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chapter 3
Which of these statements is false?
(Choose all that apply)
Group of answer choices
inclusions are membrane-bound organelles
Prokaryotes are haploid.
all cells have ribosomes
Cells that have plasmids often have hundreds of them within a single cell.
Plasmids are part of the chromosome.
Which of these statements are correct? (Choose all that apply)
Group of answer choices
The plasma membrane structure of most bacterial and eukaryotic cell types is a phospholipid bilayer
Only prokaryotic cells have a plasma membrane
some archaeal plasma membranes are lipid monolayers
The plasma membrane includes a diverse array of lipid and protein components

Answers

Answer 1

1. The false statements are inclusions are membrane-bound organelles and plasmids are part of the chromosome.

2. The correct statements are the plasma membrane structure of most bacterial and eukaryotic cell types is a phospholipid bilayer, some archaeal plasma membranes are lipid monolayers, and the plasma membrane includes a diverse array of lipid and protein components.

Thus, the correct answers are

1. A and E.

2. A, C, and D.

The statement, "Inclusions are membrane-bound organelles," is false because inclusions are not membrane-bound organelles. They are the accumulation of specific substances that are produced by the cell. In addition, the statement "Plasmids are part of the chromosome" is false because plasmids are not a part of the chromosome. Rather, they are small, circular, and double-stranded DNA molecules that are present in some cells.

The statement "The plasma membrane structure of most bacterial and eukaryotic cell types is a phospholipid bilayer" is correct because the plasma membrane structure of most bacterial and eukaryotic cell types is a phospholipid bilayer. Some archaeal membranes are lipid monolayers instead of bilayers.

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Related Questions

Please help me i am desperate, I don’t really understand this….

Answers

Answer: I got you

Explanation: Photosynthesis makes glucose which is used in cellular respiration for making ATP. Then glucose is then transformed back into carbon dioxide, which is used in photosynthesis. It helps cells to release and store energy. It maintains the atmospheric balance, of carbon dioxide and oxygen.

Stella has cystic fibrosis which is a genetic disorder that affects the digestive system and the lungs. It causes persistent lung infections and difficulty breathing. Individuals with cystic fibrosis have a mutation in the CFTR (transmembrane conductance regulator) gene which causes the CFTF protein to become dysfunctional.
Stella’s parents do not have this genetic disease. However, Stella’s paternal grandfather and maternal grandmother had cystic fibrosis. Her remaining grandparents (paternal grandmother and maternal grandfather) do not have the disease.
a. Determine if the type of inheritance is seen for this disease.
b. Draw the pedigree of this condition in Stella’s family.
c. What are the probable genotypes for Stella, her father, and her mother?
d. If Stella marries a man that is heterozygous for the CFTR mutation (Aa), what proportion of their children could have Cystic fibrosis? Determine using a Punnett square.

Answers

The type of inheritance seen for this cystic fibrosis disease is autosomal recessive inheritance. This means that an individual must inherit two copies of the mutated gene (one from each parent) in order to develop the disease.

The Answer to Question B to D


B. The pedigree of this condition in Stella's family would look like this:

             Paternal Grandfather (aa)   Paternal Grandmother (AA)

                              |                                                 |

                               ----------------------------------------------

                                                       |

                                                Father (Aa)

                                                        |

                                       ---------------------------------

                                     |                                       |

                    Mother (Aa)                                    Stella (aa)

                                      |                                      |

        Maternal Grandfather (AA)             Maternal Grandmother (aa)

C. The probable genotypes for Stella, her father, and her mother are as follows:

Stella: aa (affected with cystic fibrosis)Father: Aa (carrier of the CFTR mutation)Mother: Aa (carrier of the CFTR mutation)

D. If Stella marries a man that is heterozygous for the CFTR mutation (Aa), the proportion of their children that could have cystic fibrosis is 50%. This can be determined using a Punnett square:

        A              a

     ----------------------

  a |   Aa    |    aa  |

     ----------------------

  a |  Aa    |     aa  |

     -----------------------

The Punnett square shows that there is a 50% chance of their children inheriting the aa genotype and developing cystic fibrosis.

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Who is more important first author or last author?

Answers

Both first and last authors are important in scientific publications, but their roles differ.

The first author is typically the researcher who made the greatest contribution to the research project, and who wrote the first draft of the manuscript. As such, they deserve recognition for their hard work and dedication.

On the other hand, the last author is often the principal investigator, who provided guidance and oversight throughout the research process. They may also have secured funding for the project, and are often responsible for the overall direction of the research.

In summary, both first and last authors play critical roles in scientific publications, and their contributions should be acknowledged and valued equally.

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what are the 7 life processes​

Answers

The seven life processes, also known as the characteristics of life, are the fundamental properties that distinguish living organisms from non-living things.

Life processes explained

The seven life processes, also known as the characteristics of life, are the fundamental properties that distinguish living organisms from non-living things. The seven life processes are:

Nutrition: the process by which organisms take in and use nutrients for energy, growth, and repair.

Respiration: the process by which organisms convert nutrients into energy through chemical reactions, such as cellular respiration.

Movement: the ability of organisms to move themselves or their internal parts.

Sensitivity: the ability of organisms to detect and respond to changes in their environment, such as light, temperature, or sound.

Growth: the process by which organisms increase in size or number of cells.

Reproduction: the ability of organisms to produce offspring, either sexually or asexually.

Excretion: the process by which organisms eliminate waste products, such as carbon dioxide or urine.

Therefore, all living organisms exhibit these seven life processes to some degree, although the specific details of each process may vary depending on the organism and its environment.

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Movement
Respiration
Sensitivity

Growth
Reproduction
Excretion
Nutrition

REMEMBER MRS GREN

students are modeling transcription during the process of protein synthesis. which answer choice correctly describes the result of the DNA sequence ACGCAA being transcribed?

Answers

Answer:

the mRNA strand is complementary to the DNA template strand; however, uracil instead of thymine is paired with adenine

Explanation:

Hope this helps :)

The mRNA strand is complementary to the DNA template strand; however, uracil instead of thymine is paired with adenine

What is thymine?

Thymine is one of the four nucleotide bases that make up DNA, the other three being adenine, guanine, and cytosine. It is a pyrimidine base that pairs with adenine through hydrogen bonding in the DNA double helix structure. Thymine is specifically bonded to adenine through two hydrogen bonds. The sequence of these four nucleotide bases determines the genetic information that is encoded in DNA. Thymine is also important in the process of DNA replication, as it provides a template for the synthesis of a complementary DNA strand during cell division. In RNA, uracil replaces thymine as a complementary base to adenine.

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The________ cycle is also known as carbon________ because the
inorganic_______ dioxide is fixed, or attached, to an organic
molecule.

Answers

The Calvin cycle is also known as carbon fixation because the inorganic carbon dioxide is fixed, or attached, to an organic molecule.

What is the calvin cycle?

The Calvin cycle is a set of light-independent redox reactions that occur during photosynthesis and carbon fixation to convert carbon dioxide into glucose. It is also known as the photosynthetic carbon reduction (PCR) cycle or dark reaction. During this cycle, carbon dioxide is reduced and combined with other molecules to form sugar molecules.

The cycle is composed of three main stages: carbon fixation, reduction, and regeneration of the enzyme RuBisCO. In the first stage, carbon dioxide is converted into an intermediate compound, which is then reduced and combined with other molecules, such as phosphoglycerate, to form sugar molecules. In the second stage, the sugar molecules are then used to regenerate the enzyme RuBisCO, which starts the cycle again. The Calvin cycle is important for plants, as it allows them to convert sunlight and carbon dioxide into energy-rich sugar molecules.

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what is the smallest distance to points can be separated, and still
resolved, using light microscopy?
a. 16-20nm
b. 0.16-0.2 um
c. 1600 nm- 2um
d. 160-200 um

Answers

The smallest distance to points that can be resolved using light microscopy is 160-200 nm.

This limit is determined by the diffraction of light waves, which causes them to spread out and interfere with one another. This diffraction limit is also known as the Abbe limit and is given by the equation d=0.61λ/NA, where d is the smallest distance that can be resolved, λ is the wavelength of light used, and NA is the numerical aperture of the lens. For visible light, which has a wavelength of about 500 nm, the resolution limit is about 200 nm. However, with the use of advanced techniques such as super-resolution microscopy, it is now possible to resolve distances as small as 10-20 nm.

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Which of the following explanations describes a sound wave?


disturbance travelling generally through air as a longitudinal wave

disturbance travelling generally through liquids as a transverse wave

disturbance travelling generally through plasma as a longitudinal wave

Answers

The explanation that describes a sound wave is: "disturbance travelling generally through air as a longitudinal wave".

option A.

What is sound wave?

A sound wave is a type of mechanical wave that is created by the vibration or disturbance of matter, such as air molecules. When an object vibrates, it causes the particles around it to vibrate, creating a series of compressions and rarefactions in the medium through which the wave travels.

Thus, Sound waves are longitudinal waves that travel through various media, including gases (such as air), liquids, and solids. However, sound waves are not transverse waves and do not generally travel through plasma, which is a state of matter consisting of ionized gas.

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Does this have a good snzyme level when plotting the rate of
formation vs. the substrate con. ?

Answers

The good enzyme level would result in an increase in the rate of formation as the substrate concentration increases, but there is a limit to how much the rate can increase.

When plotting the rate of formation vs. the substrate con, a good enzyme level would result in an increase in the rate of formation as the substrate concentration increases. This is because enzymes work by catalyzing reactions, which means they increase the rate of the reaction without being consumed by the reaction.

At low substrate concentrations, there may not be enough substrate for the enzyme to work on, so the rate of formation may be low. As the substrate concentration increases, there will be more substrate available for the enzyme to work on, and the rate of formation should increase as well.

However, it's important to note that there is a limit to how much the rate of formation can increase with increasing substrate concentration. At a certain point, the enzyme may become saturated with substrate, which means all of the active sites on the enzyme are occupied and adding more substrate won't increase the rate of formation any further.

In summary, a good enzyme level would result in an increase in the rate of formation as the substrate concentration increases, but there is a limit to how much the rate can increase.

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What are 10 ways Plasmodium falciparum avoids being
killed inside humans ( do not include genetic factors like sickle
Cells)?

Answers

The 10 ways Plasmodium falciparum avoids being killed inside humans includes:

Adhesion to endothelial cellsAlteration of surface proteinsFormation of rosettesFormation of knobsResistance to complementSuppression of the immune responsePrevention of antigen presentationProduction of toxinsAlteration of cytokine productionDevelopment of drug resistance

Plasmodium falciparum is a parasite that is responsible for causing the deadliest form of malaria. The parasite has developed various mechanisms to avoid being killed inside humans.

Some of the ways in which Plasmodium falciparum avoids being killed inside humans are:

1. Adhesion to endothelial cells

Plasmodium falciparum adheres to endothelial cells, which allows it to avoid being swept away by blood flow. This adhesion also allows the parasite to enter the organs and tissues, where it can continue to multiply.

2. Alteration of surface proteins

The parasite alters the surface proteins of infected red blood cells, which makes it difficult for the immune system to recognize and destroy them.

3. Formation of rosettes

Plasmodium falciparum-infected red blood cells can form rosettes, which are clusters of red blood cells around an infected cell. This formation makes it difficult for the immune system to target infected cells.

4. Formation of knobs

The parasite can form knobs on the surface of infected red blood cells, which makes them stick to the walls of blood vessels. This allows the parasite to avoid being carried away by blood flow.

5. Resistance to complement

The parasite has developed resistance to the complement system, which is part of the immune system that attacks foreign invaders.

6. Suppression of the immune response

The parasite can suppress the immune response of the host, which allows it to evade detection and destruction by the immune system.

7. Prevention of antigen presentation

The parasite can prevent the presentation of antigens, which are molecules that are recognized by the immune system. This prevents the immune system from recognizing and destroying infected cells.

8. Production of toxins

The parasite can produce toxins that damage red blood cells and cause symptoms such as fever and anemia. These symptoms can also impair the immune response of the host.

9. Alteration of cytokine production

The parasite can alter the production of cytokines, which are molecules that regulate the immune response. This alteration can prevent the immune system from recognizing and destroying infected cells.

10. Development of drug resistance

The parasite has developed resistance to many antimalarial drugs, which makes it difficult to treat and control the disease.

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Microbiologists are fond of saying that only a tiny minority of bacteria causes disease. Are there reasons for thinking this might not be true?

Answers

It is true that only a tiny minority of bacteria cause disease but there are reasons to believe that this may not always be the case. Most bacteria are harmless or even beneficial to humans.

There are some reasons for thinking that this might not always be the case:

One reason is that there may be undiscovered or unidentified bacteria that can cause disease. With new technology and research, it is possible that more bacteria will be found to be pathogenic. Additionally, bacteria can mutate and develop new ways to cause disease, potentially increasing the number of disease-causing bacteria.Another reason is that bacteria that are normally harmless can become pathogenic under certain conditions. For example, if a person's immune system is compromised, bacteria that normally live on or in the body without causing harm can cause infections and illness.

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What are all of the different types of pathology found in people
infected with: Acanthamoeba but not found in people infected with
Giardia?

Answers

Some of the different types of pathology that can be found in people infected with Acanthamoeba but not in people infected with Giardia includes: Neurological pathology, Ocular pathology, and Skin pathology.

Pathology refers to the study of disease, especially the structural and functional changes caused by disease. It is a vast field and covers many different types of diseases and conditions. The following are some of the different types of pathology that can be found in people infected with Acanthamoeba but not in people infected with Giardia:

Neurological pathology: Acanthamoeba infections can cause a range of neurological symptoms, including headache, seizures, confusion, and coma. These symptoms are not typically seen in Giardia infections.

Ocular pathology: Acanthamoeba infections can also cause a range of eye symptoms, including redness, swelling, pain, and blurred vision. These symptoms are not typically seen in Giardia infections.

Skin pathology: Acanthamoeba infections can cause a range of skin symptoms, including rashes, itching, and ulcerations. These symptoms are not typically seen in Giardia infections.

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Three novel ideas or hypothesis specifically related to the role of
exosomes involved in insulin receptor trafficking and how this can
be targeted in disease state

Answers

these novel ideas and hypotheses suggest that exosomes could be a promising therapeutic target for diseases involving insulin receptor trafficking, such as type 2 diabetes and some forms of cancer.

1. One novel idea is that exosomes could be used as a delivery system for insulin receptors to cells that are insulin resistant, such as those found in type 2 diabetes. This could potentially help to restore insulin sensitivity and improve glucose control in these patients.
2. Another hypothesis is that exosomes could be used to selectively remove insulin receptors from cells that are overexpressing them, such as in some forms of cancer. This could potentially help to slow down or halt the growth of these cancer cells.
3. A third idea is that exosomes could be used to deliver small interfering RNA (siRNA) molecules to cells that are involved in insulin receptor trafficking, in order to knock down the expression of specific genes that are involved in this process. This could potentially help to prevent or reverse insulin resistance in disease states such as type 2 diabetes.
Further research is needed to explore these ideas and to determine their feasibility and effectiveness in treating these diseases.

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. A new prokaryote Leioid sp. has evolved a Complex IV that is different from other organisms' version of Complex IV. Complex IV can bind to and be reduced in reactions with both CytC AND with Complex I. Complex IV in Leioid sp. still binds to and reduces O. All other respiration molecules are the same as other prokaryotes. A diagram of the typical ETC is on the Reference page 7.
a. (3) Is the active site within Complex IV that binds and reacts with CytC the same site that binds and reacts with Complex I?
Yes
No
b. (4) How much ATP per glucose will be made by Leioid sp, cells, compared to other aerobic prokaryotes? Assume that Leioid's Complex IV pumps the same number of protons as Complex IV in other organisms when it is reduced with 2 e- a little bit less. -half as much
more
the same amount
Explain your choice in 1-2 sentences. Be as specific as possible!
much less

Answers

3. The active site within Complex IV that binds and reacts with CytC is the same site that binds and reacts with Complex I.b.

4. Much less ATP per glucose will be made by Leioid sp, cells, compared to other aerobic prokaryotes because Leioid's Complex IV pumps half as many protons as Complex IV in other organisms when it is reduced with 2 e-.

Thus, the correct answers are

3. Yes, it is.

4. Much less.

What is Complex IV?

Complex IV is the cytochrome c oxidase enzyme complex. It is the final protein complex in the electron transport chain that accepts electrons from cytochrome c and passes them to oxygen, resulting in the production of water. The movement of electrons through Complex IV enables the translocation of protons (H+) from the cytoplasmic side of the mitochondrial membrane to the intermembrane space.

What is ATP?

ATP (adenosine triphosphate) is a molecule that stores and supplies energy for biological processes in the cells of organisms. During cellular respiration, ATP is produced in the mitochondria and serves as the primary energy currency for cells. ATP releases energy when the bond between the second and third phosphate groups is broken, resulting in the formation of ADP (adenosine diphosphate) and a free phosphate group.

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No Service 12:50 AM Expert Q&A Done need help with these questions 1. The polytrichum spimoss belongs to the phylum---? 2. The ---- is the dominant life form of the mosses and the is the smaller portion of the life cycle
3. The ---- of the Merchantia sp is the diploid stage in the life cycle. 4. The asexual reproduction form of the marchantia sp is called ---- 5. Marchantia sp belongs to the phylum - 6. What type of cell division would occur within the capsule of the polytrichum sp. 7. Which of the following is the location where zygotes and embryos form in a moss? 8. The antheridia produces ------ whereas archegonia houses the... 9. which of the following characteristics doesnt not describe monocot plants a. has one cotyledon b. have parallel venation c. generally have a fibrous roots system d. sporophyte generation is dominant e. has netlike venation 10. Leaves on a monocots and eudicot are attach to the stem at the 11. label this diagram 76707AUS Label diagram 2 .

Answers

1. The polytrichum spimoss belongs to the phylum Bryophyta.
2. The gametophyte is the dominant life form of the mosses, and the sporophyte is the smaller portion of the life cycle.
3. The sporophyte of Merchantia sp is the diploid stage in the life cycle.
4. The asexual reproduction form of Marchantia sp. is called gemmae.
5. Marchantia sp belongs to the phylum Hepatophyta
6. Meiosis would occur within the capsule of Polytrichum sp.
7. Zygotes and embryos form within the archegonia of a moss.
8. The antheridia produces sperm whereas archegonia house the egg.
9. The characteristic that does not describe monocot plants is netlike venation.
10. Leaves on monocots are attached directly to the stem and eudicots are attached to the stem via a short stalk, the petiole.

Plants characteristicsBryophyta is a phylum of non-vascular land plants that includes mosses, liverworts, and hornworts.The gametophyte of mosses is the dominant, photosynthetic stage of their life cycle, which produces gametes for sexual reproduction.The sporophyte of Marchantia is the diploid stage of its life cycle that produces spores for asexual reproduction.Gemmae are small, asexually produced structures that are found in liverworts, such as Marchantia, and can grow into new plants.Hepatophyte is another term for liverwort, which is a type of non-vascular land plant.The archegonia of mosses are structures that produce and house the eggs for sexual reproduction.Monocot plants are a type of flowering plant that have a single cotyledon (embryonic leaf), parallel venation in their leaves, and a fibrous root system.The leaves of dicot plants usually have a branched vein system with a prominent midrib and netlike veins that spread out from it.

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Is it required to test the raw materials for microorganisms before the batch manufacturing of terminal sterilization parenteral product? Select one or more: a. Yes, because the number of microorgonism in the raw materiats effect the sterization method b. Yes. because the raw materials must be sterilized betore manufocturing c. No, becouse the number of microorganism in the raw materials don't effect the sterilization method d. No, becquse the product will be sterized in the final packaging

Answers

Yes, it is required to test the raw materials for microorganisms before the batch manufacturing of terminal sterilization parenteral product.

This is because the number of microorganisms in the raw materials can affect the sterilization method and the raw materials must be sterilized before manufacturing. Therefore, the correct answers are options A and B.

Option A is correct because the number of microorganisms in the raw materials can affect the sterilization method. If there are too many microorganisms present, the sterilization method may not be effective in killing all of them.

Therefore, it is important to test the raw materials for microorganisms before manufacturing to ensure that the sterilization method will be effective.

Option B is also correct because the raw materials must be sterilized before manufacturing. If the raw materials are not sterilized, they can introduce microorganisms into the product during manufacturing, which can lead to contamination of the final product.

Options C and D are incorrect because the number of microorganisms in the raw materials can affect the sterilization method and the product will not necessarily be sterilized in the final packaging.

Therefore, it is important to test the raw materials for microorganisms before manufacturing to ensure the safety and effectiveness of the final product.

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1. Answer the following characteristics for Ascomycota
Fungi.
A. Color
B. Texture
C. Form
D. Size
E. Starch storage (where)

Answers

Ascomycota Fungi has

Color: Generally white, gray, black, or brown Texture: Usually velvety, powdery, or fuzzy Form: Varied, includes cup-shaped, bracket-like, or disk-like Size: Can range from microscopic to several centimeters Starch storage: Generally in their cells

The Ascomycota Fungi has the following characteristics:


A. Color: Depending on the type of Ascomycota Fungi, they can range in colors from white to yellow, purple, pink, and brown.
B. Texture: Ascomycota Fungi typically have a smooth, powdery texture.
C. Form: Ascomycota Fungi have a filamentous, or tubular, form.
D. Size: Depending on the type of Ascomycota Fungi, their size can range from a few micrometers up to several centimeters.
E. Starch Storage (where): Ascomycota Fungi store their starch in the cytoplasm and not in vacuoles. They store their nutrients in the form of starch.

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The molecule formed by joining 2 fatty acids with glycerol is a
dipeptide
diglyceride
disaccharide
diglycoside
diazepam
Clear my choice

Answers

The molecule formed by joining 2 fatty acids with glycerol is a diglyceride.

What is glycerol?

Glycerol, also known as glycerine or glycerin, is a simple polyol compound. It is a colorless, odorless, viscous liquid that is used in many applications. Glycerol is made from the enzymatic or chemical hydrolysis of fats and oils, which is a significant byproduct of the biodiesel industry.Glycerol is utilized in a variety of cosmetic products such as lotions, creams, shampoos, and soaps. It is also used in the pharmaceutical industry as a medication or excipient.

Glycerol is used in the food industry as a humectant, sweetener, and preservative.The molecule formed by joining two fatty acids with glycerol is known as a diglyceride, also known as a diacylglycerol (DAG). Diglycerides are fats that include two fatty acids linked to a glycerol molecule.

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It is typically sufficient to rupture cells when the solute concentration is reduced from 0.15M to 0.001M. Calculate what transmembrane pressure this would result in.
Use that to access if the red blood cells would break. Yes or No?
Compare to the transmembrane pressure when cells are in normal saline solution (0.91%NaCl) -> 0.156M(change unit to osM)
Basically
Calculate the transmembrane pressure when the solute concentration is reduced from 0.15M to 0.001M
Determine if that transmembrane pressure would result in the breakage of red blood cells
Calculate the transmembrane pressure when cells are in a normal saline solution and compare

Answers

In order to answer this question, we first need to calculate the transmembrane pressure when the solute concentration is reduced from 0.15M to 0.001M. This can be done using the following equation:

Transmembrane Pressure (TMP) = Solute Concentration * R * T

Where R is the ideal gas constant (0.0821 liter-atmosphere/mole-Kelvin) and T is the temperature in Kelvin (273.15 K).

For a 0.15M solute concentration, the transmembrane pressure (TMP) is:

TMP = 0.15 * 0.0821 * 273.15 = 4.1585 atm.

For a 0.001M solute concentration, the transmembrane pressure (TMP) is:

TMP = 0.001 * 0.0821 * 273.15 = 0.2751 atm.

The transmembrane pressure when the solute concentration is reduced from 0.15M to 0.001M is therefore 4.1585 atm. When cells are in a normal saline solution, the transmembrane pressure is 0.156M (converted to osM). This is equivalent to 0.1297 atm. Therefore, the transmembrane pressure resulting from the reduction in solute concentration is much higher than the pressure in normal saline solution, and the red blood cells would likely break. So, the answer to the question is yes.


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1. The chances of any one individual being born with a beneficial mutation is very low. How then is it possible for mutations to play such a key role in evolution? 2. How might it be possible for a neutral mutation to play an important role in the evolution of a species? 3. Explain why harmful mutations do not accumulate over time and cause harm to populations. 4. Use the evolution of antibiotic resistance to show a mutation that is advantageous for one species can be harmful for another.5. Most wolves look quite similar. Use the domestication of dogs to illustrate the genetic diversity that is contained within the wolf population. . All domestications of plants and animals begin with humans selecting a wild species living nearby. Use the internet and other sources to research where each of the following domesticated species originated (2 plants & 1 animal):

Answers

When using two different restriction enzymes to double digest both vector and insert DNA, the sticky ends generated are no longer compatible with each other.

This means that ligation can only occur between the vector and insert, preventing self-ligation of either DNA molecule.The resulting products after ligation would be the desired vector+gene ligation, as well as possible vector-vector or gene-gene ligation products.

However, the likelihood of these products forming is reduced compared to a single enzyme digest due to the non-complementary nature of the sticky ends.

Overall, double digestion with two different restriction enzymes provides a more specific and controlled way to insert the gene into the vector, reducing the likelihood of undesirable ligation products.

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students go for a vision test. 20% of them
are found to be near-sighted. How many
students are near-sighted?​

Answers

If 20% of students are found to be near-sighted, then this means that out of every 100 students, 20 of them are near-sighted. To find the number of near-sighted students when the total number of students is unknown, we must calculate the fraction of students that are near-sighted.

For example, if there are 80 students, then 20% of the students would be 16 near-sighted students. Similarly, if there are 200 students, then 20% of the students would be 40 near-sighted students.

Therefore, the exact number of near-sighted students depends on the total number of students who went for a vision test. In conclusion, 20% of the students who went for a vision test are near-sighted, and the exact number of near-sighted students can be determined by calculating the fraction of students with near-sightedness from the total number of students who went for the vision test.

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Carbon chains are principal features of BOTH carbohydrates and lipids. What is the primary difference between these two types of biomolecules?
A.
Lipids always have longer carbon chains that carbohydrates.
B.
Carbohydrates carry hydroxyl groups on their carbon backbone. .C
Carbohydrates cannot form rings as lipids can.
D.
Lipids provide energy, but carbohydrates do not.

Answers

The main difference is that lipids have longer carbon chains than carbohydrates, as option A shows.

Why are the other options incorrect?Both lipids and carbohydrates can have hydroxyls attached to their carbon skeletons.Carbohydrates can ring and lipids cannot.Both lipids and carbohydrates provide energy to the body.

Lipids are molecules that store a lot of energy and are the main producers of fats and oil, saving energy for times of great need. Carbohydrates, on the other hand, are molecules that store energy to be used in the short term, during day-to-day life.

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These four trucks are identical. Each box loaded on the trucks has the same mass. Choose the truck that has the greatest force of gravity acting on it.

Answers

The green truck has the greatest force of gravity acting on it because it has the highest number of boxes loaded on it.

What is the relationship between mass and the force of gravity acting on an object?

The relationship between mass and the force of gravity acting on an object is given in the formula below:

Force of gravity = mass * acceleration due to gravity

The acceleration due to gravity is a constant, hence the force of gravity acting on an object increases with and increase in the mass of the object.

The mass of the boxes on the green truck is greatest since it has the highest number of boxes on it.

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You are setting up a chromatin digest with Micrococcal nuclease, and add a large amount of MNase to your tube. When you run your samples on the DNA gel, about how large would you expect your band(s) to be, and why?

Answers

When running a chromatin digest with Micrococcal nuclease (MNase), the resulting band(s) on the DNA gel should be approximately 200-400 bp in size. This is because MNase has the ability to make precise cuts between two adjacent nucleosomes, resulting in fragments of 200-400 bp.

MNase is an enzyme that breaks down DNA into smaller fragments.

The more MNase you add to your tube, the smaller the DNA fragments will be, and therefore the smaller the band(s) will be on the DNA gel.

The size of the band(s) on the DNA gel is directly related to the size of the DNA fragments, so a large amount of MNase will result in small band(s).

Hence, your band(s) are expected to be small in size.

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1. how does Forest pregnancation, human encroachment, pathogenic exchange, and human healthcare policies positively or negatively affect biodiversity?
2. how does biodiversity loss play a role in global disease

Answers

Forest fragmentation, human encroachment, pathogenic exchange, and human healthcare policies can all affect biodiversity in different ways.

Biodiversity loss can play a role in global disease by reducing the number of species that can serve as natural reservoirs for pathogens, increasing the likelihood that diseases will spread to human populations.

The Explanation to Each Answer

Forest fragmentation can negatively affect biodiversity by reducing the amount of habitat available for species and isolating populations, which can lead to reduced genetic diversity and increased risk of extinction.

Human encroachment can also negatively affect biodiversity by destroying or altering natural habitats and introducing invasive species. Pathogenic exchange can lead to the spread of diseases among species, which can negatively affect biodiversity by causing population declines or extinctions.

Human healthcare policies can positively affect biodiversity by promoting conservation and protecting natural habitats, but they can also negatively affect biodiversity if they do not adequately address the impacts of human activities on the environment.

Biodiversity loss can also lead to changes in ecosystems that can create conditions that are more conducive to the spread of diseases, such as reduced habitat quality and increased contact between humans and wildlife. Additionally, biodiversity loss can reduce the availability of natural resources.

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Suppose that the DNA sequence: 5’-TAGTACTCGTAC-3’ was replicated by a DNA polymerase. Record the replication product’s sequence in the space provided - Please use ALL CAPS, with no spaces, in proper orientation (5’- to -3’): 5'-

Answers

The replication product would have the sequence 5'-ATGATGAGCATG-3', with each base complementary to the original strand.

About DNA replication

The replication product of the DNA sequence 5'-TAGTACTCGTAC-3' would be 5'-ATGATGAGCATG-3'. This is because DNA replication occurs by the complementary base pairing of the original DNA strand with new nucleotides. The base pairing rules are as follows:

A pairs with T, and C pairs with G.

Therefore, the replication product would have the sequence 5'-ATGATGAGCATG-3', with each base complementary to the original strand.

It is important to note that the replication product is also in the 5'-to-3' orientation, as this is the direction in which DNA replication occurs.

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11. If ATP is not being produced, why would this cause an increase in respiration rate? (hint: think about how respiration is related to ATP production)
12. Describe the gas exchange surface of the respiratory system (what does it look like).
13. Now explain why this structure (previous question) helps it do its function of gas exchange.
14. Thinking about all the information presented in this test, why does DNP cause weight loss?
A) more glucose is broken down and more CO2 is exhaled
B) more glucose is excreted as feces
C) more glucose is burned and released as heat

Answers

11. If ATP is not being produced, it would lead to an increase in respiration rate because respiration is the process by which cells produce ATP. When ATP production decreases, the body tries to compensate by increasing respiration to provide more oxygen and glucose to the cells for energy production.

12. The gas exchange surface of the respiratory system is made up of millions of small, thin-walled sacs called alveoli. These sacs are surrounded by networks of blood vessels called capillaries, which allow for the exchange of gases between the lungs and the blood.

13. The structure of the gas exchange surface of the respiratory system, with its thin walls and extensive network of capillaries, facilitates the diffusion of gases between the air in the alveoli and the blood in the capillaries. Oxygen from the air diffuses through the walls of the alveoli and into the capillaries, where it binds to hemoglobin and is transported to the body's tissues. Similarly, carbon dioxide produced by the body's tissues diffuses from the capillaries into the alveoli and is then exhaled.

14. DNP (2,4-dinitrophenol) causes weight loss because it uncouples oxidative phosphorylation in the mitochondria, disrupting the production of ATP. As a result, the body's metabolism is forced to work harder to produce more ATP, leading to an increase in energy expenditure and a higher rate of respiration. This increased metabolic rate causes the body to burn more calories, resulting in weight loss. Option C, more glucose is burned and released as heat, is the correct answer.

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_______ bilayers are really good at setting up __________
gradients, meaning keeping the__________ higher on one side than
the__________ (note: the second and third blanks are the same
word.)

Answers

The words that complete the sentence are lipid and concentrations. Therefore, "Lipid bilayers are really good at setting up concentration gradients, meaning keeping the concentration higher on one side than the other."

What is a lipid bilayer?

A lipid bilayer is a thin membrane made of two layers of lipid molecules that act as a barrier between the inside and outside of a cell. Lipid bilayers are important for maintaining concentration gradients, which are differences in the concentration of a substance across a membrane.

The lipid bilayer acts as a barrier to prevent substances from moving freely across the membrane, allowing the cell to maintain a higher concentration of a substance on one side of the membrane than the other. This is important for many cellular processes, including the generation of energy and the transport of molecules across the membrane.

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Write a comprehensive note on agricultural systems

Answers

Agricultural systems refer to the methods and practices used in the production of food and fiber.

The location, climate, and resources that are available, as well as the political, social, and cultural background, all have a significant impact on these systems. Following are a few prevalent agricultural system types:

Small-scale farmers who cultivate crops and raise livestock for their own consumption are said to conduct subsistence agriculture. It is common in poor nations where farmers depend on outdated practices and scant resources to feed their family.

Large-scale farmers that engage in commercial agriculture produce crops and cattle for local or international markets. Modern inputs and technology are used in commercial agriculture to boost yields and profitability, including machinery, insecticides, and fertilizers.

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Noncompetitive Receptor AntagonistsTraditional noncompetitive antagonists bind to the active site. Binding to the active site of the receptor can be --- or with very high ---, such that binding is --- ---. Such irreversible binding can not be '--- ---' or outcompeted by the ---, therefore noncompetitive. Noncompetitive antagonists --- the --- --- or --- (---). In some receptor systems, noncompetitive antagonists can also produce a --- in --- (--- ---)

Answers

Traditional noncompetitive antagonists bind to the active site. Binding to the active site of the receptor can be irreversible or with very high affinity, such that binding is tightly coupled. Such irreversible binding can not be 'outperformed' or outcompeted by the agonists, therefore noncompetitive. Noncompetitive antagonists inhibit the agonist-induced signalling or desensitise (down-regulate) the receptor. In some receptor systems, noncompetitive antagonists can also produce a constitutive activation in receptor functionality (constitutive activation).

А receptor аntаgonist is а type of receptor ligаnd or drug thаt blocks or dаmpens а biologicаl response by binding to аnd blocking а receptor rаther thаn аctivаting it like аn аgonist. Аntаgonist drugs interfere in the nаturаl operаtion of receptor proteins.

They аre sometimes cаlled blockers; exаmples include аlphа blockers, betа blockers, аnd cаlcium chаnnel blockers. In phаrmаcology, аntаgonists hаve аffinity but no efficаcy for their cognаte receptors, аnd binding will disrupt the interаction аnd inhibit the function of аn аgonist or inverse аgonist аt receptors.

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