, causing an increase in lung volume.

This increased volume leads to a decrease in intrapulmonary pressure, creating a pressure gradient that allows air to flow into the lungs. The given value of 763 mmHg for intrapulmonary pressure is slightly higher than atmospheric pressure (760 mmHg), indicating that the pressure inside the lungs is slightly elevated during forced inspiration.

So, the provided intrapulmonary pressure of 763 mmHg represents forced inspiration.

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Please read all: (This is technically neuro-physiology so
hopefully putting this under anatomy and phys was the correct
idea)
Compare and contrast LTP, mGluR-LTD and
NMDAR-LTD.
INCLUDING:
– Inductio

Answers

LTP (Long-Term Potentiation), mGluR-LTD (Metabotropic Glutamate Receptor-Dependent Long-Term Depression), and NMDAR-LTD (N-Methyl-D-Aspartate Receptor-Dependent Long-Term Depression) are three forms of synaptic plasticity that contribute to the modulation of neural connections in the brain. Here's a comparison and contrast between these processes:

1. Induction:

- LTP: It is induced by strong and repetitive stimulation of the presynaptic neuron, leading to the activation of NMDA receptors and subsequent calcium influx.

- mGluR-LTD: It is induced by the activation of metabotropic glutamate receptors (mGluRs) located on the postsynaptic neuron.

- NMDAR-LTD: It is induced by low-frequency stimulation of the presynaptic neuron, resulting in the activation of NMDA receptors.

2. Mechanism:

- LTP: It involves the strengthening of synaptic connections through increased synaptic efficacy, primarily mediated by an increase in the number and activity of AMPA receptors.

- mGluR-LTD: It leads to the weakening of synaptic connections through the activation of intracellular signaling pathways that result in the removal of AMPA receptors from the postsynaptic membrane.

- NMDAR-LTD: It also leads to the weakening of synaptic connections, primarily by reducing the number and function of AMPA receptors.

3. Receptor Involvement:

- LTP: NMDA receptors play a crucial role in the induction of LTP, as their activation is necessary for calcium influx and subsequent signaling events.

- mGluR-LTD: Metabotropic glutamate receptors (mGluRs) are involved in the induction of mGluR-LTD, as their activation triggers intracellular cascades leading to synaptic depression.

- NMDAR-LTD: NMDA receptors are involved in the induction of NMDAR-LTD, although their activation under low-frequency stimulation leads to different signaling pathways compared to LTP.

4. Duration and Persistence:

- LTP: It is characterized by long-lasting potentiation of synaptic strength and can persist for hours to days.

- mGluR-LTD: It leads to long-term depression of synaptic strength and can persist for an extended period.

- NMDAR-LTD: It also results in long-term depression but can be reversible and transient.

In summary, LTP involves the strengthening of synaptic connections, mGluR-LTD and NMDAR-LTD involve the weakening of synaptic connections, and they differ in their induction mechanisms, receptor involvement, and persistence. These processes collectively contribute to synaptic plasticity and play a crucial role in learning, memory, and brain function.

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Question 21 Dense granules contain all of the following except: O Serotonin Calcium thrombospondin O ADP

Answers

Dense granules contain serotonin, calcium, and ADP, but do not contain thrombospondin. Dense granules are small organelles found in platelets.

Dense granules play a crucial role in hemostasis and blood clot formation. These granules contain various substances that are released upon platelet activation. Serotonin, calcium, and ADP are key components of dense granules, contributing to their physiological functions. Serotonin acts as a vasoconstrictor, helping to constrict blood vessels and reduce blood flow at the site of injury.

Calcium is involved in platelet activation and aggregation, facilitating the clotting process. ADP serves as a signaling molecule, promoting further platelet activation and aggregation. However, thrombospondin, a large glycoprotein, is not typically found in dense granules.

Thrombospondin is primarily located in the alpha granules of platelets, where it plays a role in platelet adhesion and wound healing. Therefore, the correct answer is option 3, thrombospondin.

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Auxin is a plant
nutrient required for cell wall synthesis.
nutrient required for hormone synthesis.
hormone that inhibits cell elongation.
hormone that stimulates cell elongation.

Answers

Auxin is a hormone that stimulates cell elongation. This hormone has the capacity to transport itself from the tip of a plant to the basal areas, and the action helps in the growth and development of the plant body. So, the correct option is: a hormone that stimulates cell elongation. Auxins are one of the most essential plant hormones that play crucial roles in plant growth, development, and environmental responses. These hormones are synthesized in the shoot and root apical meristem and transported from the apical region to the base to regulate diverse developmental processes, including cell elongation, division, differentiation, tissue patterning, and organogenesis.

Auxins are involved in almost all aspects of plant growth and development, such as root initiation, leaf development, shoot and root elongation, phototropism, apical dominance, gravitropism, fruit development, and senescence.

Apart from auxin, other plant hormones that regulate plant growth and development include gibberellins, cytokinins, abscisic acid, ethylene, and brassinosteroids.

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what are qualities common to plants pollinated at
night?

Answers

Plants that are pollinated at night typically have several qualities that help attract nocturnal pollinators which include: Strong Fragrances, Light-Colored Flowers, Large Flower Size, Production of Nectar, and Sturdy Structure.

1. Strong Fragrances: Flowers that release strong scents are easier for night-flying insects like moths and bats to detect. The fragrance often differs from that of day-blooming flowers, attracting the nocturnal pollinators that are more active at night.

2. Light-Colored Flowers: Insects that are active at night are usually attracted to lighter colors. Since most night-blooming plants are pollinated by nocturnal insects, they are more likely to be light-colored.

3. Large Flower Size: The size of the flowers is often larger and more complex to capture the attention of the night-flying animals.

4. Production of Nectar: Flowers that produce nectar provide an additional reward to their nocturnal pollinators. Since nectar is a good source of food for many animals, nocturnal pollinators are attracted to nectar-rich flowers.

5. Sturdy Structure: Night-blooming flowers have sturdy structures to withstand harsh winds. Wind resistance is important to ensure the flowers aren't damaged by the nightly winds.

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PLEASE ANSWER BOTH
1- All the following diseases may be associated with Claviceps purpurea, except one:
a. It produces aflatoxins.
b. It produces amatoxins.
c. It grows in the human respiratory tract.
d. It causes a specific skin rash.
e. It produces ergotism.
2 - Which one of the following characteristic signs of toxic shock syndrome is correct?
a. TSS is a self-limiting disease that resolves in a couple of days.
b. Only topical antibiotics are effective.
c. Symptoms are high temperature, vomiting, diarrhea, fainting, severe muscle aches, and peeling of the skin.
d. TSS is a fungal infection.
e. It is only occurring in children with weakened immune system.

Answers

It grows in the human respiratory tract. Claviceps purpurea is a parasitic fungus that attacks the ovaries of cereals and grasses, causing the disease known as ergot. Hence option C is correct.

It produces ergotism (a disease resulting from prolonged ingestion of ergot-contaminated grains) which can cause hallucinations, severe gastrointestinal upset, gangrene, and death. Aflatoxins and amatoxins are produced by fungi other than Claviceps purpurea. 2. The correct characteristic sign of toxic shock syndrome is c. Symptoms are high temperature, vomiting, diarrhea, fainting, severe muscle aches, and peeling of the skin.

Toxic shock syndrome (TSS) is a rare but life-threatening disease caused by toxins produced by bacteria such as Staphylococcus aureus and Streptococcus pyogenes. It can cause high fever, rash, low blood pressure, and organ failure. Treatment includes antibiotics and supportive care.

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Briefly, what is the difference between Metaphase I during Meiosis I and Metaphase Il during Meiosis II?

Answers

During meiosis, the chromosome number is reduced to half by two consecutive divisions, meiosis I and meiosis II. There are a few differences between metaphase I and metaphase II of meiosis.

The metaphase of meiosis is characterized by the alignment of chromosomes along the spindle equator, which is the area where they will split during anaphase. During metaphase I, chromosomes align in homologous pairs that are tetrads, each made up of four chromatids from two different homologous chromosomes. During metaphase II, chromosomes align individually along the spindle equator, each having only two chromatids. Metaphase I of meiosis is the phase in which the homologous chromosomes line up at the metaphase plate and are ready for segregation. Metaphase I is the longest phase of meiosis I.

During metaphase I, spindle fibers attach to the kinetochores of the homologous chromosomes and align them along the cell's equator. The spindle fibers are the organelles responsible for moving the chromosomes during mitosis and meiosis. They're responsible for moving the chromosomes to the poles of the cell in an orderly and organized manner. When the spindle fibers are pulling the chromosomes, they will also align themselves with each other at the metaphase plate. Each homologous pair of chromosomes is positioned at a point known as the metaphase plate during metaphase I, and each chromosome's two kinetochores are attached to spindle fibers from opposing poles.

In meiosis II, the spindle fibers attach to the sister chromatids of each chromosome, causing them to align along the cell's equator. When the spindle fibers are done pulling the chromosomes, they are separated into individual chromatids during the process of cytokinesis.The major difference between metaphase I and metaphase II is that in the former, homologous chromosomes line up as pairs, whereas in the latter, individual chromosomes line up. Chromosomes align at the metaphase plate during both phases. Meiosis II proceeds more quickly than meiosis I because the second division does not have an interphase stage. The whole process of meiosis results in four haploid daughter cells.

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Please redesign the shaft groove to accommodate that. Please use conservative estimates where needed 1. The Kuba king introduced a wooden sculpture called ______________.2. Kuba is in __________ (Congo, Egypt, Sudan).3. Hand-carved crocodile masks were from __________ (Nigeria, Ghana, Liberia)4. The earliest known stone art carving of giraffes is found in ____________5. _______________ is a process that separates metals from rocks dug out of the earth.6. Igbo Ukwu is an Albanian iron age archaeological site T/F7. The Yoruba civilization was in present-day Namibia T/F8. Meaning of indigenous _____________________ Collateral sprouting is an intercellular mechanism in responseto CNS injury. 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