Answer:
Explanation:
Nucleotide found in prokaryotes only is a nucleus-like irregularly shaped region within the prokaryotic cell that contains all or most of the genetic material.
DNA replication and gene expression can occur simulateously in prokaryotes only this is because prokaryotic cells lack a membrane bound nucleus, transcription and translation occurs simultaneously in the cytoplasm.
nucleus with nuclear membrane is a characteristics of eukaryotic cell only. The eukaryotic cell contain a membrane bound nucleus separated from the cytoplasm of the cell.
DNA is genetic material is the both prokaryotes and eukaryotes.
Diploid- Eukaryotes only are diploid; most eukaryotes have two sets of their genetic information, their DNA is organized into multiple linear chromosomes found in the nucleus.
DNA replication and gene expression are separated: This is a characteristics of eukaryotic cells only. DNA replication occurs in the membrane bound nucleus and mRNA transcript is transported to the cytoplasm for translation (gene expression)
Haploid- Prokaryotes are typically haploid, usually having a single circular chromosome found in the nucleoid.
The following F2 phenotypic data was obtained from a Drosophila testcross using an F1 offspring. Assume red eye and brown body are dominant wild-type phenotypes, and white eye and yellow body are mutant phenotypes.White-eye, brown body 670Red-eye, yellow body 650White-eye, yellow body 38Red-eye, brown body 561. What is the genotype of the F1? How do you know this?
Answer:
The genotype of the F1 was wy+/w+y.
Explanation:
One of the given options has a typo: the red eye-brown body offspring count should be 56 instead of 561.
We have two genes with two alleles each:
Red eyes (w+) is dominant over white eyes (w).
Brown body (y+) is dominant over yellow body (y).
The phenotypes of the F2 tesulting from a test cross (F1 x wy/wy) are:
wy+/ey (white-eye, brown body): 670w+y/wy (red-eye, yellow body): 650wy/wy (white-eye, yellow body): 38w+y+/wy (red-eye, brown body 56If the genes w and y are linked, two phenotypes in the F2 will be much more abundant than the other two. Recombination during meiosis is a rare event, so the most abundant phenotypes are the parentals (the ones present in the F1 parent).
Every individual in the offpsring has a wy chromosome, as this was the gamete inherited from the test cross individual.
In this case, the most abundant gametes are wy+ and w+y, so the genotype of the F1 was wy+/w+y.
Notice how when recombination occurs in the F1 parent, the recombinant gametes appear: wy and w+y+, which are the less abundant in the F2 progeny.
The F1 female fruit fly's genotype is likely XWXwCc, based on the 1:1:1:1 phenotypic ratio observed in the F2 generation from a test cross, indicating heterozygosity for X-linked eye color and autosomal body color traits.
To determine the genotype of the F1 Drosophila in the given question, we need to analyze the F2 phenotypic data from a test cross. The data shows a 1:1:1:1 ratio of phenotypes (white-eye, brown body: red-eye, yellow body: white-eye, yellow body: red-eye, brown body). This suggests that the F1 female was heterozygous for both traits. Considering that eye color is an X-linked trait, and the body color follows autosomal inheritance, the most likely genotype for the F1 female would be XWXwCc, where XW represents the wild-type allele for red eyes, Xw represents the mutant allele for white eyes, and Cc denotes the heterozygous condition for body color, with C being wild-type (brown body) and c being mutant (yellow body). This genotype would produce the observed F2 phenotypes when crossed with a male that is phenotypically mutant for both traits (XwYcc).
Valerie Homework. Unanswered Valerie is a studious 15 year old who is also a valuable member of her school's field hockey team. They are 15-1 this year and have a chance at winning the state championship. However, Valerie recently began feeling ill with noticeable signs of a high fever. After several days, her parents decided to take her to the ER and the attending physician admitted her to the hospital with an ongoing bacterial infection. After several more days, doctors are perplexed that her body is unable to fight off the infection. They decide to take a blood sample and have it sent to the lab for analysis. Lab results indicate that the vacuoles in her white blood cells are successfully trapping the bacteria. However, the bacteria are not being digested by the cell. Which organelle is likely disrupted?
А. Peroxisome
B. Mitochondria
C. Lysosome.
D. Smooth ER
E. Rough ER
F. Golgi Apparatus
Answer:
C) Lysosome.
Explanation:
Lysosomes are the membrane-enclosed small vesicles formed from the Golgi apparatus. The lysosomes contain enzymes like RNAase, DNAase, protease and other enzymes which can digest the biomolecules and maintains the pH at 5.0.
The acidic environment and the digestive enzymes help lysosomes digest the foreign material and the cellular debris.
In the given question, since the cell is not able to digest the bacteria and its components therefore the affected organ is the lysosomes.
Thus, Option-C is the correct answer.
What is the smallest unit of matter
Answer: Atom
Explanation: The atom is the smallest unit and most basic unit of matter. The atom is made up three subatomic particles and they are called the protons, neutrons, and electrons. The protons and neutrons are stored in the core of the atom and the electrons are located in the energy levels.
The smallest unit of matter is an atom. Atoms are the smallest quantity of an element that retains the unique properties of that element. They are made up of subatomic particles such as protons, electrons, and neutrons.
Explanation:The smallest unit of matter is an atom. Atoms are the smallest quantity of an element that retains the unique properties of that element. They are made up of subatomic particles such as protons, electrons, and neutrons. For example, a molecule of water consists of two hydrogen atoms and one oxygen atom bonded together.
Learn more about smallest unit of matter here:https://brainly.com/question/32286191
#SPJ6
Water is composed of one oxygen atom covalently boded to two hydrogen atoms. Select all of the
following statements that are true concerning water.
A: The hydrogen atoms are surrounded by fewer electrons than the oxygen atom.
B: The oxygen atom is slightly negatively charged.
C: The hydrogen atoms in the molecule are slightly negatively charged.
D: The electrons in the covalent bonds will be shared unequally between the hydrogen and oxygen atoms.
E: The molecule contains two nonpolar covalent bonds.
F: The oxygen atom has a greater attraction for the electrons in the surrounding covalent bonds than the
hydrogen atoms.
Answer:
F: The oxygen atom has a greater attraction for the electrons in the surrounding covalent bonds than the
hydrogen atoms.
Explanation:
Between hydrogen and oxygen atoms exist a covalent bond. Oxygen is more electronegative than hydrogen atom hence its ability to get attracted to itself more electrons than hydrogen.
Answer: Option A, B, D and F.
Explanation:
Water is a universal solvent I.e everything dissolves in it. Water molecules consist of an oxygen atom and two hydrogen atoms linked by covalent bond. The oxygen atom is electronegative and it tends to attract more electrons to itself, there is unequal sharing of electrons which make the oxygen slightly negatively charged and hydrogen slightly positively charged, making it polar molecules I.e unlike charges attract each other.Electrons in the covalent bond spend more time around oxygen atom and less around hydrogen atoms.
Lysosomes are membranous organelles that contain digestive enzymes. Lysosomes can function inside the cell, where their enzymes digest particles taken in by endocytosis or worn‑out cell components. Lysosomes can also release their enzymes outside the cell, where the enzymes break down extracellular material. I‑cell disease is a lysosomal storage disease that results in the buildup of carbohydrates, lipids, and proteins as inclusion bodies within the cell. Which is the probable cause?
Answer:
I cell disease is caused by a mutation in GNPTA gene that leads to deficiency in the enzyme UDP-N-acetylglucoseamine-1-phosphotransferase.
Explanation:
I cell disease (mucolipidosis) is a rare inherited lysosomal storage disease that results in the buildup of carbohydrates, lipids, and proteins as inclusion bodies within the cell GlcNAc-1-phosphotransferase catalyzes the N-linked glycosylation of asparagine residues with a molecule called mannose-6-phosphate (M6P). M6P acts as an indicator of whether a hydrolase should be transported to the lysosome or not. Once a hydrolase indicates an M6P, it can be transported to a lysosome. Mutation in this gene causes this disease.
I-cell disease is caused by a defect in lysosomal enzyme transport that prevents these enzymes from reaching lysosomes. As a result, undesirable substances build up within the cells instead of being broken down and disposed of, leading to the manifestation of I-cell disease symptoms.
Explanation:I-cell disease is caused by a defect in the process of lysosomal enzyme production or transport, preventing these enzymes from reaching the lysosome. These enzymes are key in breaking down carbohydrates, lipids, and proteins. If they cannot reach the intended location within the lysosome, these substances build up as inclusion bodies within the cell.
Normally, lysosomes function much like a 'garbage disposal' within the cell. They take in damaged and unneeded cellular components, and with the help of their enzymes, break down these materials. This cleanup process is crucial for removing foreign invaders such as bacteria as well. This is achieved by endocytosis and autophagy processes.
In the case of I-cell disease, the failure of digestive enzymes to reach lysosomes results in the ineffective breakdown of foreign and unwanted intracellular materials, causing a buildup of these substances within the lysosome organelle. This 'traffic jam’ of materials within cells eventually leads to the symptoms observed in I-cell disease.
Learn more about Lysosomal Dysfunction here:https://brainly.com/question/14086696
#SPJ11
Give an example of a trait that may have evolved as a result of the handicap principle and explain your reasoning.
Answer:
The handicap principle may be defined as the evolution on the basis of the reliable or honest signalling between the animals. This principle was given by the scientist Amotz Zahavi.
The trait that follow the handicap principle is peacock tail. The attractive peacock tail lure the predators and reduces the chances of escape of the peacock. This may acts as disadvantage but only chose the fittest males among the population. The tails attract the females as the tail is honest and sign of quality. In return male has more chances of mating and shows high reproductive success.
The Handicap Principle suggests evolution favors traits that are disadvantageous because they demonstrate the individual's overall fitness. An example of this is the large antlers of male elk, which show physical fitness because they require energy to grow and maintain, and lead to reproductive success.
Explanation:The Handicap Principle in biology refers to the idea that evolution favors traits that may have a disadvantage or impose a cost on an individual, as these traits serve as demonstrations of the individual's overall fitness. One example of such a trait could be the elaborately large antlers of a male elk.
The large antlers are a handicap in a sense that they require substantial energy to grow, and they may also impede movement through dense forests. Therefore, an elk with large antlers must be very fit and have access to plenty of food, since it can afford to spend energy in growing large antlers. Furthermore, large antlers impress females during mating season and intimidate other males, leading to greater reproductive success. Therefore, the antlers have evolved as a result of the Handicap Principle.
Learn more about Handicap Principle here:https://brainly.com/question/34278838
#SPJ3
how does dehydration sythesis relate to macromolucules and getting energy from food?
Answer:
Dehydration synthesis deals with the combination of two molecules, or compounds, with the loss of water to form a large molecule (macromolecule). This process uses up energy.
Hydrolysis, the opposite of dehydration synthesis, is a chemical process that degrades another molecules with the aid of water to form a molecule with a different make up. This process gives energy. Organic hydrolysis pairs water with neutral molecules, while inorganic hydrolysis combines water with ionic molecules.
Hydrolysis degrades the raw nutrients in the macromolecules such as protein etc., dehydration synthesis consumes raw nutrients to build up this macromolecules .
Answer: Dehydration synthesis is the process involved in making macromolecules
Explanation: Macromolecules are polymers, they have smaller subunits known as monomers that are joined together by bonds to create one single unit, the Macromolecule. The process involved in this creation is the dehydration synthesis reaction in which the monomers are linked with the loss of water. These reactions can be described based on the energy macromolecule formed and from with monomers are involved in the linking. Example, carbohydrates from monosaccharides, proteins from amino acids and fatty acids from acetyl-CoA
An experiment was performed to study the progress of cells through the mitotic cell cycle. The compounds listed were used one at a time to study their effects on the cell cycle: cytochalasin: an inhibitor of actin microfilament assembly colchicine: an inhibitor of microtubule formation aphidicolin: an inhibitor of DNA polymerase activity emetine: an inhibitor of ribosome activity (blocks protein synthesis) Which of these compounds would be most likely to arrest cells in S Phase?
a. emetine: an inhibitor of ribosome activity (and therefore protein synthesis)
b. cytochalasin: an inhibitor of actin microfilament assembly
c. aphidicolin: an inhibitor of DNA polymerase activity
d. colchicine: an inhibitor of microtubule formation
The colchicine: an inhibitor of microtubule formation, will be most likely to arrest cells in S Phase. The correct option is D.
What is inhibitor?A substance that slows or stops a chemical process is known as a reaction inhibitor. In contrast, a catalyst is something that quickens a chemical reaction.
Enzyme inhibitors are substances that alter the enzyme's catalytic capabilities.
As a result, they reduce the enzyme's ability to catalyse reactions or, in extreme cases, completely stop them. These inhibitors function by obstructing or changing the active site.
As part of an investigation to look at how cells go through the mitotic cell cycle.
Colchicine, a drug that prevents the synthesis of microtubules, is most likely to cause cell arrest in the S Phase.
Thus, the correct option is D.
For more details regarding enzyme inhibitor, visit:
https://brainly.com/question/17320375
#SPJ6
If you observed two features on a slide with your naked eye that were 0.5 mm, how far apart would they appear to be if you observed them with the microscope in front of you, using the second objective?
Answer:
The correct answer is - 50 mm apart.
Explanation:
A microscope have objective lens to magnify the distance to look far apart. These objectives lenses are 3 to 4 in number in the microscope. The magnification power of these lenses are 4x, 10x, 40x and 100x powers in order.
As it is mention above the second objective lens has magnification of 10x power which means the reflection would be 10 times. With coupled with 10x of eye piece.
So the total magnification = 10x10 = 100x
thus, it will appear, 0.5 x 100 = 50mm apart.
Thus, the correct answer is - 50 mm apart.
Describe the actual and relative sizes of a virus, a bacterium, and a plant or animal cell.
In general, viruses are the smallest, ranging from 20 to 250 nanometers. Bacteria are larger than viruses, typically measuring up to a couple of micrometers. Eukaryotic cells, such as plant and animal cells, are the largest, ranging from 10 to 100 micrometers.
Explanation:The sizes of a virus, a bacterium, and a plant or animal cell differ greatly. Viruses are typically the smallest of the three, ranging from approximately 20 to 250 nanometers in diameter, although some can be as large as 900 nanometers. However, these measurements can indeed vary, as some viruses discovered recently have sizes approaching that of a bacterium.
Bacteria, generally larger than viruses, can extend up to a couple of micrometers in size. Like viruses, bacterial sizes can also vary, but they are typically larger than most viruses.
Plant or animal cells, or eukaryotic cells, are generally much larger than both viruses and bacteria. These eukaryotic cells can range from 10 to 100 micrometers in size, which is substantially larger compared to both viruses and bacteria.
These measurements are generally unseen to the eye and therefore, require tools like an electron microscope for precise observations and comparisons.
Learn more about Cell Sizes here:https://brainly.com/question/35705956
#SPJ3
In the presence of cocaine, the maximal rate of transport (Vmax) is unaffected, but the apparent affinity of the transporter for dopamine is reduced
True/False
Answer:
True
Explanation:
Neurotransmission is modulated by the dopamine transporter. Dopamine transporter leads to activation of protein kinase which in turn reduces the uptake of dopamine. Due to this, the Vmax reduces and the affinity of transporter remains unchanged.
However, in the presence of cocaine activity of dopamine transporter is inhibited as cocaine and dopamine-binding sites are more or less similar to each other. Dopamine stops blocks the dopamine transporter and reduces the uptake of dopamine. This leads to constant Vmax but reduced affinity of transporter
Hence, the given statement is true
How how many carbon and hydrogen atoms would be contained within this molecule?
Answer:
The answer to your question is This molecule has 10 carbons and 20 hydrogens.
Explanation:
- Number of carbons
You can see the picture below, all the edges are carbons and they are numbered, there are 10 carbons.
- Number of hydrogens
Carbon 1 has 3 hydrogens
Carbon 2 has 2 hydrogens
Carbon 3 has 1 hydrogen
Carbon 4 has 2 hydrogens
Carbon 5 has 3 hydrogens
Carbon 6 has 2 hydrogens
Carbon 7 has 1 hydrogen
Carbon 8 has 1 hydrogen
Carbon 9 has 2 hydrogens
Carbon 10 has 3 hydrogens
Total number of hydrogens = 20
Answer:
The molecule has ten (10) carbon atoms and twenty (20) hydrogen atoms.
Explanation:
Increasing the amount of action potentials that arrive at the presynaptic terminal will:
A) Increase the amount of depolarization in the terminal
B) Increase the amount of Ca2+ that enters the presynaptic terminal.
C) Increase the number of vesicles that fuse with the presynaptic membrane
D) Increase the amount of neurotransmitter released.
Answer:
Increasing amount of action potentials that arrive at pre synaptic terminal will increase the amount of Ca2+ that enters the presynaptic terminal.
Explanation:
Neuromuscular junction is called as the site of signal exchange which occur in step wise manner.As action potential reaches the axon terminal, it opens axon terminal through which calcium enters.Then fusion takes place between presynaptic membrane that causes release of ACh in synaptic cleft through exocytosis.Binding of ACh with postsynaptic receptor cause opening of ion channels which allow the entry of sodium and potassium.This flow generates action potential that travels to the myofibril and causes muscle contraction.The ________ structure of a protein is created by hydrogen bonds between the hydrogen atom on the amine group and the oxygen atom on the carboxyl group.
Answer: The secondary structure of protein is creates by hydrogen bonds between the hydrogen atoms on the amine group and the oxygen atom on the carboxyl group.
Explanation:
Secondary structure of protein is a three dimensional type of protein that is formed by hydrogen bonds between the hydrogen atoms on the amino group and oxygen atom on the carboxyl group. The secondary structure of protein chains are organised into two regular structures which are alpha helixes and beta pleated sheets. In alpha helix the chain is like a loosely coiled spring and beta pleated sheets, the chains are folded. Meanwhile, protein have three types of structures which are primary, secondary and Tertiary.
When the heart contracts the _______ receives its blood supply. When the heart relaxes the _______ receives its blood supply.
Answer:
When the heart contracts the body tissue receives its blood supply. When the heart relaxes the heart ventricles receive its blood supply.
Explanation:
The cardiac cycle consists of contraction and relaxation of the heart. The contraction is called systolic pressure and relaxation is called diastolic pressure.
The systole and diastole are mainly concerned with the ventricles, still, articular contraction and relaxation also occur. The blood from the auricles pushes into the ventricles. This is because of the auricular contraction. This initiates by SA node with a "Lub" sound.
The contraction means the pressure of the blood is more in the auricle and the mitral valve and tricuspid valves open. The blood enters the ventricles. Then the pressure of auricles becomes less.
Now the ventricles are filled with blood and the pressure of the blood is more in the ventricles. This is called ventricular systole, and it leads to open the semilunar valves.
The blood is ejected out by aorta and supply to different parts of the body. After the blood ejection, the ventricles are in the relax stage, which is called the diastolic stage.
One systolic pressure is followed by one diastolic phase. Due to systolic pressure/ contraction, the blood pumps out of the heart.
As discussed in Investigating Life 15.1, why might bioluminescence be an advantage for some species of dinoflagellates?
Final answer:
The correct option is: Bioluminescent dinoflagellates avoid predation by attracting predators of copepods. Bioluminescent dinoflagellates avoid predation by attracting larger predators of their would-be attackers, serving as an evolutionary defense mechanism.
Explanation:
The correct answer is that bioluminescent dinoflagellates avoid predation by attracting predators of copepods. Bioluminescence in dinoflagellates serves as a defense mechanism. When these organisms are disturbed, they emit light, which can deter predators directly, or more interestingly, attract larger predators that feed on their would-be attackers, such as copepods. This inadvertent 'call for backup' can reduce the predation pressure on the bioluminescent dinoflagellates, thus offering them an evolutionary advantage. Additionally, the phenomenon of bioluminescence contributes to the visually stunning marine phenomena known as 'red tides' when dinoflagellate populations boom, although this is more related to their pigment than their light-emitting capabilities.
How do you adjust the focus of the microscope when observing organisms at high magnifications?
Answer:
The microscope is an instrument used for the visualization of the cell and its other component. Different types of microscope are compound microscope, electron microscope and binocular microscope.
The adjustment of the slide is important to visualize the focused object. A proper source of light is required to focus the image. At high magnification, the fine focus control is used for the adjustment. At 100 X the oil is used to visualize the object.
Imagine that you live in the midlatitudes—say, Virginia—and a cyclone passes by. What is the correct sequence of events?
two successive warm fronts
first a cold front and then a warm front
first a warm front then a cold front
two successive cold fronts
first a warm front, then the eye, then a storm surge
Answer:
First a warm front then a cold front.
Explanation:
Cyclone may be defined as the amount of the large air and move around the area that has low atmospheric pressure. The cyclone can cause the large disturbances to the living organisms.
The cyclone formation occurs at the low polar vertices. The two main types of front are related to the cyclone. The warm front first form for the air movement. This warm front is later replaced by the formation of the cold front of that cyclone.
Thus, the correct answer is option (3).
In Avery's experiment, he used DNase, RNase or proteinase to treat the heat killed S strain to see what component in the debris is the genetic material. When he used RNase to treat it, which colony formed on the plate?
A. R strains
B. Half R and half S strains
C. Neither R nor S strains
D. S strains
Answer:
S strain
Explanation:
The Avery experiment demonstrated DNA is the genetic material. It expanded upon the findings made by Griffith.
They used Pneumococcus; Smooth strain which was virulent and the Rough which was not.
Cultures of heat killed smooth strain were prepared after which it was treated with DNases ,RNases and Proteinases to remove DNA, RNA, and proteins respectively. It will then be introduced to living Rough strain.
When treated with RNases only the RNA will be destroyed and transformation will take place leading to colonies of S stains being formed.
Only when treated with DNase did the colonies S strain fail to be formed.
What is the term that refers to the range of animal and plant species and the genetic variability among those species?
Answer:
Biodervasity.
Explanation:
The term refers to the variety of life on earth.It demonstrates the extent of variation at different levels of species,genetic and ecosystem in a community of organisms.
It not equal in its distribution on earth;but rather varies with the type of ecosystem available and the distribution of biotic and abiotic factors in the ecosystem. For example;
It is highest in the Tropics,due to the species richness and diversity of the region. However, it varies in other ecosystem based on species diversity peculiar to parts of the region .e.g it is highest in the marine ecosystem along the coast of abundant temperature for primary producers.
Who though that aging was attributable to a loss of irritability in nervous and muscular tissue?
Answer:
The answer is Erasmus Darwin
Explanation:
In the 19th century, Erasmus Darwin, Charles Darwin's grandfather proposed the theory that loss of irritability in the nervous system and a decreased response to sensation or stimuli was associated with ageing.
This is one of the first theories of biological ageing; the others being Hippocrates' and Eli Metchnikoff's.
Hippocrates attributed ageing to the loss of body heat whereas, Eli Metchnikoff associated it with autointoxication. A state caused by the poisoning or intoxication of the body by toxins produced by the body itself.
Why might deuterostomes differentiate their embryonic cells later than the protostomes?
Answer:
The Deuterostomes shows indeterminate development, in which each of the cells of the eight-cell embryo if separated remain capable of developing as complete organisms. This contrasts to protostomes determinate development, in which the development fate of each cell in the adult organism has already been determined.
Explanation:
In this image, what letters represent nonpolar covalent bonds. Please select them by going by the blue letters.
Answer:
The answer to your question is C, D, E, F, G, H, I, J
Explanation:
To know if a bond is nonpolar covalent we must calculate the difference of electronegativity if it is between 0 and 0.5, the bond is nonpolar.
A) N and H 3.04 - 2.2 = 0.84
B) N and H 3.04 - 2.2 = 0.84
C) C and H 2.55 - 2.2 = 0.35 Nonpolar covalent
D) C and N 3.04 - 2.55 = 0.49 Nonpolar covalent
E) C and C 2.55 - 2.55 = 0 Nonpolar covalent
F) C and C 2.55 - 2.55 = 0 Nonpolar covalent
G) C and H 2.55 - 2.2 = 0.35 Nonpolar covalent
H) C and H 2.55 - 2.2 = 0.35 Nonpolar covalent
I) C and H 2.55 - 2.2 = 0.35 Nonpolar covalent
J) C and H 2.55 - 2.2 = 0.35 Nonpolar covalent
Place each of the labels in the box designating which plane or section is being referred to.
Note: Question is incomplete i have added full question with answers in picture format as attached.
Answer:
Frontal:
1. Which section could not display the sternum and the vertebrae simultaneously
2 . Which section would be necessary to display the length of both femurs simultaneously?
3. Which section would be necessary to see the full length of the roots of the two front teeth simultaneously?
4. Which section divides the body into front and back?
Sagittal:
1. Which section could not produce a view of both kidneys simultaneously?
2. Which section divides the body into right and left?
Transverse:
1. Which section could not display the abdominal and thoracic organs simultaneously?
2. Which section divides the body into top and bottom?
3. Which section allows circumferential comparisons between arms?
Place the labels in the box.
The labels that are provided in the box include the frontal, sagittal, and transverse all relate to the human body parts and have a specific function. These labels are marked as per their according values. They are given in the designated section of the box.
The answer refers to the Frontal (front part of the human brain), Sagittal plan that is dividing the body into vertical and horizontal positions. The transverse is in the longitudinal waveform.
The frontal label includes the section that could not display in the sternum and vertebrae at the same time. The part to display the length of both femurs. The part to see the full length of the roots of the two front teethSagittal is the section that could not produce a view of both kidneys, divides the body into halves.Transverse is the section that could not display the abdominal, divides the body into top and bottom allows comparisons between the arms.Learn more about the put each of the labels.
brainly.com/question/10150057.
Many proteins have a structure that allows them to change shape in order to accomplish their function. How could the motor protein kinesin twist, bend or otherwise change shape to accomplish its function?
Answer:
Motor protein kinesin can "walk" along a microtubule while carrying vesicles and by changing its shape the kinesin.
Explanation:
Motor protein kinesin is an important microtubule based motor protein that is conserved among all eukaryotic organisms. Its movement along the microtubule is ATP powered. Most of the kinesins walk towards the plus end of the microtubule.
Due to its ability it is responsible for unidirectional transportation of cargos including membranous organelles and mRNA.
Motor protein kinesin can change shape through the movement of ATP-binding domains and a flexible neck region, which allows it to transport cargo along microtubules in cells.
Motor protein kinesin can change shape in order to accomplish its function of transporting cargo along microtubules within a cell. One key mechanism that allows kinesin to change shape is through the movement of ATP-binding domains, which provide the energy for kinesin's movement. These domains undergo conformational changes, twisting and bending to enable kinesin to move along the microtubule track.
Additionally, kinesin has a flexible neck region that allows it to adopt different conformations. This flexibility allows kinesin to adjust its position and orientation as it moves along the microtubule, accommodating different shapes and sizes of cargo.
The ability of kinesin to twist, bend, and change shape is essential for its function as a motor protein, enabling it to efficiently transport cargo within cells.
Learn more about Motor protein kinesin here:https://brainly.com/question/31413601
#SPJ3
Scientists were monitoring the biochemical properties of a proton transporter by comparing the pH inside & outside the cell in the presence or absence of ATP. They noticed that the pH of the solution outside of the cell went down when ATP was added. Explain (a) whether the transporter used active or passive transport, (b) the direction of movement of the proton (H+), and (c) whether the reaction was endergonic or exergonic
Answer:
Explanation:
a; The transporter used Active transport. This is an endegonic process that requires the use of ATPs as sources of energy,. and abundant oxygen to transport ions from region of lower concentration gradient to region of high concentration gradient , that is against their individual concentration gradient. It is a transport mechanism that is inhibited by hypoxia, cyanide, mercury compounds.
Therefore if the pH of the outside is lowered (more acidity),when ATP was introduced, then protons must have been pumped outwards from the internal cell layers, that is more protons are now outside than inside of the cells. Thus the outside should have higher concentration gradient than inside.
Consequently, protons are pumped ,against their concentration gradients, by active transports, and the fact that this occurred when ATPs was introduced showed that active transporter is the answer because energy is needed for it operates.
b. pH decreases with increase in acidity of a medium. The drop in the extracellular pH showed that the active transporter pumps protons from inwards to outward solution of the cells environments.
c. Endergonic because it involved use or consumption of energy as ATP as described above; an not the release.
The proton transporter is using active transport to move protons against their concentration gradient, resulting in a decrease in pH outside the cell. The direction of movement of the protons is from outside the cell to the inside. The reaction is exergonic and requires ATP hydrolysis to provide the energy for the active transport.
Explanation:The movement of protons (H+) from outside the cell to the inside when ATP is added indicates that the proton transporter is using active transport. Active transport requires energy to move particles against a concentration or electrochemical gradient. In this case, ATP is providing the energy for the transporter to move protons against their concentration gradient, resulting in a decrease in pH outside the cell.
The direction of movement of the protons is from outside the cell to the inside. The addition of ATP allows the transporter to actively move the protons against their concentration gradient, resulting in a decrease in pH outside the cell.
The reaction is exergonic because the movement of protons against their concentration gradient is coupled with ATP hydrolysis, which releases energy.
One technique some MI clinicians use to increase client discrepancy is the ________ approach.
Answer:
Colombo approach
Explanation:
This is a questioning techniques formulated and used by the character Lieutenant Colombo played by Peter Falk in the Television series Colombo of the 70s.
The techniques involved 2 methods of ; engaging the individual under interrogation in a long casual talk to gain bonding,
AND
surreptitiously introduce the intended question;so that the subject in relax state is caught unaware and therefore find it difficult to hide the truth.
The techniques involved combination of the questioning approaches with a non-aristocratic demeanor of harmless, and non intimidating fellow, thus sealing the bonds between the interrogator and the subject; for ambience of trust with weak resistance to obtain needed answers.
what fluids are considered to be isotonic and appropriate in the treatment of fluid loss due to a surgical procedure
Answer: Isotonic crystalloid is are considered appropriate I treating fluid losses due to surgical procedure.
Explanation:
Isotonic crystalloids are intravenous fluid administered to patients to ensure initial resuscitation and restore isotonic fluid loss like blood. They are considered appropriate because they ensure initial resuscitation and to expand intravascular and interstitial spaces and also maintain hydration.
Examples of these isotonic fluid are normal saline(NaCl) and lactated ringers(RL).
Most scientists consider the Human Genome Project (HGP) to be the most significant scientific project of the 21st century. From the list below, choose the statements that describe the key findings of the Human Genome Project.
a. DNA exists in a double helical form.
b. The genetic information of a cell is stored in the form of DNA.
c. The human genome contains approximately 25,000 genes.
d. There are 23 pairs of chromosomes that make up the human genome.
e. There are approximately three billion base pairs in the human genome.
Answer: c) the human genome contains approximately 25,000 genes
e) there are approximately three billion base pairs in the human genome
Statements C, D, and E are correct descriptions of the key findings of the Human Genome Project: the human genome contains about 25,000 genes, is made up of 23 chromosome pairs, and includes around three billion base pairs. Statements A and B are also true, but they were discovered before the Human Genome Project took place.
Explanation:The Human Genome Project (HGP) led to several significant findings. Out of the options you've given, statements c, d, and e are correct. Under the HGP, it was discovered that the human genome contains approximately 25,000 genes (c), and these are arranged across 23 pairs of chromosomes (d) - these chromosomes make up the human genome.
In total, the human genome consists of approximately three billion base pairs (e). However, it's worth noting that the discovery of the double helical form of DNA (a) and the understanding that the genetic information of a cell is stored in the form of DNA (b) predates the HGP.
Learn more about Human Genome Project here:https://brainly.com/question/34217329
#SPJ6
Explain why NADH synthesized in the cytosol of liver cells and used for mitochondrial energy conversion generates 2.5 ATP per NADH, whereas NADH synthesize in muscle cell cytosol only generates 1.5 ATP per NADH
Answer: FAD is being reduced rather than NAD+
Explanation: This occurs in the G-3-P shuttle system wherein oxidation of transported NADH by respiration generates 1 ATP lesser in myocytes (1.5) than the 2.5 ATP generated as is the case with hepatocytes. This reduction in ATP is as a result of FAD (which enables electrons from cytosolic NADH to be moved against an NADH concentration gradient) rather than NAD+ being the electron acceptor G-3-P dehydrogenase of the mitochondria. The resultant effect of this transport is one molecule of ATP generated for every two electrons reduced.