CN VII in the periphery. This condition produces paralysis of the upper and lower face, which is false.
CN VII in the periphery. This condition produces paralysis of the upper and lower face, which is false. CN VII, also known as the facial nerve, is responsible for facial movements. When there is peripheral damage to CN VII, it typically produces paralysis of the entire side of the face, not just the upper and lower face. A lesion or deficit affecting CN VII can result in several issues, including facial weakness or paralysis on the affected side (Bell's palsy), loss of taste sensation in the anterior two-thirds of the tongue, and reduced tear and saliva production. Treatment and recovery depend on the cause and severity of the lesion or deficit.
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Which of the following statement is NOT TRUE regarding the Yersinia pestis type 3 secretion system (T3SS)? O The injectosome (aka Yop secretion machinary) directly delivers effector Yop proteins into target eukaryotic cells O Expression of genes encoding the Yop secretion machinary (aka "injectosome") is induced at 28 C. O Yop proteins do not require Sec-dependent secretion signals to be delivered to host cells by the T3SS. O Secretion of Yop proteins through the injectosome is energy- dependent Secreted acyl-homoserine lactones (AHLs) are hydrophilic and therefore need to be actively transported outside of the bacterial cell. O True O False
The statement that is NOT TRUE regarding the Yersinia pestis type 3 secretion system (T3SS) is: O Expression of genes encoding the Yop secretion machinery (aka "injectosome") is induced at 28 C. This statement is false because the expression of genes encoding the Yop secretion machinery (injectosome) is actually induced at 37°C, not 28°C.
The other statements are true. The injectosome, also known as the Yop secretion machinery, directly delivers effector Yop proteins into target eukaryotic cells. Yop proteins do not require Sec-dependent secretion signals to be delivered to host cells by the T3SS, and the secretion of Yop proteins through the injectosome is energy-dependent. The statement about secreted acyl-homoserine lactones (AHLs) being hydrophilic and needing to be actively transported outside of the bacterial cell is also true.
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The imaginary, vertical plane that cuts through the body in such a manner as to divide it into two symmetrical halves is referred to as the
The imaginary, vertical plane that cuts through the body in such a manner as to divide it into two symmetrical halves is referred to as the midsagittal plane. This plane is also known as the median plane and it runs down the center of the body, dividing it into left and right symmetrical halves.
The midsagittal plane is essential in anatomical and medical studies as it helps to understand the structure and function of different body parts. It is used as a reference point to describe the position of different organs, bones, and tissues in the body. Moreover, medical professionals use this plane to locate the precise location of abnormalities, injuries, and diseases. In summary, the midsagittal plane is a vital component of the body's anatomy and is essential in various medical applications. Answer in 200 words vertical plane, body, and symmetrical halves.
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With the exception of the teeth, the hardest tissue in the body is:
A) bone
B) muscle
C) the brain
D) marrow
The awnser is A hope that helps Lol Its the hardest because a bone is atleast One of the hardest things in your body and your teeth have bones
What things made of atoms do you see in the video
Things made of atoms do you see in the video Both the blue ocean and the dolphin are comprised of molecules.
The smallest particle of a substance that has the potential to participate in a chemical reaction is the subject of the term "atom." Atoms make up every substance in the universe, as we are aware. The food we eat, the clothes we wear, and even the air we breathe all contain atoms.
Therefore, there are atoms everywhere in the universe. Additionally, living things are made up of a variety of atoms in varying proportions. Viewing the picture from our perspective, there is the blue ocean and there is a dolphin. Atoms makeup both the dolphin and the blue sea.
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Q-What things made of atoms do you see in the video?
Think about what you learned about atoms and elements so far in the unit and the information you already know.
the ability of oxygen to bind hemoglobin changes with altitude as shown. which statement, if true, best explains this observation?
The ability of oxygen to bind hemoglobin changes with altitude due to the phenomenon of altitude-induced hypoxia. As altitude increases, the atmospheric pressure decreases, resulting in a decrease in the partial pressure of oxygen (PO2). This decrease in PO2 leads to a reduced availability of oxygen for binding to hemoglobin in the blood.
At lower altitudes, where the atmospheric pressure is higher, hemoglobin has a higher affinity for oxygen and readily binds to oxygen molecules, allowing for efficient oxygen transport to tissues and organs. However, at higher altitudes, where the atmospheric pressure is lower, the reduced PO2 leads to a decreased affinity of hemoglobin for oxygen. This means that hemoglobin is less likely to bind to oxygen molecules, resulting in reduced oxygen transport to tissues and organs.
This phenomenon is an adaptive response of the body to low oxygen availability at high altitudes. It helps to facilitate oxygen unloading from hemoglobin in tissues with lower oxygen tension, ensuring that oxygen is delivered to cells that need it the most. This is important for acclimatization to high altitudes, allowing the body to cope with reduced oxygen availability by adjusting the binding affinity of hemoglobin to oxygen.
Overall, the ability of oxygen to bind hemoglobin changes with altitude due to the altered atmospheric pressure and oxygen tension, and this adaptive response helps the body adapt to high altitudes and maintain oxygen delivery to tissues and organs.
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choose the answer that best completes the blanks of this sentence in order. The two major groups of parasitic flatworms include the ____ with a long, ribbon-like body and the ____ with a flat, ovoid body.
The two major groups of parasitic flatworms include the cestodes with a long, ribbon-like body and the trematodes with a flat, ovoid body.
Cestodes, also known as tapeworms, have a long, ribbon-like body divided into segments called proglottids. They typically live in the intestines of their host and absorb nutrients through their body surface. Cestodes have a specialized structure called a scolex, which allows them to attach to the host's intestinal wall. Some common examples of cestodes are Taenia solium (pork tapeworm) and Taenia saginata (beef tapeworm).
Trematodes, commonly referred to as flukes, have a flat, ovoid body and are generally smaller than cestodes. They have a complex life cycle involving multiple hosts, usually including a snail as an intermediate host. Trematodes can be found in various organs of their host, such as the liver, lungs, or blood vessels. A well-known example of a trematode is Schistosoma, which causes the disease schistosomiasis.
In summary, the two major groups of parasitic flatworms are cestodes, which have a long, ribbon-like body, and trematodes, which have a flat, ovoid body. These organisms can cause various diseases in humans and animals, and understanding their biology is crucial for effective treatment and prevention.
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To which of the major classes of enzymes does the enzyme that catalyzes each of the following reactions belong?.
a) A cis double bond is converted to a trans double bond.
b) An alcohol is dehydrated to form a compound with a double bond.
c) An amino group is transferred from one substrate to another.
d) An ester linkage is hydrolyzed.
a) The enzyme that catalyzes the conversion of a cis double bond to a trans double bond belongs to the class of enzymes called isomerases. Isomerases catalyze the rearrangement of molecules to form isomers, which are molecules with the same chemical formula but a different arrangement of atoms.
b) The enzyme that catalyzes the dehydration of an alcohol to form a compound with a double bond belongs to the class of enzymes called dehydratases. Dehydratases catalyze the removal of water molecules from molecules.
c) The enzyme that catalyzes the transfer of an amino group from one substrate to another belongs to the class of enzymes called transferases. Transferases catalyze the transfer of functional groups, such as amino groups, from one molecule to another.
d) The enzyme that catalyzes the hydrolysis of an ester linkage belongs to the class of enzymes called hydrolases. Hydrolases catalyze the cleavage of bonds using water molecules.
a) The enzyme that converts a cis double bond to a trans double bond belongs to the class of enzymes called isomerases. Specifically, this type of reaction is catalyzed by cis-trans isomerases.
b) The enzyme that dehydrates an alcohol to form a compound with a double bond belongs to the class of enzymes called lyases. In this case, the specific enzyme responsible for this reaction is a dehydration synthase, which falls under the lyase category.
c) The enzyme that transfers an amino group from one substrate to another belongs to the class of enzymes called transferases. A specific type of transferase that performs this reaction is an aminotransferase, also known as a transaminase.
d) The enzyme that hydrolyzes an ester linkage belongs to the class of enzymes called hydrolases. Esterases, a subcategory of hydrolases, are specifically responsible for breaking ester linkages in these reactions.
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reset if during anaphase 1, n1 segregates with n2 and t1 segregates with t2, the gametes will be because each gamete will have . the n1 n2 and t1 t2 pairs are called the pairs because of their orientation in the cross-shaped diagram. other segregation patterns are possible. n1 may segregate with t2, and n2 may segregate with t1. these are called pairs because of their orientation in the diagram. less frequently, n1 may segregate with t1, and n2 may segregate with t2. these are termed pairs. the adjacent-1 and adjacent-2 combinations produce gametes that are . in each combination, and two copies of other genes exist. when this translocation heterozygote is mated with a normal individual, one can expect that about half of the zygotes will be inviable. this condition is termed .
If during anaphase 1, n1 segregates with n2 and t1 segregates with t2, the gametes will be heterozygous for the translocation. This is because each gamete will have one normal chromosome (containing n1 and t1 or n2 and t2) and one translocated chromosome (containing n2 and t1 or n1 and t2).
The n1n2 and t1t2 pairs are called the alternate pairs because of their orientation in the cross-shaped diagram. Other segregation patterns are possible, such as n1 segregating with t2 and n2 segregating with t1, which are also termed alternate pairs. Less frequently, n1 may segregate with t1 and n2 may segregate with t2, which are called adjacent pairs. The adjacent-1 and adjacent-2 combinations produce gametes that are imbalanced in each combination, and two copies of other genes exist. When this translocation heterozygote is mated with a normal individual, one can expect that about half of the zygotes will be inviable. This condition is termed semisterility.
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At any time t 0, in days, the rate of growth of a bacteria population is given by y' = ky, where y is the number of bacteria present and k is a constant. The initial population is 1,500 and the population is quadrupled during the first 2 days. By what factor will the population have increased during the first 3 days? (A) 4 (B) 5 (C) 6 (D) 8 (E) 10
The rate of growth of the bacteria population will increase by a factor of 64 during the first 3 days, so 8 is the factor value. Option D is the correct answer.
Since the population is quadrupled during the first 2 days, the population at the end of the 2nd day will be 1,500 x 4 = 6,000.
Using the given differential equation, we can solve for y:
y' = ky
dy/y = k dt
Integrating both sides:
ln|y| = kt + C
where C is a constant of integration.
Using the initial condition y(0) = 1,500, we get:
ln|1,500| = C
C = ln|1,500|
Therefore, the population equation is:
ln|y| = kt + ln|1,500|
ln|y/1,500| = kt
y/1,500 = [tex]e^{(kt)}[/tex]
To find the factor by which the population will increase during the first 3 days, we need to find y(3):
y/1,500 = [tex]e^{(kt)}[/tex]
y/1,500 = [tex]e^{(k*3)}[/tex]
y/1,500 = ([tex]e^k[/tex])³
y/1,500 = (y/1,500)_0 × 4³
y/1,500 = (y/1,500)_0 × 64
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Binary fission results in four daughter cells from one parent cell
True or False
Answer:
false
Explanation:
Binary fission results in four daughter cells from one parent cell is False.
Thus, Prokaryotes (bacteria and archaea) (a type of asexual reproduction) frequently exhibit binary fission.
A single parent cell divides into two daughter cells in a process known as binary fission, producing two daughter cells from a single parent cell.
It is a cell division process that results in two identical daughter cells, each of which has a copy of the genetic material of the parent cell. Binary fission produces two daughter cells rather than four as a result.
Thus, Binary fission results in four daughter cells from one parent cell is False.
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The ability of water to rise inside a tube of narrow diameter against the pull of gravity is called
The ability of water to rise inside a tube of narrow diameter against the pull of gravity is called Capillary action
Capillary action is the ability of a liquid to flow in narrow spaces without the assistance of, or even in opposition to, external forces like gravity. When a tube with a narrow diameter, such as a thin straw or a thin glass tube, is inserted into a liquid, the liquid rises inside the tube due to capillary action. This is because of the attraction between the liquid molecules and the walls of the tube. The narrower the tube, the higher the liquid will rise. Capillary action is important in many natural processes, such as the movement of water in plants and the formation of blood clots in the human body.
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Order the steps required to analyze gene expression from a particular cell type using a DNA microarray. First step Last step Answer Bank reverse transcribe mRNA to cDNA add cDNA to microarray and incubate wash away unbound cDNA
label cDNA with a chemiluminescent molecule extract mRNA from cells
The correct order of steps required to analyze gene expression from a particular cell type using a DNA microarray is as follows:
1. Extract mRNA from cells.
2. Reverse transcribe mRNA to cDNA.
3. Label cDNA with a chemiluminescent molecule.
4. Add cDNA to microarray and incubate.
5. Wash away unbound cDNA.
The first step is to extract mRNA from the cells of interest. Next, the extracted mRNA is reverse transcribed into cDNA using a reverse transcriptase enzyme. The cDNA is then labeled with a chemiluminescent molecule to allow detection on the microarray. The labeled cDNA is added to the microarray and incubated, allowing for hybridization with the complementary DNA probes on the microarray. Finally, any unbound cDNA is washed away to allow for accurate detection and analysis of the gene expression levels on the microarray.
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In the 1970s, Fernando Nottebohm and colleagues identified the major structures controlling birdsong as:
In the 1970s, Fernando Nottebohm and colleagues identified the major structures controlling birdsong as the high vocal center (HVC), the robust nucleus of the arcopallium (RA), and the lateral magnocellular nucleus of the anterior nidopallium (LMAN).
These structures are involved in the learning, production, and modulation of birdsong. The HVC is responsible for coordinating the timing and sequence of song syllables, while the RA is involved in the motor control of song production. The LMAN plays a role in song learning and plasticity. Together, these structures form a complex neural network that allows birds to produce and modify their songs.
Hi! In the 1970s, Fernando Nottebohm and his colleagues identified the major structures controlling birdsong as the song control system, which consists of interconnected brain nuclei such as the High Vocal Center (HVC), the Robust nucleus of the Arcopallium (RA), and the Anterior Forebrain Pathway (AFP). These structures play crucial roles in learning, producing, and maintaining complex vocalizations in birds.
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A paper company located on the banks of a river discharges its treated wastewater into the river. Which of the following would be the best control group to evaluate the treated wastewater from the paper company? - A sample of water downstream from the same river - A sample of water upstream from the same river - A sample of distilled water - A sample of water from a nearby river
The best control group to evaluate the treated wastewater from the paper company would be a sample of water upstream from the same river. This allows for a comparison between the water before it's affected by the company's wastewater discharge and the water after the discharge, giving you an accurate assessment of the impact of the treated wastewater on the river's water quality.
The best control group to evaluate the treated wastewater from the paper company would be a sample of water upstream from the same river. This is because it would provide a baseline for the natural state of the river and any changes or impacts from the discharged wastewater can be compared to it. A sample of water downstream from the same river would be affected by other sources of pollution and may not provide an accurate comparison. A sample of distilled water or water from a nearby river would not be relevant as they do not reflect the conditions of the river in question.
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acoustic energy (ear) to mechanical energy (bones) to hydrolic energy
what is the transmission path?
The transmission path is from the ear, to the bones in the middle ear, to the fluid in the cochlea, where the mechanical energy is transformed into hydraulic energy.
The ear is responsible for detecting sound waves in the environment and transforming them into neural signals that can be interpreted by the brain. The first step in this process is the conversion of acoustic energy (sound waves) into mechanical energy, which occurs when the sound waves reach the ear and cause the eardrum to vibrate. These vibrations are then transmitted through the middle ear via a chain of three small bones (the malleus, incus, and stapes) that amplify the vibrations and transmit them to the inner ear.In the inner ear, the mechanical energy is converted into hydraulic energy, as the vibrations of the middle ear bones cause fluid in the cochlea to move. This movement stimulates tiny hair cells in the cochlea, which convert the hydraulic energy into electrical signals that are transmitted to the brain via the auditory nerve. The brain then interprets these signals as sound.
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A bowling ball rolled directly toward the pins at a constant velocity for 2.4 seconds. It moved 18 meters in that time. What was its velocity?
Answer:
Explanation:
The velocity of the bowling ball is given by the formula:
velocity = distance / time
In this case, the distance covered by the ball is 18 meters and the time taken is 2.4 seconds. Therefore, the velocity of the bowling ball is:
velocity = 18 / 2.4
velocity = 7.5 m/s
So, the velocity of the bowling ball is 7.5 meters per second.
The rigid tube that descends from the larynx is called the _____.
The rigid tube that descends from the larynx is called the trachea, also known as the windpipe. It is a vital part of the respiratory system, which connects the larynx (voice box) to the bronchi of the lungs.
The trachea is a flexible, yet sturdy tube made up of cartilage rings stacked one on top of the other. The rings provide support and prevent the trachea from collapsing or closing during inhalation and exhalation. The trachea also has a lining of mucus-producing cells and tiny hair-like structures called cilia, which work together to help trap and remove debris and microorganisms from the air we breathe.
The trachea serves as a pathway for air to travel from the nose and mouth to the lungs, where oxygen is exchanged for carbon dioxide through tiny air sacs called alveoli. The trachea also plays a crucial role in protecting the lower respiratory system by preventing foreign particles from entering the lungs. Overall, the trachea is an essential component of the respiratory system that helps us breathe and maintain a healthy body.
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Bacteria that live attached to surfaces are described as ___________, whereas bacteria that are free-living, floating with a fluid environment are described as _________.
Bacteria that live attached to surfaces are described as biofilm, whereas bacteria that are free-living, floating with a fluid environment are described as planktonic.
Biofilms are communities of microorganisms that stick to a surface and are surrounded by a matrix of extracellular polymeric substances (EPS). Biofilms can form on a variety of surfaces, such as medical implants, teeth, and pipes, and can be beneficial or harmful to the host.
In contrast, planktonic bacteria are free-living and are not attached to any surface. They can be found in the water column, soil, or air and are more vulnerable to environmental stressors, such as antibiotics or disinfectants.
In summary, bacteria can exist as either biofilm or planktonic, depending on their mode of growth and lifestyle. Understanding the differences between these two forms of bacteria is crucial for developing effective strategies to control bacterial infections and biofouling.
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When saccharomyces was prepared for the budding slides, the yeast was mixed with warm water and sugar. Why was sugar added?
Sugar was added to the Saccharomyces preparation because yeast requires sugar as a source of energy to undergo fermentation and produce carbon dioxide, which causes the yeast cells to bud.
Sugar was added to the saccharomyces when preparing for budding slides to provide a source of energy for the yeast cells. Yeasts are unicellular fungi that primarily rely on glucose as their energy source for cellular respiration. In the absence of a readily available carbon source, yeast cells can enter into a dormant state, which may affect their ability to undergo budding and reproduce.
By adding sugar to the water, the yeast cells have access to a readily available source of glucose, which they can metabolize to produce energy and carry out essential cellular processes, including budding.
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Fluconazole is the prototype for antifungals. Which are characteristics of fluconazole? (Select all that apply.)
-Binds to sterols in the fungal cell membrane
-Treatment of oropharyngeal, esophageal, and vaginal candidiasis
-Metabolized in the liver and excreted in the urine
Answer:
All of the above characteristics are true of fluconazole. Fluconazole binds to sterols in the fungal cell membrane, making it an effective antifungal agent. It is commonly used to treat oropharyngeal, esophageal, and vaginal candidiasis. Fluconazole is metabolized in the liver and excreted in the urine.
Explanation:
The characteristics of fluconazole are:
Treatment of oropharyngeal, esophageal, and vaginal candidiasis: Fluconazole is an antifungal medication that is commonly used to treat fungal infections such as thrush, esophageal candidiasis, and vaginal candidiasis. It works by inhibiting the growth of the fungi responsible for the infection.Binds to sterols in the fungal cell membrane: Fluconazole belongs to the class of azole antifungal medications, which work by inhibiting the synthesis of ergosterol, an essential component of the fungal cell membrane. By doing so, fluconazole disrupts the integrity of the fungal cell membrane, leading to the death of the fungal cells.Metabolized in the liver and excreted in the urine: Fluconazole is primarily metabolized in the liver by the cytochrome P450 enzyme system and is excreted in the urine. It has a long half-life, which allows for once-daily dosing in many cases.Therefore, the correct characteristics of fluconazole are that it is used for the treatment of oropharyngeal, esophageal, and vaginal candidiasis, binds to sterols in the fungal cell membrane, and is metabolized in the liver and excreted in the urine.
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Place the locations of the heart's conducting system in order, beginning with the location where the heartbeat is initiated.
Sinoatrial (SA) node
Left and right bundles
Atrioventricular (AV) bundle
Purkinje cells
Atrioventricular (AV) node
1
4
3
5
2
The order of the heart's conducting system is: SA node, AV node, AV bundle, left and right bundles, and Purkinje cells.
The conducting system of the heart begins with the initiation of the heartbeat at the Sinoatrial (SA) node, which is located in the upper right atrium. From there, the electrical impulses travel to the Atrioventricular (AV) node, which is located in the lower part of the right atrium. The impulses are then passed to the Atrioventricular (AV) bundle, also known as the bundle of His, which is a group of specialized fibers located in the septum of the heart that divides the left and right ventricles.
From the AV bundle, the impulses are then transmitted to the left and right bundles, which are branches of the AV bundle that extend throughout the ventricles. Finally, the impulses are spread to the Purkinje cells, which are specialized cells located in the walls of the ventricles that rapidly transmit the electrical impulses to the rest of the ventricular muscle cells, causing them to contract and pump blood out of the heart.
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A plant leaf is constructed from a variety of cell types with specialized structures and functions. Many of the properties of leaf cells facilitate some aspect of water transport.The diagram below shows a cross section through a leaf.Drag the labels to the appropriate targets to match the function with the structure indicated in the diagram. Labels may be used once, more than once, or not at all.A. cells that transport water from the roots to the leavesB. a group of different cell types involved in long-distance transport of water and nutrientsC. cells with a coating that prevents evaporation of waterD. cells with a coating that prevents evaporation of waterE. cells where most evaporation of water in the leaf occursF. cells that control the rate of water loss from the leaf
A plant leaf is constructed from a variety of cell types with specialized structures and functions. Many of the properties of leaf cells facilitate some aspects of water transport.
A. Xylem cells
B. Vascular bundle
C. Cuticle
D. Epidermis
E. Mesophyll cells
F. Stomata
A. Xylem cells - these cells transport water from the roots to the leaves.
B. Vascular bundle - this is a group of different cell types (xylem and phloem) involved in long-distance transport of water and nutrients.
C. Epidermal cells - these cells have a coating called the cuticle that prevents evaporation of water.
D. (Same as C)
E. Mesophyll cells - specifically, spongy mesophyll cells are where most evaporation of water in the leaf occurs.
F. Guard cells - these cells control the rate of water loss from the leaf by regulating the opening and closing of stomata.
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A(n) ________ can be described as a broad tendinous sheet.
a. aponeurosis
b. interstitium
c. fasciae
e. retinaculum
d. tympanum
The answer is "aponeurosis". It is a broad tendinous sheet that connects muscles to other muscles or to bones.
Aponeurosis is a strong, fibrous tissue that serves as a flat, broad attachment site for muscles and helps to distribute the forces generated by muscular contraction. It is composed of collagen fibres and forms a sheet-like structure that covers or encloses muscles, organs, or other structures. Aponeurosis also acts as a protective layer for underlying tissues and helps to maintain the shape of the body.
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PLEASE HELP
Calculate the average
distance that the DNA fragments traveled. How
much farther than the average distance did the
smallest DNA fragment travel?
Where the information on the histogram is considered,
the average distances travelled = 1.733cmThe smallest DNA fragment travelled 0.933 cm less thatn the average distance.How is this so?Average distance travelled is: (A + B + C + D + E + F)/6
Given:
Distance for A = 2.75cm
Distance for B = 1.25cm
Distance for C = 0.8cm
Distance for D = 1.75cm
Distance for E = 2.25cm
Distance for F = 1.4cm
Average distance = ( 2.75 + 1.25 + 0 .8 + 1.75 + 2.25 + 1.4) /6
Average distance = 1.733 cm
Smallest fragment distance = Distance of C - Average distance
= 0.8 - 1.733
= -0.933
Thus, the smallest DNA Fragment travelled farther than the average distance by 0.933cm.
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CN X, contols muscles in the pharynx and larynx, influences function of hte abdominal and thoracic visera, and carries sensory information from the abdomen and thorax to the brain. It's fxn can be assessed by inducing gag reflex.
true or fasle
True. CN X (Vagus nerve) controls muscles in the pharynx and larynx, influences the function of the abdominal and thoracic viscera, and carries sensory information from the abdomen and thorax to the brain. Its function can be assessed by inducing the gag reflex.
CN X, also known as the vagus nerve, controls muscles in the pharynx and larynx, influences the function of abdominal and thoracic viscera, and carries sensory information from the abdomen and thorax to the brain. Its function can be assessed by inducing the gag reflex.
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when an action potential reaches a synaptic knoc and the membran depolarizes ion channels open and calcium ions enter the cell, ion channels open and calcium clows out of the cell
When an action potential reaches a synaptic knob, the membrane depolarizes, causing ion channels to open. As a result, calcium ions enter the cell. This influx of calcium ions plays a crucial role in the process of neurotransmitter release and the propagation of the signal to the next neuron.
When an action potential reaches a synaptic knob, the depolarization of the membrane causes ion channels to open, allowing calcium ions to enter the cell. This influx of calcium triggers the release of neurotransmitters from the synaptic vesicles into the synaptic cleft. Once the neurotransmitters bind to their receptors on the post-synaptic neuron, ion channels open and ions such as sodium or potassium flow in or out of the cell, causing a change in the membrane potential. Eventually, ion channels open and calcium flows out of the cell, which helps to reset the system and prepare for the next action potential.
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What property allows O2 and CO2 to cross a lipid bilayer without the help of membrane protiens?
CC 7.2
The property that allows O2 and CO2 to cross a lipid bilayer without the help of membrane proteins is their small size and lipid solubility. Both O2 and CO2 are small and non-polar molecules that can easily dissolve in the hydrophobic interior of the lipid bilayer.
This allows them to diffuse across the membrane without the need for specialized transport proteins. The lipid bilayer acts as a barrier to larger or charged molecules, but small non-polar molecules like O2 and CO2 can easily pass through. This property is critical for the exchange of gases in living cells, where O2 is needed for cellular respiration and CO2 is produced as a waste product.
O2 and co2 are both nonpolar milecules that can easily pass through the hydrophobic interior of membrane.
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Throughout life we come into contact with microbes. The initial contact is during birth. As we grow and age, some of those contacts:
a. become normal flora
b. never colonize and are lost
c. result in infection
d. result in disease
e. all of the above
The answer is e. All of the above. Throughout our life, we come into contact with microbes, with the initial contact occurring during birth. As we grow and age, some of those contacts become normal flora, meaning they establish a beneficial relationship with our body.
This means that some microbes:
a. become normal flora, which are the harmless or beneficial microorganisms that colonize our body;
b. never colonize and are lost, meaning they do not establish a long-term presence on or within our body;
c. result in infection, which occurs when harmful microorganisms invade and multiply within the host; and
d. result in disease, when these infections cause negative effects on the host's health.
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just medial to the inferior part of the scapula lies the_____over which lung sounds can be heard.
Just medial to the inferior part of the scapula lies the "lateral thoracic wall" over which lung sounds can be heard.
1. Identify the location: The question refers to an area just medial (toward the midline) to the inferior part (lower portion) of the scapula (shoulder blade).
2. Determine the structure: The structure in this area is the lateral thoracic wall, which consists of the ribcage and the muscles covering it.
3. Lung sounds: Since the ribcage houses the lungs, this is the area where lung sounds can be heard using a stethoscope during auscultation.
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The green roof on Chicago's city hall is a great example of ___?
A. Urban Greening
B. Environmental Engineering
C. Carbon Promotion
D. Building for plants
The green roof on Chicago's city hall is a great example of urban greening. The correct option is A
What is Urban Greening ?
Urban greening is the practice of adding plants and greenery to urban areas. This can have a positive impact on the environment by lowering the urban heat island effect improving air and water quality and creating habitat for wildlife.
By increasing access to nature, offering recreational possibilities, improving air quality, lowering the urban heat island effect and enhancing the aesthetic appeal of the cityscape, urban greening seeks to improve the quality of life for urban dwellers.
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