This athlete's karyotype reveals a genetic anomaly caused by nondisjunction event that occurred in the gamete cells of the athlete's mother or father.
Nondisjunction is a genetic anomaly that occurs during cell division, where the chromosomes fail to separate properly, leading to an abnormal distribution of chromosomes in the resulting gametes. As a result, the athlete's karyotype will show an abnormal number of chromosomes, either too many or too few.
Nondisjunction is a random event that can occur in either the mother or the father's gamete cells, and it is not related to any environmental factors or lifestyle choices. Deletion, on the other hand, is another type of genetic anomaly that occurs when a part of a chromosome is lost during cell division. This can result in the loss of critical genetic material, leading to various physical and developmental abnormalities. However, in this athlete's case, the karyotype revealed a nondisjunction event, which is the cause of the genetic anomaly.
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what types of organisms must an ecosystem have at a minimum to recycle nutrients
The 30S and 50S ribosomal subunits are so designated to indicate.:________
The 30S and 50S ribosomal subunits are designated based on their sedimentation coefficients, which reflect their sizes and densities.
The sedimentation coefficient is a measure of how fast a particle, such as a ribosome, settles in a centrifugal field. It is influenced by factors such as the size, shape, and density of the particle, as well as the viscosity and temperature of the medium.
In the case of ribosomes, the 30S and 50S subunits are separated based on their sedimentation coefficients in an ultracentrifuge. The "S" in the designations stands for Svedberg units, which is a unit of sedimentation coefficient.
The 30S subunit is smaller and less dense, with a sedimentation coefficient of approximately 30S, while the 50S subunit is larger and more dense, with a sedimentation coefficient of approximately 50S. Together, they form the 70S ribosome, which is the ribosome found in prokaryotes.
In eukaryotes, the ribosome is larger and is composed of a 40S small subunit and a 60S large subunit, which together form the 80S ribosome. The designations of 30S and 50S are therefore specific to prokaryotic ribosomes.
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The key to molecular-motor function involving the core structures of myosin, kinesin, and dyneins is A. their ability to respond to a change in conformation of bound nucleotide triphosphates. B. their ability to hydrolyze nucleotide triphosphates. C. their ability to change conormations in response to binding and hydrolysis of nucleotide triphosphates. D. their ability to bind nucleotide triphosphates.
The key to molecular-motor function involving the core structures of myosin, kinesin, and dynein is C. their ability to change conformations in response to binding and hydrolysis of nucleotide triphosphates.
The correct answer is C. The key to molecular-motor function involving the core structures of myosin, kinesin, and dynein is their ability to change conformation in response to binding and hydrolysis of nucleotide triphosphates. This ability allows them to carry out their functions of moving along cytoskeletal filaments, such as microtubules, and transporting cargo within cells. Kinesin, in particular, is a molecular motor that uses the energy from hydrolysis of ATP (adenosine triphosphate) to move along microtubules, while myosin and dynein use a similar mechanism involving other nucleotide triphosphates.
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A student creates a study guide describing each of the five phases of a neuronal action potential. How should she fill in the blanks for the phase indicated above?A. A = Rapid diffusion into cell; B = Equal movement in and out of cell; C = Most positive value
B. A = Slow diffusion out of cell; B = No movement; C = Most positive value
C. A = Equal movement in and out of cell; B = No movement; C = Most negative value
D. A = Rapid diffusion out of cell; B = Equal movement in and out of cell; C = Most negative value
Action potentials go through four phases: depolarization, overshoot or peak phase, repolarization, and refractory period. The three phases of an action potential are depolarization, overshoot, and repolarization. Hence (a) is the correct option.
In relation to the action potential, the membrane potential has two additional states. First, there is hypopolarization, which comes before depolarization, and then there is hyperpolarization, which comes after repolarization. The three primary phases of a neural action potential are depolarization, repolarization, and hyperpolarization. The threshold voltage of the cell, or the membrane potential at which voltage-gated sodium channels (Nav) open to let an inflow of sodium ions occur, controls how depolarized the cell becomes initially.
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what type of elution study should be performed to identify most antibodies from adult patients in the blood bank laboratory?
The Adsorption-Elution technique is highly effective for identifying antibodies in adult patients' blood samples, as it helps to isolate and concentrate the specific antibodies of interest.
In a blood bank laboratory, the most suitable elution study to identify most antibodies from adult patients is the Adsorption-Elution technique. This method consists of the following steps:
1. Collect the patient's blood sample.
2. Separate plasma from the blood cells by centrifugation.
3. Perform adsorption by incubating the patient's plasma with appropriate donor red blood cells (RBCs) known to express the corresponding antigens. During this step, the antibodies in the plasma will bind to the antigens present on the donor RBCs.
4. Centrifuge the mixture to separate antibody-bound RBCs from unbound plasma.
5. Perform elution by incubating the antibody-bound RBCs with an elution reagent, such as acid, heat, or a specific eluting solution. This process will release the bound antibodies from the RBCs into the eluate.
6. Collect the eluate, which contains the eluted antibodies, and perform antibody identification testing using an antigram or other serologic techniques.
The Adsorption-Elution technique is highly effective for identifying antibodies in adult patients' blood samples, as it helps to isolate and concentrate the specific antibodies of interest. This method enables accurate and reliable identification of antibodies, facilitating efficient blood transfusion and blood bank management.
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[T/F] Opioids, such as hydrocodone (Vicodin), usually suppress nausea and vomiting.
True. Opioids, including hydrocodone (Vicodin), are often used to suppress nausea and vomiting. Opioids, including hydrocodone (Vicodin), can suppress nausea and vomiting by acting on the central nervous system (CNS).
They inhibit the activity of the vomiting center in the brainstem and also reduce the sensitivity of the gastrointestinal tract to emetic stimuli. However, this effect may not be universal and can vary depending on the individual and the specific situation. In some cases, opioids can actually cause nausea and vomiting as a side effect. They are commonly used to manage pain, including chronic pain, acute pain, and pain associated with cancer or surgery. In addition to their pain-relieving effects, opioids can also cause a range of side effects, including constipation, dizziness, confusion, respiratory depression, and nausea/vomiting.
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What kinds of changes to the atmosphere could affect how much energy is absorbed by Earth's surface? Use ALL of the words in the word bank below for full credit.
carbon dioxide
methane
temperature
energy
increase
decrease
absorb
surface
atmosphere
pls help
Much of the Sun's energy is prevented from leaving via Earth's atmosphere and into space. The planet's temperature is maintained at a level where life may survive thanks to a phenomenon known as the greenhouse effect.
How much energy does the surface and atmosphere of the Earth absorb?Around 30% of the radiation that enters the Earth's atmosphere and surface is reflected back to space and does not heat the surface, leaving the other 70% to be absorbed by these two surfaces and the atmosphere. Due to its lower temperature, the Earth emits radiation in wavelengths that are far longer than those of the Sun.
Infrared radiation is emitted from the atmosphere in two directions: towards the surface of the Earth and out into space. This is the principal mechanism that results in energy loss from the atmosphere.
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during in situ hybridization, a dna probe made from a cloned gene binds to ______.
During in situ hybridization, a DNA probe made from a cloned gene binds to the complementary target DNA sequence present in the sample being analyzed.
In situ hybridization (ISH) is a technique used in molecular biology to localize specific nucleic acid sequences, such as DNA or RNA, within cells or tissues. The technique involves hybridizing a labeled complementary nucleic acid probe to the target sequence in the cells or tissues of interest.
There are two main types of ISH: DNA ISH and RNA ISH. In DNA ISH, a labeled DNA probe is used to detect the presence of a specific DNA sequence in cells or tissues. In RNA ISH, a labeled RNA probe is used to detect the presence of a specific RNA transcript in cells or tissues.This binding occurs due to the specific base pairing between the probe and the target sequence, allowing for the detection and localization of specific DNA sequences within a biological sample.
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Mutant gene alleles associated with known genetic disorders can be detected using DNA microarray analysis. Select one: True False
True. DNA microarrays detect mutations in DNA, including those associated with genetic disorders, by comparing an individual's genome to a reference.
Valid. Freak quality alleles related with realized hereditary problems can be identified utilizing DNA microarray examination. DNA microarrays can distinguish contrasts in quality articulation and changes in DNA groupings, including single nucleotide polymorphisms (SNPs) and duplicate number varieties (CNVs). By contrasting the DNA grouping of a singular's genome to a reference genome, scientists can distinguish transformations related with hereditary issues. DNA microarray examination is generally utilized in research and clinical settings to analyze hereditary problems and guide customized treatment.
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what are the advantages of using a restriction enzyme whose recognition site is relatively rare? when would you use such enzymes?
Select all the correct answer
- will produce large fragments
- will produce smaller fragments
- useful for isolating intact genes or centromeres
The advantages of using a restriction enzyme whose recognition site is relatively rare are that it will produce large fragments and is useful for isolating intact genes or centromeres.
You would use such enzymes when you need to analyze large genomic regions or preserve the integrity of specific genes during DNA manipulation.
1. A restriction enzyme with a rare recognition site will cut DNA less frequently.
2. This results in larger DNA fragments being produced.
3. These large fragments are beneficial when isolating intact genes or centromeres, as they preserve their structure and function.
4. Use such enzymes when you need to analyze large genomic regions or maintain the integrity of specific genes during DNA manipulation.
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which sequence correctly orders psychological disorders from earliest to latest age of onset?
The sequence that correctly orders psychological disorders from earliest to latest age of onset is typically: neurodevelopmental disorders, anxiety disorders, mood disorders, personality disorders, and psychotic disorders. However, it's important to note that the age of onset can vary greatly depending on the individual and the specific disorder.
The correct sequence for ordering psychological disorders from earliest to latest age of onset is:
1. Autism Spectrum Disorder (ASD)
2. Attention-Deficit/Hyperactivity Disorder (ADHD)
3. Anxiety Disorders
4. Mood Disorders (such as Major Depressive Disorder and Bipolar Disorder)
5. Schizophrenia
This sequence is based on the typical age of onset for each disorder, although individual cases may vary.
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how can a signal molecule from one cell alter gene expression in a target cell without entering the target cell?
A signal molecule can alter gene expression in a target cell without entering the cell by binding to a receptor on the surface of the cell.
This binding triggers a signaling cascade inside the cell that ultimately leads to changes in gene expression. When a signal molecule binds to a receptor on the cell surface, it can cause a conformational change in the receptor, leading to the activation of intracellular signaling pathways. These signaling pathways can involve the activation of intracellular enzymes or the recruitment of cytoplasmic proteins, ultimately leading to the activation of transcription factors that regulate gene expression.
For example, in the case of steroid hormones, which are lipophilic (fat-soluble) molecules that cannot easily pass through the cell membrane, they bind to intracellular receptors that are typically located in the cytoplasm or nucleus of the target cell. Once bound to the receptor, the hormone-receptor complex can enter the nucleus and directly bind to specific DNA sequences, leading to changes in gene expression.
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Which of the following statements is/are true?
Select one:
a. Homologous motor structures across related species tend to evolve independent neural circuitry to facilitate their control.
b. Homologous neurons across related species are known to control homologous motor structures.
c. Within vertebrates, the correlation between the raw brain size and evolutionary age of a species is known to be approximately inversely proportional.
d. Within vertebrates, the ratio between the brain and body size of a species is known to scale in an approximately linear fashion.
The correct response is option b. It is well established that homologous neurons in related species control identical motor structures in one another. Because of this, the answer that should be chosen is option B.
This is due to the fact that homologous neurons are neurons that are similar to one another and can be found in different species but serve the same purpose, such as controlling homologous motor structures. The statement makes no direct reference to the connection between evolution and the development of homologous neuronal and motor structures, despite the fact that evolution plays a part in the process. The other assertions are either false or missing important information.
Therefore, the correct answer is option B.
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_________ is the breakdown product formed when one phosphate group is removed from atp.
Adenosine diphosphate (ADP) is the breakdown product formed when one phosphate group is removed from adenosine triphosphate (ATP).
ATP is a high-energy molecule that provides energy to fuel various cellular processes. When a cell needs to perform work, such as muscle contraction or the synthesis of molecules, ATP releases one of its phosphate groups through a process called hydrolysis.
This breakdown of ATP releases energy, and ADP is formed as a result. ADP can be further broken down to adenosine monophosphate (AMP) by the removal of another phosphate group. The energy released during the breakdown of ATP and ADP is essential for various cellular processes, including metabolism and signal transduction.
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the most powerful flexor of the forearm at the elbow is the:_____.
At the elbow, the forearm's strongest flexor is the: Brachialis. The correct answer is (D).
In the absence of supination, the brachialis muscle is the strongest elbow flexor. However, when the elbow is in supination or flexion, the brachialis muscle loses more mechanical momentum than the biceps brachialis muscle.
The major flexor of the elbow is the brachialis. It has more strength than the coracobrachialis and biceps brachii due to its larger cross-sectional area. Due to the biceps brachia's function as a supinator and flexor, the forearm must be in pronation in order to isolate the brachialis muscle.
Although there are a few muscles that cross the elbow that are involved in the movement of the wrist and fingers, only a small number of them move the elbow joint. The primary flexors of the elbow are the biceps brachii, brachialis, and brachioradialis. The primary extensor of the elbow is the triceps brachia.
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Q-The most powerful flexor of the forearm at the elbow is the _____.
a. Anconeus
b. Biceps brachii
c. Brachioradialis
d. Brachialis
what systems interacts with the digestive system and how do they work together?
Answer:
Circulatory System
Explanation:
The Digestive System takes the nutrients from the food you intake and hands it off to the Circulatory system. Once this handoff has occurred, the Circulatory System then brings the nutrients to where it is needed the most.
Answer:
Circulatory System
Explanation:
I did the test
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true or false? the sex chromatin body (barr body) represents an inactivated x chromosome and is only encountered in persons having two x chromosomes and lacking a y chromosome.
The given statement "The sex chromatin body (barr body) represents an inactivated X chromosome and is only encountered in persons having two X chromosomes and lacking a Y chromosome is true because the presence of a Barr body is a useful diagnostic tool in identifying individuals with specific chromosomal abnormalities or disorders.
This process of inactivation, known as Lyonization, occurs randomly in females during embryonic development, and one of the two X chromosomes becomes inactivated in each cell. This mechanism ensures that the dosage of X-linked genes is balanced between males (who have one X chromosome) and females (who have two X chromosomes).
The inactivated X chromosome condenses into a visible Barr body, which can be identified by staining cells with specific dyes. Although the Barr body is typically associated with females, some males with certain genetic conditions can also have a Barr body due to X chromosome abnormalities.
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A cloning host has the following desirable features EXCEPT:A. fast growth rateB. pathogenicityC. well-mapped genomeD. keeps foreign genes intact
The correct answer is B. Pathogenicity. A cloning host typically has a fast growth rate, which allows for the quick and efficient production of large amounts of cloned DNA.
It may also have a well-mapped genome, which can make it easier to manipulate and study genetic material. Additionally, a desirable cloning host will keep foreign genes intact, meaning that the cloned DNA is not degraded or lost during the cloning process. However, pathogenicity is not a desirable feature in a cloning host, as it can pose a risk to the safety of those working with the host organism.
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Question 52 The fertilization of ovules from plant Q by pollen from plant results in the production of seeds. What percent of the genes in each offspring's chloroplasts will have been inherited from plant R?
In plants, chloroplasts contain their own DNA and are inherited maternally through ovules. Therefore, the genes in the offspring's chloroplasts after fertilisation will only come from plant Q.
In plants, the chloroplasts are inherited from the female parent. Therefore, in this scenario, the chloroplasts in the offspring's cells will have been inherited from plant Q, which is the female parent. The pollen from plant R, which is the male parent, does not contribute any chloroplasts to the offspring.
Therefore, all of the chloroplast genes in the offspring will have been inherited from plant Q. The percentage of genes that will have been inherited from plant R will depend on the specific genes in the offspring's nuclear DNA.
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Which is the primary purpose of using restriction enzymes in gel electrophoresis?
The primary purpose of using restriction enzymes in gel electrophoresis is to cut DNA into smaller fragments of specific sizes. Restriction enzymes are enzymes that recognize and cut DNA at specific nucleotide sequences, called restriction sites.
By using different restriction enzymes, DNA can be cut into different sizes, allowing for the separation of DNA fragments by size in gel electrophoresis. Gel electrophoresis is a laboratory technique that separates DNA molecules based on their size and charge, allowing scientists to analyze and visualize DNA fragments of different sizes. By cutting DNA with restriction enzymes before running gel electrophoresis, scientists can generate DNA fragments of known sizes and compare them to other DNA samples to identify similarities or differences. This is a common technique used in DNA fingerprinting, genetic engineering, and other areas of molecular biology.
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if a food-borne disease requires ingestion of the pathogen, growth of the organism, and release of toxins in the intestine, it is referred to as a food-borne ;
If a food-borne disease requires ingestion of the pathogen, growth of the organism, and release of toxins in the intestine, it is referred to as a food-borne intoxication.
Food-borne intoxication is an illness brought on by eating food that has been contaminated with bacteria, viruses, parasites, or toxins.
The most frequent causes of food poisoning are infectious organisms or their poisons. Symptoms of food poisoning can include cramps, nausea, vomiting, or diarrhea.
Most cases of food poisoning are minor and get better on their own. The most crucial part of treatment is making sure the patient is well hydrated.
This type of food-borne illness occurs when a person consumes food containing toxins produced by the bacteria or pathogens present in the food. These toxins then cause symptoms of the illness.
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an octogenarian ask the nurse practitioner if it's okay for him to have an alcoholic beverages in the evening
Because lean body mass and total body water diminish with age, alcohol consumption should be restricted or avoided. An elderly man asks the nurse practitioner (NP) if it's okay for him to drink alcohol in the evenings. Hence (b) is the correct option.
There are no restrictions.The youngster should be able to sketch a human with a body and at least six body parts by the time they are five years old. She should also be learning to tie her shoelaces, be able to imitate triangles and other geometric designs, and be able to dress and undress herself.As an inactive vaccine, Shingrix can be given with other inert or live vaccines.
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An octogenarian asks the nurse practitioner if its OK for him to have an alcoholic beverage in the evenings. There is no obvious contraindication. How should the nurse practitioner respond?
a. yes, but not more than 4 days/week
b. yes, but not more then 1-2 drinks per day
c. no, you will increase your risk of falling and injury
d. it depends of the type of alcohol you would like to consume
The osmolarity of the filtrate is virtually the same at the entrance and exit of the ____________ .
The osmolarity of the filtrate is virtually the same at the entrance and exit of the proximal tubule in the nephron of the kidney.
The proximal tubule is the first segment of the renal tubule, and it plays a crucial role in the reabsorption of water and solutes from the filtrate. The cells lining the proximal tubule are highly permeable to water and solutes, and they actively transport sodium, glucose, amino acids, and other molecules from the filtrate into the interstitial fluid.
As a result of this active transport, the osmolarity of the filtrate is reduced, and the composition of the filtrate is altered. However, because the reabsorption is isotonic, the osmolarity of the filtrate remains essentially unchanged from the entrance to the exit of the proximal tubule.
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the orifice between the esophagus and the stomach is bounded by a thin band of circular ______ muscle, the ______ esophageal sphincter.
The orifice between the esophagus and the stomach is bounded by a thin band of circular smooth muscle, the lower esophageal sphincter (LES). The orifice between the esophagus .
the stomach is surrounded by a circular band of muscle known as the lower esophageal sphincter (LES). This muscle acts as a valve, opening to allow food and liquid to pass from the esophagus into the stomach, and then closing to prevent stomach contents from refluxing back into the esophagus. The LES is a complex structure consisting of smooth muscle fibers and a layer of specialized cells called the intrinsic esophageal sphincter cells. Dysfunction of the LES can lead to conditions such as gastroesophageal reflux disease (GERD) and hiatal hernias, which can cause symptoms such as heartburn, chest pain, and difficulty swallowing.
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Locate the sticklebacks (a small fish) in this food web. Sticklebacks are a consumer of euphausiids. If the gray whales were gone, would consumption of euphausiids by sticklebacks be likely to compensate for the diminished predation pressure on the euphausiids? Why or why not?
Answer: Probably not.
Explanation: I think there are two main arguments for why not.
1) the euphausiids population is adjusted to predation from both sticklebacks AND gray whales, loss of one of these predators would affect the euphausiid population (assuming the population of sticklebacks is does not grow suddenly and massively after the disappearance of gray whales.
2) the amount of euphausiids consumed by gray whales versus sticklebacks is vastly different. Sticklebacks are small fish, it would take quite a few of them to eat as much as one gray whale would.
The sticklebacks (a small fish) in this food web. Sticklebacks are a consumer of euphausiids, if the gray whales were gone it is possible that sticklebacks would consume more euphausiids for the diminished predation pressure because they face predation pressure from other predators
Sticklebacks are a consumer in the food web, and they feed on euphausiids, which are a type of small crustacean. If gray whales were to disappear from the food web, it is possible that sticklebacks would consume more euphausiids to compensate for the reduced predation pressure on them. However, whether or not this would happen depends on a variety of factors. For example, the abundance of euphausiids could be limited by other factors such as competition or environmental conditions, which would make it difficult for sticklebacks to increase their consumption.
Additionally, the sticklebacks themselves may face predation pressure from other predators in the food web, which could limit their ability to increase their consumption of euphausiids. Ultimately, the impact of the disappearance of gray whales on the consumption of euphausiids by sticklebacks is difficult to predict and would depend on many different factors.
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what is the most likely cause for cloudy centers appearing in lambda plaques on an escherichia coli-covered plate?
The most likely cause for cloudy centers appearing in lambda plaques on an Escherichia coli-covered plate is the presence of a lysogenic phage.
A lysogenic phage is a bacteriophage that incorporates its DNA into the host bacterial genome instead of immediately killing the host bacterium. This results in a non-lytic infection, where the host bacterium continues to grow and divide, carrying the prophage (integrated phage DNA) within its genome. The cloudy centers in lambda plaques are indicative of this non-lytic infection, as the bacterial population in the center of the plaque remains partially intact and continues to grow, creating a hazy appearance.
Another possible cause could be the presence of bacterial contamination. The contamination could be caused by other bacteria or viruses that are present in the sample or on the plate, leading to a mixed population of cells that can affect the growth of lambda phage and result in cloudy centers in the plaques.
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salmon eggs hatch in fresh water. the fish then migrate to the ocean and, after several years of feeding and growing, return to fresh water to breed. how can these organisms make the transition from fresh water to ocean water and back to fresh water?
Salmon are able to make the transition from fresh water to ocean water and back again due to their unique physiology and behavior.
In general , salmon are able to make the transition from fresh water to ocean water and back again due to their unique ability to regulate salt and water balance within their bodies and undergo physiological changes to adapt to changing environments.
During this transition, they undergo physiological changes that allow them to adapt once again to low-salinity environments. They stop drinking seawater and instead rely on freshwater sources. They also stop excreting excess salt and instead retain it in their bodies to help them adapt to the low-salt environment.
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sand dollars, sea stars, brittle stars, sea cucumbers, and sea urchins are all examples of , the group of invertebrates most closely related to chordates.
The group of invertebrates that are most closely related to chordates echinoderms includes sea urchins, sand dollars, sea stars, brittle stars, sea cucumbers, and sea cucumbers.
What is echinoderms?A species of marine invertebrate known as an echinoderm has an internal endoskeleton and a hard, spiky skin. They can be discovered in every ocean, usually at depths of under 200 metres. Echinoderms, which include sea stars, sea urchins, sand dollars, sea cucumbers, and brittle stars, are thought to number over 7,000 different species. These creatures contribute to the wellbeing of their surroundings by feeding larger predators and keeping the ocean floor clean.
Echinoderms have a number of distinctive traits. The majority feature five-fold symmetry, which makes them simple to identify. Additionally, they have a water circulatory system that facilitates movement and feeding. The animals' bodies are lined with tiny tube feet and canals that make up this system.
Echinoderms mostly consume tiny particles like plankton and algae. Some species may even catch small prey using their tube feet to catch food. Additionally, they rely on their strong shells and spines to protect them from predators.
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The complete question is,
The class of invertebrates most closely related to chordates includes, among others, sand dollars, sea stars, brittle stars, sea cucumbers, and sea urchins.
The counterclockwise rotation of flagella propels a bacterium forward through a solution. What happens if the bacterium reverses the direction of rotation?
The bacterium stops briefly and then continues to move in the same direction.
The bacterium moves backwards.
The bacterium tumbles with no definite direction.
The bacterium remains in place until the direction of rotation is reversed again.
If the bacterium reverses the direction of rotation the bacterium tumbles with no definite direction, the correct option is (c).
The counterclockwise rotation of flagella propels a bacterium forward through a solution. However, if the direction of rotation is reversed clockwise, the flagella push against the fluid in the opposite direction, causing the bacterium to change its direction.
As a result, the bacterium undergoes a random change in its direction and orientation, known as "tumbling." During tumbling, the flagella rotate randomly, which prevents the bacterium from moving in a particular direction. Tumbling allows the bacterium to explore its environment and find a more favorable direction to move towards, the correct option is (c).
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The complete question is:
The counterclockwise rotation of flagella propels a bacterium forward through a solution. What happens if the bacterium reverses the direction of rotation?
a. The bacterium stops briefly and then continues to move in the same direction.
b. The bacterium moves backward.
c. The bacterium tumbles with no definite direction.
d. The bacterium remains in place until the direction of rotation is reversed again.
A group of celery stalks is immersed in 100% pure water for several hours. This group of celery becomes stiff. A second group of celery stalks is left in a salt solution. The second group becomes soft. In terms of osmosis, why are there two different outcomes for the celery?
The two different outcomes for the celery stalks are due to the process of osmosis. Osmosis is the movement of water molecules from an area of higher concentration to an area of lower concentration across a semipermeable membrane.
In the first group of celery stalks immersed in pure water, the concentration of water molecules outside the cells was higher than inside the cells, causing water to move into the cells through the cell membrane. This caused the cells to swell and become stiff. In the second group of celery stalks left in a salt solution, the concentration of salt molecules outside the cells was higher than inside the cells, causing water to move out of the cells through the cell membrane. This caused the cells to shrink and become soft. Therefore, the different outcomes for the celery are due to the differences in the concentration of water and salt molecules outside and inside the cells, which affected the direction of water movement through the cell membrane.
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The osmosis mechanism is what causes the two distinct results for the celery stalks. Osmosis is the transfer of water molecules through a semipermeable membrane from a region of greater concentration to an area of lower concentration.
The 100% pure water concentration of water molecules outside the cells in the first set of celery stalks submerged in pure water was greater than within the cells, allowing water to enter the cells through the cell membrane.
The cells grew larger and stiffer as a result of this. The concentration of salt molecules outside the cells in the second set of celery stalks kept in a salt solution was greater than within the cells, leading water to pass through the cell membrane and leave the cells.
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