Name three differences (as discussed in class and listed in your lab manual) between a plant and animal cell. Be able to recognize these on the plant model.

Answers

Answer 1

The three main differences between a plant cell and an animal cell, as discussed in class and listed in the lab manual, are:

A plant cell has a cell wall and an animal cell does not.A plant cell has a central vacuole, while an animal cell does not.A plant cell has chloroplasts, while an animal cell does not.




1. Cell Wall: Plant cells have a rigid cell wall made of cellulose, which provides structural support and protection. Animal cells do not have a cell wall, and instead have a flexible cell membrane.

2. Chloroplasts: Plant cells contain chloroplasts, which are the site of photosynthesis and allow the plant to produce its own food. Animal cells do not have chloroplasts and must obtain their food from other sources.

3. Vacuoles: Plant cells have a large central vacuole that stores water and nutrients, and helps maintain the cell's shape. Animal cells have smaller vacuoles that are used for storage and waste removal.

These differences can be easily recognized on the plant model. The cell wall will be visible as a thick outer layer, the chloroplasts will be visible as small green structures within the cell, and the central vacuole will be visible as a large, clear structure in the center of the cell.

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Related Questions

Suzie has a sore throat, fever, difficulty swallowing and a rash that covers neck and chest. A gram positive cocci bacteria, found in chains, is isolated from a swab. It has the following virulence factors: a. M proteins - b. hyaluronic acid capsule - c. Streptokinases \& streptolysins- d. Cs peptidase e. pyrogenic toxins carried ny a prophage-
f. identify this pathogen -

Answers

The pathogen that is causing Suzie's symptoms is Streptococcus pyogenes. This gram-positive cocci bacteria is known for its ability to cause a wide range of infections, including strep throat, scarlet fever, and impetigo.

The virulence factors that are associated with Streptococcus pyogenes include M proteins, which help the bacteria to adhere to host cells and avoid phagocytosis; a hyaluronic acid capsule, which also helps the bacteria to avoid phagocytosis; streptokinases, which break down blood clots and help the bacteria to spread; streptolysins, which destroy red blood cells and other host cells; C5 peptidase, which degrades complement proteins and helps the bacteria to evade the immune system; and pyrogenic toxins, which are carried by a prophage and can cause fever and other symptoms.

Based on Suzie's symptoms and the presence of these virulence factors, it is likely that she is suffering from an infection with Streptococcus pyogenes. Treatment may include antibiotics to kill the bacteria and alleviate symptoms.

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The ability to roll the tongue is caused by an autosomal dominant gene. Penny does not roll her tongue, and both of her parents do. What is the genotype of Penny, her father and her mother?
A. Penny: aa Penny's father: Aa Penny's mother: Aa
B. Penny: aa Penny's father: Aa Penny's mother: aa
C. Penny: Aa Penny's father: Aa Penny's mother: Aa
D. Penny: AA Penny's father: aa Penny's mother: Aa
E. Penny: xaxa Penny's father: xAy Penny's mother: xAxa

Answers

The genotypes of Penny, her father, and her mother are Penny: aa Penny's father: Aa Penny's mother: Aa

The correct option is A.

What is an autosomal dominant gene?

An autosomal dominant gene is a type of genetic inheritance pattern in which a single copy of the mutated gene, inherited from either parent, is sufficient to cause the expression of the trait or disorder associated with the gene. This means that individuals who inherit the mutated gene will have the trait or disorder, regardless of whether the other copy of the gene is normal or mutated.

The ability to roll the tongue is caused by an autosomal dominant gene.

Since Penny is not able to roll her tongue and her parents can, it follows that she is homozygous for the inability to roll the tongue while her parents are heterozygous for tongue rolling.

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How do you get floppy ears in Phet simulation?

Answers

To get floppy ears in the Phet simulation, click on the Human tab, select the Ears section, and use the draggable slider to adjust the size of the ears.

To get floppy ears in Phet simulation, follow the steps below:Open the Phet Simulation software that is installed on your computer.

Find the "Rabbit Population by Natural Selection" simulation and launch it on your system.

After the simulation is launched, you'll see that it has many different customizable options. These include different color options, shapes, sizes, and other features like floppy ears, long ears, short ears, and more.

To get floppy ears in the Phet Simulation software, you'll need to select the right combination of features that will enable you to customize your rabbit and give it the floppy ears you want.

There are two different options to select from when customizing your rabbit's ears: the long ear and the floppy ear. If you want floppy ears, choose this option from the menu.

After choosing the floppy ear option, you can further customize the ears by adjusting their size, shape, and other features. This will allow you to create a custom rabbit with floppy ears that is unique to your preferences and specifications.

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The rate of migration of a protein in SDS-PAGE is NOT influenced by
Size of the protein
Strength of the electric field
Charge of the protein
Pore size of the gel

Answers

The rate of migration of a protein in SDS-PAGE is not influenced by the charge of the protein.

Thus, the correct answer is the charge of the protein (C).

SDS polyаcrylаmide gel (SDS-PAGE) is а type of аnаlyticаl technique thаt is used to sepаrаte proteins. SDS pаge is а type of electrophoresis technique thаt sepаrаtes substаnces on the bаsis of size. The rаte of migrаtion through in SDS polyаcrylаmide gel will not be influenced by chаrge in the protein. SDS, thаt's аdded in the gel is responsible for negаting the chаrge of аll proteins.

Even though the ionic detergents SDS hydrolyzes аnd binds to molecules to mаke them evenly negаtively chаrged, SDS-PАGE predominаntly sepаrаtes proteins by mаss. Аll SDS-bound molecules in а sаmple will therefore move аcross the gel аnd towаrds the positive chаrged electrode whenever а current is delivered.

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During which phase would penicillin, an antibiotic that inhibits cell-wall synthesis, be most effective?
A. death phase
B. lag phase
C. log phase
D. stationary phase

Answers

Penicillin, an antibiotic that inhibits cell-wall synthesis, would be most effective during the Log Phase. Option B.

Penicillin is an antibiotic that is used to prevent or treat bacterial infections. Penicillin, an antibiotic that inhibits cell-wall synthesis, would be most effective during the Log Phase.

In the log phase, the growth rate of the bacteria is constant, and the cells are rapidly dividing. In the presence of nutrients, bacteria continue to grow exponentially, and the population doubles at a constant rate. During this phase, bacteria are more susceptible to antibiotics and are easily killed because they are actively growing and dividing.

Because penicillin inhibits cell-wall synthesis, it is more effective in preventing bacterial cell growth and division during the Log Phase. Therefore, penicillin, an antibiotic that inhibits cell-wall synthesis, would be most effective during the Log Phase.

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T/F This beta 1 selective catecholamine has + inotropic effects and leads to increased cardiac output-it is used to treat heart failure.

Answers

True. The beta 1 selective catecholamine that has positive inotropic effects and leads to increased cardiac output is called Dobutamine. It is used to treat heart failure by improving the heart's ability to pump blood effectively.

. A drug called dobutamine is used in the ICU to treat low blood pressure. Although the medication is safe, use of it needs to be closely watched because it has the potential to increase blood pressure and cause arrhythmia. Dobutamine is administered intravenously and is typically used in hospital settings for patients with acute heart failure. It works by stimulating the beta 1 receptors in the heart, which increases the contractility of the heart muscle and leads to increased cardiac output.

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57. Which particular hormone does BPA mimic? A. Testosterone B. Adrenocorticotropic hormone C. Human growth hormone D. Estrogen

Answers

The particular hormone that BPA mimics is D. Estrogen.

BPA, or Bisphenol A, is a chemical that is used in the production of polycarbonate plastics and epoxy resins. It is found in many common household items, such as plastic containers, canned foods, and even cash register receipts.

Studies have shown that BPA can mimic the hormone estrogen, which is responsible for the development of female secondary sexual characteristics and the regulation of the menstrual cycle. This can lead to a variety of health problems, including reproductive disorders, obesity, and an increased risk of cancer.

It is important to be aware of the potential risks associated with BPA exposure and to take steps to reduce your exposure, such as avoiding canned foods and using glass or stainless steel containers instead of plastic.

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illness in which red bloods cells, normally round, take on crescent shape and clog blood flow, leading to fever, pain etc. is called?

Answers

The illness you are describing is called Sickle Cell Anemia. This is a genetic disorder in which the body produces abnormal hemoglobin, causing red blood cells to become rigid and take on a crescent shape. These abnormal cells can clog blood flow, leading to symptoms such as fever, pain, and organ damage.

Hemoglobin, a protein found in red blood cells that transports oxygen throughout the body, is produced under conditions known as sickle cell anemia, a hereditary illness. Hemoglobin in people with sickle cell anemia creates aberrant, crescent-shaped red blood cells, which can block tiny blood capillaries and result in a number of issues.

Sickle cell anemia is presently incurable, thus therapy focuses on symptom management and avoiding complications. This can need frequent blood transfusions to boost the body's supply of healthy red blood cells, pain medication to lessen periods of discomfort brought on by sickled red blood cells, and antibiotic treatment to avoid infections.

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Use the Nernst equation to calculate the equilibrium membrane potential for K, Na, Ca, and Cl, at room temperature, where:
a) [K]i = 100, [K]o = 5
b) [Na]i = 20, [Na]o = 120
c) [Ca]i = 0.1, [Ca]o = 10
d) [Cl]i = 5, [Cl]o = 120
e) Explain or Define what the Nernst Equilibrium Potential is (in words).

Answers

The equilibrium membrane potential for a)K, b)Na, c)Ca, and d)Cl, at room temperature using Nernst equation is a)-77.1 mV, b)62.7 mV, c)-128.4 mV and d)-69.3 mV respectively. The Nernst equation is a mathematical equation used to calculate the equilibrium membrane potential (Em) of a given ion, based on its permeability, its concentration inside and outside the cell, and the temperature.

The Nernst equation is expressed as: Em = (RT/zF) ln ([X]i/[X]o)

Where:

R is the gas constantT is the temperature in Kelvinz is the valence of the ionF is the Faraday constant[X]i is the intracellular concentration of the ion[X]o is the extracellular concentration of the ion

In order to calculate the equilibrium membrane potential for K, Na, Ca, and Cl at room temperature, we will use the Nernst equation above and the given concentrations as follows:

K: Em = (8.314 J/Kmol)(293 K)/(1)(96485 C/mol) ln ([100]i/[5]o) = -77.1 mV

Na: Em = (8.314 J/Kmol)(293 K)/(1)(96485 C/mol) ln ([20]i/[120]o) = 62.7 mV

Ca: Em = (8.314 J/Kmol)(293 K)/(2)(96485 C/mol) ln ([0.1]i/[10]o) = -128.4 mV

Cl: Em = (8.314 J/Kmol)(293 K)/(-1)(96485 C/mol) ln ([5]i/[120]o) = -69.3 mV

The Nernst Equilibrium Potential is a measure of the membrane potential at which the net flow of ions across a membrane is zero, resulting in a state of equilibrium. At equilibrium, the concentrations of the ions on both sides of the membrane are equal and the membrane potential remains constant. The Nernst equation can be used to calculate the equilibrium membrane potential for a given ion.

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1) What is the genotype for box A?
a) GG
b) Gg
c) gg
d) G
2) What is the phenotype of box A?
a) Gray
b) White
c) White with grey spots
d) Gray with white spots
3) What is the percentage of heterozygous rabbits?
a) 25%
b) 50%
c) 75%
d) 100%
4) What is the percentage of white rabbits?
a) 0%
b) 25%
c) 50%
d) 100%
5) What is the phenotype of box D?
a) Gray
b) White
c) White with gray spots
d) Gray with white spots

Answers

The genotype for box A is b) GgThe phenotype of box A is a) GrayThe percentage of heterozygous rabbits is b) 50%The percentage of white rabbits is b) 25%The phenotype of box D is b) White

Genotype and phenotype in box A

We can see box A in the picture above.

The genotype for box A is b) Gg, as it is a heterozygous rabbit with one dominant allele (G) and one recessive allele (g).

The phenotype of box A is a) Gray, as the dominant allele (G) masks the recessive allele (g) and results in a gray phenotype.

The percentage of heterozygous rabbits is b) 50%, as there are two heterozygous rabbits (Gg) out of a total of four rabbits.

The percentage of white rabbits is b) 25%, as there is one white rabbit (gg) out of a total of four rabbits.

The phenotype of box D is b) White, as it is a homozygous recessive rabbit with two recessive alleles (gg) that result in a white phenotype.

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Your friend decides to set up an experiment with four different groups with ten mice each. Each group of mice will have their respiration rate tested at a specific time of the day - Moening ( 6 AM), Noon (12 PM), Evening (6 PM) and Night (12 AM). Fach mouse was placed in the respirometer for a penod of five minutes. 5. If your friend's predietion (and hypothesis) is supported, which group of mice will have the highest respiratory rate?' (Circle ONE)' A. The Morning group B. The Day group C. The Evening group D. The Night group

Answers

Answer: A

Explanation:

A. The Morning group. Mice typically exhibit higher levels of physical activity during the morning, so it is likely that their respiratory rate will be the highest at this time.

Corn grain contains 7% protein, 85% carbohydrate, and 5% fat.
Assuming digestibility of protein is 92%, carbohydrate is 88.2%,
and fat is 87.3%, what is the TDN.

Answers

The TDN of corn grain contains 7% protein, 85% carbohydrate, and 5% fat and fat 87.3% is: 0.97kg.

Corn grain contains 7% protein, 85% carbohydrate, and 5% fat, which translates to 0.07 kg protein, 0.85 kg carbohydrate, and 0.05 kg fat per kg of corn grain. The Total Digestible Nutrients (TDN) is calculated by the sum of digestible protein, fat, and carbohydrates. Therefore, the TDN of corn grain is 0.07 + 0.05 + 0.85 = 0.97 kg.

The TDN is a measure of the energy value of feed and is expressed as a percentage of the feed dry matter. The TDN is an important measure when formulating rations for livestock and is determined by the type and quality of feed consumed. The digestible protein, fat, and carbohydrates all contribute to the total energy in the feed, which is important for animal health and performance.

To calculate the TDN of corn grain, you need to first determine the percentage of protein, fat, and carbohydrates in the grain. You then multiply each of these percentages by the weight of the grain to get the amount of each nutrient per kilogram of corn grain. Finally, you add up the total amounts of each nutrient and this gives you the TDN.

In conclusion, the TDN of corn grain is 0.97 kg per kg of feed. This is calculated by adding the amounts of digestible protein, fat, and carbohydrates in the grain. The TDN is a useful measure to consider when formulating rations for livestock, as it provides a measure of the total energy content of the feed.

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Occurs from a combination of immaturity of the hematopoietic system combined with the destruction of RBC because of low levels of Vit. E

Answers

This condition you are describing is called Hemolytic Anemia, and is caused by a combination of the immaturity of the hematopoietic (blood) system, coupled with the destruction of red blood cells (RBCs) due to low levels of Vitamin E in the body.


Hemolytic anemia occurs when the red blood cells (RBC) are destroyed faster than they can be produced by the hematopoietic system, leading to a decrease in the number of RBC in the body. One of the causes of hemolytic anemia is a deficiency in vitamin E, which is necessary for the proper functioning of the hematopoietic system. Without sufficient levels of vitamin E, the hematopoietic system cannot produce enough RBC to replace those that are destroyed, leading to anemia.

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Where would a frameshift mutation cause the most damage?
A.Near the very end of the gene
B.Near the beginning of the gene
C.Near the center of the gene
D.All frameshift mutations are equal
Down syndrome is a genetic disorder that is also called trisomy 21. A person with Down Syndrome has an extra copy of chromosome 21. What can you infer is most likely the genetic mutation that results in Down syndrome?
A.Nondisjunction during meiosis, resulting in uneven distribution of chromosomes
B.Complete duplication of chromosomes during polyploidy
C.Translocation during genetic replication, creating nonhomologous chromosomes
D.Crossing over during meiosis, leading to an exchange of genetic information
Which of the following is an example of a silent mutation?
A.A codon mutated from CCG to CCA, both coding for glycine
B.A substitution of glycine amino acid for a stop codon
C.A substitution of a serine amino acid for a glycine amino acid
D. An insertion or deletion of nucleotides that is not a multiple of three

Answers

1 - A frameshift mutation causes the most damage B. Near the beginning of the gene.

2- The most likely genetic mutation that results in Down syndrome is A. Nondisjunction during meiosis, resulting in uneven distribution of chromosomes.

3 - An example of a silent mutation is A. A codon mutated from CCG to CCA, both coding for glycine.



1 - A frameshift mutation is a type of genetic mutation that occurs when one or more nucleotides are inserted or deleted from a gene. This causes the reading frame of the gene to shift, leading to changes in the sequence of amino acids that are encoded by the gene. If a frameshift mutation occurs near the beginning of the gene, it can have a greater impact on the protein that is produced, as it will affect a larger portion of the protein's sequence. Therefore, the correct answer is B. Near the beginning of the gene.


2 - Down syndrome is caused by an extra copy of chromosome 21, which is typically the result of nondisjunction during meiosis. Nondisjunction is a type of genetic mutation that occurs when chromosomes do not separate properly during cell division, leading to an uneven distribution of chromosomes in the resulting cells. This can result in an extra copy of a chromosome, as is the case with Down syndrome. Therefore, the correct answer is A. Nondisjunction during meiosis, resulting in uneven distribution of chromosomes.

3 - A silent mutation is a type of genetic mutation that does not result in a change in the amino acid sequence of a protein. This can occur when a mutation changes one of the nucleotides in a codon, but the new codon still codes for the same amino acid. In the example provided, the codon CCG is mutated to CCA, but both codons code for the amino acid glycine. Therefore, this is an example of a silent mutation, and the correct answer is A. A codon mutated from CCG to CCA, both coding for glycine.

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What happens to the air pressure in an area if the air temperature increases? (2 points) a The air pressure will increase because the air particles are packed together. b The air pressure will decrease because the air particles are spread out. c The air pressure will remain the same because temperature does not affect the air particles. d The air pressure will remain the same because the hot air will travel to different areas.

Answers

The correct answer is option b. The air pressure will decrease because the air particles are spread out.

What is air pressure ?

Air pressure is the force exerted by the weight of air molecules on the surface of the Earth or any other object immersed in the atmosphere. It is the weight of the air molecules above a given point and is typically measured in units of pressure such as pounds per square inch (psi), millibars (mb), or pascals (Pa). Air pressure is affected by a number of factors, including altitude, temperature, humidity, and weather patterns. Changes in air pressure can have significant impacts on weather patterns and can also affect human health and well-being.

Therefore , When the air temperature increases, the air particles gain energy and move faster, increasing the distance between the particles. This results in a decrease in air density, which in turn causes a decrease in air pressure. Conversely, a decrease in temperature would cause the air particles to move more slowly and become more densely packed, resulting in an increase in air pressure.

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This text is asking about the relationship between air temperature and air pressure. Air pressure refers to the force exerted by the weight of air molecules in the Earth's atmosphere, while air temperature refers to the measure of how hot or cold the air is.

The text provides four options for what happens to air pressure in an area if the air temperature increases. The first option suggests that air pressure will increase because the air particles are packed together. This is a correct statement, as an increase in temperature causes the air molecules to move faster and spread out, which results in a decrease in air pressure. Conversely, a decrease in temperature causes the air molecules to move slower and pack closer together, which results in an increase in air pressure. Therefore, the correct answer to this question is "The air pressure will decrease because the air particles are spread out."

Question 1: What serious health consequence of prolonged anorexia nervosa in a teen-aged girl will be likely in the later years of her life? Explain why this is the case.
Question 3: Compare and contrast bulimia nervosa with binge-eating disorder. Identify behavioral characteristics that are shared between these two disorders, and behavioral characteristics that are unique to each.

Answers

Answer 1: One serious health consequence of prolonged anorexia nervosa in a teenage girl can be infertility later in life.

Answer 3: Both bulimia nervosa and binge-eating disorder are characterized by episodes of consuming large amounts of food in a short period of time, followed by feelings of guilt and shame.

Prolonged anorexia can lead to infertility because anorexia nervosa can disrupt hormonal balance, leading to amenorrhea (absence of menstrual periods) and potentially causing damage to the reproductive system.

Both bulimia nervosa and binge-eating disorder are characterized by episodes of consuming large amounts of food in a short period of time, followed by feelings of guilt and shame. However, individuals with bulimia nervosa will also engage in compensatory behaviors, such as self-induced vomiting or excessive exercise, in an attempt to prevent weight gain.

On the other hand, individuals with binge-eating disorders do not engage in compensatory behaviors and may experience weight gain as a result. Additionally, individuals with bulimia nervosa may have a distorted body image and place a greater emphasis on their weight and appearance, while individuals with binge-eating disorder may not have these concerns.

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How the three ducks' wing skeletons differ from one another will be the subject of a project.
The three groups of ducks will be compared in terms of their skeletal morphology as part of PROJECT. By creating digital skeletal models of a mallard or a hybrid duck using CT scans, you may compare their anatomy to that of an Indian Runner Duck.
Please come up with a hypothesis and a question.

Answers

One possible hypothesis for this project could be: "The wing skeletons of the three ducks will have significant differences in terms of their skeletal morphology, with the Indian Runner Duck having the most unique anatomy."

This hypothesis is based on the assumption that the Indian Runner Duck, which is known for its unique upright posture and running gait, will also have unique skeletal features in its wings.

A question that could be used to guide the project is: "How do the skeletal morphologies of the mallard, hybrid duck, and Indian Runner Duck compare in terms of their wing skeletons?" This question will help focus the project on the specific goal of comparing the wing skeletons of the three ducks and identifying any significant differences in their anatomy.

In order to test the hypothesis and answer the question, the project could involve creating digital skeletal models of the three ducks using CT scans and then comparing the anatomy of their wing skeletons. This could include measuring and comparing the size and shape of the bones, as well as the arrangement and orientation of the bones in the wing skeletons. By analyzing and comparing these features, the project could determine whether there are significant differences in the skeletal morphology of the three ducks and whether the Indian Runner Duck has the most unique anatomy.

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What type of tandem repeat locus is this: core DNA sequence CGT
repeated 40 times?

Answers

The core DNA sequence CGT repeated 40 times is a tetranucleotide tandem repeat locus.

A tandem repeat locus is a section of DNA where specific DNA sequences repeat one after the other. Tandem repeat loci can be used in genetic studies to determine the genetic similarity between individuals. There are three types of tandem repeats, which are minisatellites, microsatellites, and satellite DNA.

Minisatellites are usually 10-100 nucleotides long, while microsatellites consist of 1-6 nucleotides. In contrast, satellite DNA is a type of DNA that is organized into tandem repeats, similar to the other two types. However, satellite DNA is different from the other two types in that it is found in centromeres, telomeres, and heterochromatin.

Therefore, the core DNA sequence CGT repeated 40 times is a tetranucleotide tandem repeat locus.

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What is gene expression what processes are involved in gene expression?

Answers

Gene expression refers to the process in which the genetic information present in the DNA sequence of a gene is used to direct the synthesis of a protein or RNA molecule. It involves several processes that help in converting genetic information into a functional product that performs a specific function.

The processes involved in gene expression are given below:

Transcription: It is the first step of gene expression that involves the synthesis of RNA molecules from DNA. RNA polymerase enzyme binds to the promoter region of the DNA and starts synthesizing the complementary RNA strand using the template strand of DNA. RNA processing: The newly synthesized RNA molecule undergoes various modifications to form a mature and functional RNA molecule. These modifications include capping, splicing, and polyadenylation. Translation: It is the process in which the genetic information present in the RNA sequence is used to synthesize a protein. It involves the participation of ribosomes, tRNA molecules, and amino acids.

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Nutrition, nursing, biology
1. Intestinal malabsorption syndromes or pancreatic insufficiency may result in deficiency of any of the following vitamins except
a. D
b. A
c. C
d. E
e. K
2. Chronic alcoholism results in deficiency of all of these vitamins except
a. B1
b. B2
c. B3
d. Folic Acid
e. None of the above
3. Diarrhea, dermatitis, and dementia are characteristics of deficiency of
a. B1
b. B2
c. B3
d. B6
e. Folic Acid
4. High cardiac output failure suggests deficiency of which vitamins?
a. Riboflavin
b. Pyridoxine
c. Niacin
d. Thiamine
e. Ascorbic acid

Answers

1. C. Vitamin C is not typically affected by these conditions.

2. e. None of the above. Alcoholism can lead to deficiencies in all of the listed vitamins, including B1 (thiamine), B2 (riboflavin), B3 (niacin), and folic acid.

3.  c. B3. These are common symptoms of niacin deficiency, also known as pellagra.

4.  d. Thiamine. Thiamine (vitamin B1) deficiency can lead to a condition called beriberi, which is characterized by high cardiac output failure, among other symptoms.

Nutrition is an important aspect of maintaining good health. Vitamins play a crucial role in this regard, and a deficiency or insufficiency of certain vitamins can lead to various health problems, including cardiac issues. Here are the answers to the questions:


Intestinal malabsorption syndromes or pancreatic insufficiency may result in deficiency of any of the following vitamins except c. C. Vitamin C is not typically affected by these conditions.


Chronic alcoholism results in deficiency of all of these vitamins except e. None of the above. Alcoholism can lead to deficiencies in all of the listed vitamins, including B1 (thiamine), B2 (riboflavin), B3 (niacin), and folic acid.


Diarrhea, dermatitis, and dementia are characteristics of deficiency of c. B3. These are common symptoms of niacin deficiency, also known as pellagra.


High cardiac output failure suggests deficiency of which vitamins? d. Thiamine. Thiamine (vitamin B1) deficiency can lead to a condition called beriberi, which is characterized by high cardiac output failure, among other symptoms.

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What can move, grow react, protect them selves, repair damage, as well as regulate life processes, and reproduce?

Answers

The organisms that can move, grow, react, protect themselves, repair damage, regulate life processes, and reproduce are living organisms. These are the characteristics that differentiate living organisms from non-living things.

Living organisms include animals, plants, fungi, and microorganisms. All living organisms have the ability to move, whether it is through locomotion or movement of substances within their bodies. They can also grow and develop, reacting to their environment in order to protect themselves and repair any damage. Living organisms also have the ability to regulate their life processes, such as metabolism and homeostasis, in order to maintain proper functioning. Lastly, living organisms have the ability to reproduce, creating offspring in order to ensure the continuation of their species.

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Mus musculus may have black, brown or golden coloured fur. House mice with black fur of identical genotypes were mated and produced the following offspring:
25 mice with black fur, 15 mice with golden fur, and 2 mice with brown fur
What epistatic ratio is approximated by these offspring?
Explain how you determined this ratio (verify).
What type of epistasis is seen in the offspring?

Answers

The epistatic ratio approximated by these offspring is 25:15:2. This ratio was determined by counting the number of mice with each fur color and comparing them to each other.

For example, there are 25 mice with black fur, 15 mice with golden fur, and 2 mice with brown fur. These numbers can be simplified to a ratio of 25:15:2.

The type of epistasis seen in the offspring is recessive epistasis. This is because the recessive allele of one gene masks the expression of the other gene. In this case, the recessive allele for black fur masks the expression of the other genes that determine fur color, resulting in the majority of the offspring having black fur. This is evident in the fact that 25 out of the 42 offspring have black fur, while only 15 have golden fur and 2 have brown fur.

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in the RSS separating sequences palindromes are found? a. only in 1 turn spacer sequences b. only i 2 turn spacer sequences c. in both 1 and 2 turn spacer sequences d. flanking the spacers specially i heptamer regions e. in the heptamers and 1 turn spacer sequence regions of 1 turn RSS

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in the RSS separating sequences palindromes are found is c. in both 1 and 2 turn spacer sequences.

Palindromes are sequences of DNA that read the same forwards and backwards. These sequences are found in both 1 and 2 turn spacer sequences in the RSS (recombination signal sequence) of the DNA. The RSS is a specific sequence of DNA that is recognized by the RAG (recombination activating gene) proteins and is important for the process of V(D)J recombination in the immune system.

The presence of palindromes in the RSS allows for the formation of hairpin structures, which are important for the recombination process. Therefore, palindromes are found in both 1 and 2 turn spacer sequences of the RSS.

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How do otters impact CO2 levels in the ocean?

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Answer: Sea otters help ecosystems capture carbon from the atmosphere and store it as biomass and deep-sea detritus, preventing it from being converted back to carbon dioxide and contributing to climate change. (write this in your own words)

Explanation:

In humans, the "smeller"-tasting locus has two alleles, S and s, where S is dominant and is the "smeller" allele; s is recessive and is the "non-smeller" allele. Suppose in a population of 500 people, 180 have the "smeller" phenotype. The population is in Hardy-Weinberg equilibrium with respect to this locus (6 points) a. What is the frequency of the S allele? b. What is the frequency of the s allele? c. How many people have the SS genotype? d. How many people have the Ss genotype? e. How many people have the ss genotype? f. Suppose that the next generation has 600 people. How many are predicted to be "smellers"?

Answers

a. The frequency of the S allele can be calculated by taking the square root of the proportion of "smeller" individuals in the population. Therefore, the frequency of the S allele is √(180/500) = 0.6.

b. The frequency of the s allele can be calculated by subtracting the frequency of the S allele from 1. Therefore, the frequency of the s allele is 1 - 0.6 = 0.4.

c. The frequency of the SS genotype can be calculated by squaring the frequency of the S allele. Therefore, the frequency of the SS genotype is (0.6)² × 500 = 180.

d. The frequency of the Ss genotype can be calculated by multiplying the frequencies of the S and s alleles and then doubling the result (since there are two possible ways to obtain the Ss genotype). Therefore, the frequency of the Ss genotype is 2 × 0.6 × 0.4 × 500 = 240.

e. The frequency of the ss genotype can be calculated by squaring the frequency of the s allele. Therefore, the frequency of the ss genotype is (0.4)² × 500 = 80.

f. If the population is still in Hardy-Weinberg equilibrium, the frequency of the S allele will remain the same. Therefore, the predicted number of "smellers" in the next generation can be calculated by multiplying the total population size by the frequency of the SS and Ss genotypes, which are 180/600 and 240/600, respectively.

Thus, the predicted number of "smellers" in the next generation is (180/600 + 240/600) × 600 = 420.

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By law of --- ---, the relationship between free (unoccupied) and bound receptors can be described as follows: _____or re-arranged:_____If we let Ro = total number of receptors, then ____Replace [R] by [Ro] - [LR] and re-arrange above equations:____ where [LR]/Ro is the --- of --- --- --- by ---.

Answers

By the law of mass action, the relationship between free (unoccupied) and bound receptors can be described as follows: [L][R] = Kd[LR] or re-arranged: [LR] = ([L]/Kd)[R]. If we let Ro = total number of receptors, then [R] = Ro - [LR]. Replace [R] by [Ro] - [LR] and re-arrange above equations: [LR] = ([L]/Kd)(Ro - [LR]) or [LR] = (Ro[L])/(Kd + [L]) where [LR]/Ro is the fraction of total receptors occupied by ligand.

The law of mass action explains the equilibrium between a ligand (L) and its receptor (R), where [L] is the concentration of the ligand, [R] is the concentration of free (unoccupied) receptors, and [LR] is the concentration of ligand-bound receptors. The amount of ligand at which half of the receptors are bound is indicated by the dissociation constant, or Kd.

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Explain:When
there is little or no variation in fitness over wide range of
behaviors. Example Screech owls

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When there is little or no variation in fitness over a wide range of behaviors, it means that there is little or no difference in the ability of individuals to survive and reproduce, regardless of the behaviors they exhibit. This can occur when the environment is relatively stable and there are no strong selective pressures favoring one behavior over another.


For example, screech owls have a wide range of behaviors, including different hunting strategies and mating behaviors. However, there is little or no variation in fitness among screech owls exhibiting different behaviors.

This means that screech owls with different behaviors are equally likely to survive and reproduce, and there is no selective pressure favoring one behavior over another.


In summary, when there is little or no variation in fitness over a wide range of behaviors, it means that there is no selective pressure favoring one behavior over another, and individuals with different behaviors are equally likely to survive and reproduce.

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HELPP PLEASE ‼️‼️‼️
I have so much other work to get done and it’s due tomorrow!!
HELPPPPPP
WILL MARK BRAINILEST!!!

Answers

Au. africanus and Au. robustus has been discovered in a number of South African limestone caverns. Eastern Africa: Sedimentary basins and river drainage systems.

What is Great Rift Valley?

The Great Rift Valley is home to numerous australopithecine fossils, including the renowned Au. afarensis. The most well-known Australopithecus fossil discoveries in East and South Africa are probably "Lucy" and "Mrs Ples."

As opposed to what some have previously claimed, Australopithecus fossils from the richest hominin-bearing stratum (Member 4) at Sterkfontein in South Africa are much older and are contemporaneous with Australopithecus afarensis in East Africa. Afarensis exhibited traits common to both apes and humans.

Therefore, Au. africanus and Au. robustus has been discovered in a number of South African limestone caverns. Eastern Africa: Sedimentary basins and river drainage systems.

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Provide a brief intro to protozoa or non-fungal eukaryotic
microbe.
Give a specific example of protozoa or a non-fungal eukaryotic
microbe.

Answers

Protozoa are a diverse group of single-celled eukaryotic organisms that are found in various environments, including freshwater, marine, and terrestrial habitats.

They are classified into several groups based on their morphology and mode of movement, including flagellates, ciliates, and amoebae. Protozoa are important members of the microbial community and play important roles in nutrient cycling, as well as serving as food sources for other organisms.

An example of a protozoan is the genus Giardia, which includes several species that are parasitic and can cause infections in humans and other animals. Giardia species are flagellates, meaning they move using whip-like structures called flagella. They are typically transmitted through contaminated water or food, and can cause symptoms such as diarrhea, abdominal cramps, and nausea. Treatment typically involves the use of antibiotics to eliminate the infection.

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Suppose you had a plant cell with a chromosome number of 2n=4 and you knew that the gene for leaf colour was on one pair of chromosomes and the gene for bark smoothness was on a different pair of chromosomes. Use the letters G and H to represent the genes. Draw a chromosome diagram to accurately represent this plant cell during metaphase I of meiosis. Assume that all of the alleles for leaf colour and bark smoothness are recessive.

Answers

The chromosome diagram for the plant cell during metaphase I of meiosis would consist of two pairs of homologous chromosomes, with G and H genes located on separate pairs.

During metaphase I of meiosis, the homologous chromosomes pair up and align at the equatorial plane of the cell. In this scenario, the plant cell has a chromosome number of 2n=4, meaning it has two pairs of homologous chromosomes. The gene for leaf color (G) is located on one pair of chromosomes, while the gene for bark smoothness (H) is located on the other pair of chromosomes. Since both genes have recessive alleles, they would be represented by lowercase letters (g and h).

The resulting chromosome diagram would show the two pairs of homologous chromosomes, each with two chromatids. One pair of chromosomes would have the genes G and g, while the other pair would have the genes H and h. The chromosomes would be arranged in a way that the maternal and paternal copies of each chromosome would be adjacent to each other, ready for segregation during meiosis I.

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