Answer:
Explanation:
The parasites are those organisms which invade the host cell machinery to obtain nutrition, shelter and to reproduce their own progeny.
The fungi are the saprophytic organisms which obtain their food from the dead and decaying organic matter. But the fungi can also be parasitic in nature. The rust fungi produce spores which get attached to the plant surface. These spores germinate over the plant surface. They produce their germ tubes in the stomata of the leaves.
Some of the fungal strains damage the cuticle, and enter into the cells of the host plant. Some of the fungal strain also produce special structures called as haustoria which helps in deriving the nutrients from the plants.
Final answer:
Parasitic fungi and bacteria possess adaptations such as specialized structures called haustoria and appressoria to penetrate plant tissues and evade plant defenses. They infect through natural openings, wounds, and directly through plant cuticles, utilizing the plant's nutrients for their own survival.
Explanation:
Specific Adaptations of Parasitic Fungi and Bacteria
Bacteria and fungi have evolved various adaptations to parasitize and infect plant hosts. Notably, parasitic fungi like rust fungi produce specialized structures known as haustoria which penetrate the cell walls of their host plants. These haustoria make it possible for the fungi to access the cytosol materials such as sugars and amino acids, essential for their survival. The Uromyces species is a prime example of rust fungi that utilizes such mechanisms. Furthermore, these organisms can enter plant tissue through natural openings, wounds, or even directly through the cuticle. In addition to haustoria, pathogens and herbivores may have specialized cells called appressoria which help the fungi to attach and breach the epidermal cells of the plant. Additionally, certain bacteria enter through natural plant openings like stomata or wounds and multiply rapidly within the plant tissues causing cell death. Parasitic adaptation is an evolutionary process, which allows these organisms to resist the host's defense mechanisms and thrive within their living tissues.
Taylor had a serious condition that required removal of the upper third of his small intestine. Based on this information, is Taylor at risk for developing multiple nutrient deficiencies? Explain why or why not.
Answer:
Yes, Taylor is at risk for developing multiple nutrient deficiencies since most of the nutrients are absorbed in the upper small intestine.
Explanation:
The small intestine in the gastrointestinal tract has three parts: duodenum, jejunum, and ileum. Most of the digestion and the absorption of nutrients from the digested food occurs in the small intestine.
The removal of the upper part of the small intestine during surgical procedures such as gastric bypass surgery often results in multiple nutrient deficiencies as it alters the natural absorption of nutrients. The common nutrient deficiencies include vitamin B12, calcium, vitamin D, iron, folate, zinc, copper, etc. These deficiencies may lead to other problems like anemia, osteoporosis, encephalopathy, peripheral neuropathy, etc.
Vitamin B12 is required for the functioning of the nervous system and for the growth and replication of cells. The absorption process of vitamin B12 primarily takes place in the duodenum and ileum of the small intestine. The removal of that part of the small intestine interferes with the natural absorption of vitamin B12 and it results in complications like cognitive difficulties, anemia, neuropathy, etc. The duodenum and jejunum of the small intestine have an important role in iron and copper absorption. Iron deficiency can cause anemia, fatigue and copper deficiency may cause increased muscle tone, difficulty in walking, neuropathy, skin changes, psychiatric disorders, etc. Most of the ingested calcium is absorbed primarily in the duodenum of the small intestine and its deficiency often results in osteoporosis, reduced bone growth, etc. The folate and zinc absorption takes place across the intestinal walls. The reduced zinc absorption causes folate deficiency and it leads to birth defects, decreased erythropoiesis, megaloblastic anemia, neurologic problems, etc.
For most folks, their knowledge of the importance of microbes is limited to just those microbes that cause sickness. However, you know better! Which of the following describes a true critical feature of microbe biology illustrating their importance? (there can be more than one answer)a.) microbes are very small, and make up a very small portion of the Earth's biomass.b.) they influence our neurobiology and are critical components of ruminant animal digestive systems.c.) some produce oxygen.d.) they are critical for the process of decay.e.) some can fix atmospheric nitrogen which is important for plant growth.f.) they are used in the food and beverage production industry they are important food source for aquatic life
Answer:
Option (b), (c), (d), (e) and (f).
Explanation:
Microbes are the small living organisms that cannot be visible with the naked eye. Microbes plays an important role on earth and has positive as well as negative effect on the other living organisms.
Some microbes can act as pathogen as well. They cover most part of the earth and can be used in food to enhance the texture and taste. The methanogens are present in the gut of ruminant animals and are also useful in the research. Some microorganism has the ability to decompose the biodegradable waste and can even produce oxygen.
Thus, the answer is option (b), (c), (d), (e) and (f).
The importance of the microbe biology critical features statements involved
b.) they influence our neurobiology and are critical components of ruminant animal digestive systems.
c.) some produce oxygen.
d.) they are critical for the process of decay.
e.) some can fix atmospheric nitrogen which is important for plant growth.
f.) they are used in the food and beverage production industry they are an important food source for aquatic life
What is a microbe?Microbes refer to small living organisms that is invisible with the eye. It played a vital role on earth and has positive as well as negative effects on other living organisms. It should act as a pathogen as well. The methanogens should be present in the gut of ruminant animals and are also useful in the research. Some microorganisms should contain the capability to decompose the biodegradable waste and can even generate oxygen.
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You count 47 cfu on a spread plate. The plate was prepared by spreading 0.2 ml of a 1:10,000 dilution of the original sample. What is was the concentration of the original culture (in cells/ml)?
Answer: 2.35 x 10^6 cfu/ml
Explanation:
Concentration of the original culture: ?
Dilution factor: 10^4
Colony count per plate: 47cfu
Volume of cultured plate: 0.2ml
Concentration of the original culture= Colony count per plate x Dilution factor
Volume of cultured plate
Concentration of the original culture = 47 x 10^4
0.2
Concentration of the original culture = 2.35 x 10^6 cfu/ml
You can further take the log of your answer which would be:
Log(2.35 x 10^6) = 6.371
Final answer:
To find the concentration of the original culture in cells/ml from a plate count of 47 CFU using a 1:10,000 dilution and 0.2 ml sample volume, apply the formula CFU/mL = (Number of colonies × dilution factor) / volume of culture plate in ml, resulting in 2,350,000 CFU/mL.
Explanation:
To calculate the concentration of the original culture in cells/ml when you count 47 colony-forming units (CFU) on a spread plate made by spreading 0.2 ml of a 1:10,000 dilution of the original sample, you need to take into account both the dilution factor and the volume of the sample spread on the plate. The formula for this calculation is CFU/mL = (Number of colonies × dilution factor) / volume of culture plate in ml. In this case, the dilution factor is 10,000, and the volume spread on the plate is 0.2 ml. Therefore, the original concentration is calculated as follows:
CFU/mL = (47 colonies × 10,000) / 0.2 ml = 2,350,000 CFU/mL.
This means the original culture had a concentration of 2,350,000 cells per milliliter. This calculation method is commonly used in microbiology to estimate the number of viable cells in a sample.
21.
Part 1: Is it appropriate to describe a human heart as a "double pump?' (explain the term double pump and provide a detailed explanation that supports your answer)
Part 2: How is the systemic circuit different from the pulmonary circuit? Provide and explain at least 4 differences between these circuits
Part 3: A patient has an opening between the right and left atria. How, if at all, would this impact the state of oxygenation of the pulmonary or systemic circuit blood?
Answer:
Part 1. Yes, the term is appropriate. The heart pumps oxygenated blood to the head and the body and the deoxygenated blood to the lungs.
Part 2. The main difference is that the pulmonary circuit carries the blood to the lungs and back to the heart and the systemic circuit carries the blood from the heart to the body and back.
Part 3. It wouldn't impact the state of oxygenation by much.
Explanation:
Part 1. Deoxygenated blood from the body enters the heart through the vena cava to the right atrium, through the right ventricle to the pulmonary artery to the lungs. It gets oxygenated in the lungs and sent back to the heart by the pulmonary vein, through the left atrium, left ventricle to the aorta and back to the body. The process requires the heart to pump blood to the lungs and to the body; hence the "double pump".
Part 2. The bloodflow in the systemic circuit would be slower the oxygenation process from the lungs would continue and the impact wouldn't do too much damage. Some animals actually do have this opening. (e.g. crocodiles).
Two characteristics of natural wines are that they have a maximum alcohol content of 14% and are ""sparkling"" wines. Apply your understanding of alcoholic fermentation to explain these characteristics.
Explanation:
Two attributes of regular wines are that they have a most maximum alcohol content of 14% and are "sparkling" wines.In natural wines, after the ethyl alcohol and are "shimmering" wines. In regular wines, after the ethyl liquor fixation arrives at 14%, the liquor murders the yeast cells, making aging stop. The bubble or shimmer in wines is the collection of CO2. Yeast keep on aging sugars until they either come up short on nourishment or the liquor content gets sufficiently high to slaughter them. The most wine yeasts, 14% is higher than they can tolerate. There are, be that as it may, yeasts that the merchants guarantee can tolerate higher alcoholic rates.Natural wines achieve a maximum alcohol content of around 14% due to the toxicity of ethanol to yeast beyond this concentration. Their sparkling nature is due to the carbon dioxide produced during fermentation, which remains in the wine until opened. Higher alcohol concentrations in beverages like spirits are attained through distillation, beyond what is possible with fermentation alone.
Explanation:The production of natural wines is closely tied to the process of alcoholic fermentation. This natural process, which involves yeast converting sugars into alcohol and carbon dioxide, has limitations, especially in the context of alcohol concentration. A key characteristic of natural wines is that they typically have a maximum alcohol content of around 14%. This limitation arises because the common yeast species used in winemaking, Saccharomyces cerevisiae, cannot survive in environments with ethanol concentrations above 12-14%. Consequently, as the accumulation of ethanol in the wine reaches this level, the yeast dies, preventing further fermentation and higher alcohol content.
Regarding their "sparkling" quality, natural sparkling wines are created by trapping carbon dioxide within the wine. In certain methods, such as in the production of champagne, this is achieved by letting the yeast ferment inside a sealed bottle, generating carbon dioxide that dissolves into the wine under pressure. Upon opening the bottle, the pressure is released, causing the beverage to bubble, creating the sparkling effect. This effervescence is due to the carbon dioxide produced during fermentation, which remains dissolved in the liquid until released.
To produce beverages with higher alcohol content than what natural fermentation yields, a process called distillation is used. Distillation involves heating the fermented solution to evaporate the alcohol, which is then condensed back into liquid form, resulting in a higher concentration of alcohol than achievable by fermentation alone. Spirits such as gin, vodka, and rum are produced through this process, typically containing about 40% alcohol by volume.
Assume that you have an illuminated suspension of Chlorella cells carrying out photosynthesis in the presence of 0.1% carbon dioxide and 20% oxygen.
What will be the short-term effects of the following changes in conditions on the levels of 3-phosphoglycerate (PGA) and ribulose-1,5-bisphosphate (RuBP)?
Carbon dioxide concentration is suddenly reduced 1000-fold.
a. PGA down, RuBP up
b. PGA up, RuBP down
c. PGA up, RuBP unchanged
d. PGA unchanged, RuBP up
Answer:
a. PGA down, RuBP up
Explanation:
RuBP, also known as Ribulose-1,5-bisphosphate is a molecule that assist in he fixation of carbon dioxide in the Calvin Cycle or light independent reaction of photosynthesis.
RuBP is combined with carbon dioxide using the enzyme known as rubisco to form an short-lived intermediate product which divides to form two 3-carbon molecule structure known as 3-phosphoglycerate.
If the concentration of carbon dioxide is suddenly reduced, it thus means that the rate of production of 3-phosphoglycerate will reduce as carbon dioxide becomes a limiting factor. Hence, 3-phosphoglycerate's concentration will reduce while more RuBP will become available as a result.
The correct option is a.
A P. bursaria cell that has lost its zoochlorellae is said to be "aposymbiotic." It might be able to replenish its contingent of zoochlorellae by ingesting them without subsequently digesting them. Which of these situations would be most favorable to the re-establishment of resident zoochlorellae, assuming compatible Chlorella are present in P. bursaria's habitat?
A. abundant light, no bacterial prey
B. abundant light, abundant bacterial prey
C. no light, no bacterial prey
D. no light, abundant bacterial prey
The correct answer is B) abundant light, abundant bacterial prey.
The situations that would be most favorable to the re-establishment of resident zoochlorellae, assuming compatible Chlorella are present in P. bursaria's habitat is "abundant light, abundant bacterial prey."
This element needs these conditions of abundant light and bacterial prey to have the correct metabolism and live. The bacterial prey will help the element to grow. This is going to increase the preservation of the green algae.
What is primary production?a. Water injection b. Natural flow due to a pressurized reservoir c. Heating for viscosity reduction d. Pumping (horse head) e. Carbon dioxide flooding
Answer: Option B.
Natural flow due to pressurized reservoir.
Explanation:
Primary production is the process where natural materials is extracted from the Earth. It refers to natural way of producing things. It is also defined as the process where raw materials are extracted or gotten.
Natural flow due to pressurized reservoir shows that it was done naturally without artificial support or effort for the water to flow.
Consider a pendulum swinging. Which type(s) of energy is/are associated with the pendulum in the following instances: i. the moment at which it completes one cycle, just before it begins to fall back towards the other end, ii. The moment that it is in the middle between the two ends, and iii, just before it reaches the end of one cycle (just before instant i.).
a. i. potential and kinetic, ii. potential and kinetic, iii. Kinetic
b. i. potential, ii. potential and kinetic, iii. potential and kinetic
c. i. potential, ii. kinetic, iii. potential and kinetic
d. i. potential and kinetic, ii. kinetic iii. kinetic
Answer: C. i. potential, ii. kinetic, iii. potential and kinetic
Explanation: Potential energy is the energy that is possessed or exhibited by an object or a body that is static or stable. THE FORMULA FOR POTENTIAL ENERGY IS P.E = MASS OF AN OBJECT* HEIGHT * FORCE OF GRAVITY
Kinetic energy is the energy of an object or a body that is in motion or it is moving from one point to another.
THE FORMULAR FOR KINETIC ENERGY IS 1/2*MASS OF AN OBJECT *VELOCITY OF THE OBJECT SQUARE.
The interaction between people and the wild plants and animals they eventually domesticated took place mainly in the well-watered Fertile Crescent, a vernacular region referring to an arc of land stretching from Israel to western Iran.
a. True
b. False
Answer:
true
Explanation:
Nucleotide hydrolysis dehydration monosaccharide polypeptide amino acid polysaccharide polynucleotide
1. A reaction builds polymers from monomers.
2. A reaction breaks down polymers into monomers.
3. A(n) protein hemoglobin.
4. A(n) is a building block of polypeptides, such as the is a polymer of amino acids.
5. A(n) is a building block of polysaccharides such as starch.
6. A(n) is a polymer of monosaccharides. is a building block of polynucleotides such as
7. A(n) is a building block of polynucleotides such as DNA
8. A(n) is a polymer of nucleotides.
A reaction builds polymers from monomers.
Dehydration synthesis reactions build molecules up and require energy, while hydrolysis reactions break molecules down and release energy.
A reaction breaks down polymers into monomers is known as Hydrolysis
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Are most of the cells in our body differentiated or undifferentiated?
Answer: Most cells in the body differentiate.
Explanation:
Cells differentiation is the ability of cells to change form and develop into specialized cells to perform specific functions. These cells in the body retain their ability to divide and form different cells types. Examples include somatic cells such as muscles and skin cells,stem cells and germ line cells which differentiate into gamete cells. Some cells do not differentiate in the body.
A biology student accidentally loses the labels of two prepared slides she is studying. One is a slide of an intestine, the other of an esophagus. You volunteer to help her sort them out. How would you decide which slide is which?
Answer:
esophagus has stratified squamous
Explanation:
Stratified squamous is a tissue that is found covering and lining parts of the body such as the esophagus. It is known to be many layers of flattened cells.
It performs some functions like the provision of protection from abrasion, pathogens, and chemicals.
It can be found in some parts of the body like surface of skin, linings of mouth, esophagus, rectum,and vagina.
In this case, the presence of stratified squamous in a slide will definitely show that the content in the slide is esophagus. And the slides can then be easily separated and labelled accordingly.
To determine the identity of the slides, you can look for specific features and structures unique to each organ such as villi and goblet cells in the intestine, stratified squamous epithelium, and skeletal muscle fibers in the esophagus.
Explanation:To determine which slide is the intestine and which is the esophagus, you can look for specific features on each slide. The intestine slide may show the presence of villi, which are small finger-like projections that increase the surface area for nutrient absorption. The esophagus slide, on the other hand, may show the presence of stratified squamous epithelium, which is a type of tissue that can withstand mechanical stress.
Another way to differentiate the slides is to look for specific structures associated with each organ. For example, the intestine may have the presence of goblet cells, which secrete mucus, while the esophagus may have the presence of skeletal muscle fibers to aid in peristalsis.
By carefully examining the slides under a microscope and comparing them to known images or descriptions of the structures, you can determine which slide corresponds to the intestine and which corresponds to the esophagus.
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A researcher claims that all living cells must be able to synthesize proteins in order to maintain homeostasis and perform basic functions. What statement supports the scientist's claim? A) Ribosomes are required for protein synthesis, and all four cell types can synthesize proteins. B) Nuclei house the DNA, which is the blueprint for proteins. Cell types I, II, and III can synthesize proteins. C) Cell walls are required for cells to maintain their shape to manufacture proteins. Cell types II, III, and IV can synthesize proteins. D) Cells require lysosomes to digest cellular components to convert them into proteins. Only cell types I and III can synthesize proteins.
Ribosomes are required for protein synthesis, and all four cell types can synthesize proteins. Thus the option A is correct.
What are the levels of protein structure?The Primary structure of proteins include the linear structure of amino acids, bound by Covalent, peptide bonds to form primary structure, most functional form.
Secondary structure refers to folded structures of primary protein which is formed by hydrogen bonding between the amine and carboxyl group by a polypeptide chain.
These secondary structure are present in two forms like α – helix form which forms hydrogen bonds by twisting into a right-handed screw with the -NH group of each amino acid residue.
The β – pleated sheet structures are the peptide chains which are stretched out and present side by side manner together by intermolecular hydrogen bonds.
Tertiary Structure of Protein include the folding of the secondary structure by H-bonds, electrostatic forces, disulphide linkages, and Vander Waals forces where as Quaternary Structure of Protein refers to spatial arrangement of tertiary structures.
Thus the option A is correct.
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The following statement is false. Select the rewording that makes the statement true.
In gymnosperms, seed formation is the process of transferring pollen from a male cone to a female cone.
A) In gymnosperms, pollination is the process of transferring pollen from a male cone to a female cone.
B) In gymnosperms, seed dispersal is the process of transferring pollen from a male cone to a female cone.
C) In gymnosperms, gravitropism is the process of transferring pollen from a male cone to a female cone.
D) In gymnosperms, fertilization is the process of transferring pollen from a male cone to a female cone.
The correct statement is that in gymnosperms, pollination is the process of transferring pollen from a male cone to a female cone, which is necessary for fertilization to occur leading to seed formation.
The original statement is incorrect and needs to be reworded to accurately describe the process in gymnosperms. The correct reworded statement is A) In gymnosperms, pollination is the process of transferring pollen from a male cone to a female cone. Pollination involves the movement of pollen grains from the male reproductive structure to the female reproductive structure. Once the pollen lands on the female cone, it germinates, forming a pollen tube and eventually fertilizing the egg, leading to seed formation. Fertilization is the actual process of the sperm from the pollen uniting with the egg, occurring after pollination has already taken place.
On your first attempt to run a PCR, you realize you forgot to add one of the two primers. The graduate student you work with suggests you throw away your reaction and start over. Why
Explanation:
During PCR, we use two primers one is forward primer and the other one is reverse primer they match the sequence of one one of the two complementary strands of the target DNA, they flank the target region (that the region which we has to be copied). if we add only one primer it copies only one strand of the DNA in multiple copies. Usually we call this as Asymmetric PCR.
Generally Primers are synthetic short stretch of oligonucleotides that are complementary to the target DNA. They act as a foundation for the amplification process of DNA to form multiple copies
Animal cells have a membrane that separates the interior of the cell from the outside environment. Typically, an electric potential difference exists between the inner and outer surfaces of the membrane. Consider one such cell where the magnitude of the potential difference is 66 mV, and the outer surface of the membrane is at a higher potential than the inner surface. A sodium ion (Na+) is initially just inside the cell membrane (initially at rest). How much work (in J) is required for a cell to eject the ion, so that it moves from the interior of the cell to the exterior?
The work required to move a sodium ion from inside a cell to the exterior is 1.056 * 10^-21 Joules, using the formula for work where Work = Charge * Potential Difference.
Explanation:This is a question of physics, particularly involving concepts of electricity and work. In order to find the work done to move an ion from the inside of a cell to the exterior, we use the physics formula Work = Charge * Potential Difference.
The charge of a sodium ion (Na+) is equal to the elementary charge, which is approximately 1.6 * 10^-19 Coulombs. The potential difference provided in this problem is 66 mV (millivolts), but we should render this value into volts for the calculation, which would be 0.066 V (volts).
Now, multiplying these values give us:
Work = (1.6 * 10^-19 C) * (0.066 V) = 1.056 * 10^-21 Joules.
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Another name for an oxidizing agent is a(n) Group of answer choices electron acceptor. hydride transfer reagent. electron donor. electropositive metal. reductant.
Explanation:
an oxidizing agent is an agent that can oxidize other substances, therefore, accepting their electrons. common oxidizing agents are oxygen and hydrogen peroxide...
Answer: Option A.
Hybride transfer.
Explanation:
An oxidizing agent is a compound that accept electron and transfer oxygen atom during a Redbox reaction . An oxidizing agents oxidize others in a chemical reaction. Oxidizing agents are reduced in a chemical reaction. Oxidizing agents are electronegative substance. Oxidizing agents are hybrid transfer, they transfer oxygen atoms to other substances. Examples of oxidizing agents are; chlorine, flourine,oxygen e.t.c.
The P and Q genes of a plant are 20 m.u. apart. A PP qq and a pp QQ individual cross, and the F1 cross with each other to produce the F2. The P and p, and Q and q alleles are completely dominant and recessive, and there are no interactions between genes P and Q. What phenotypic ratio would you expect in the F2
Answer:
66 P_ Q_: 9 P_ qq: 9 pp Q_: 16 pp qq
Explanation:
As the genes are 20 m.u apart, recombination frequency between P & Q genes will be 20% (As 1% recombination = 1 m.u).
Parental cross \rightarrow PP QQ x pp qq
Parental gametes \rightarrow PQ, pq
F1 \rightarrow Pp Qq
F1 gametes \rightarrow PQ (Frequency = 0.4), pq (Frequency = 0.4), Pq (Frequency = 0.1), pQ (Frequency = 0.1) (As non-recombinant gametes frequency will be 0.8 & recombinant gametes frequency will be 0.2)
See the attached table.
The phenotypic ratio in the F2 generation would be 3 P Q : 3 p Q or 1:1.
Explanation:In the F2 generation, when individuals from the F1 generation cross with each other, the phenotypic ratio can be determined by using a Punnett square. In this case, the cross is between a PP qq individual and a pp QQ individual.
The gametes produced by the PP qq individual are P q and p Q. The gametes produced by the pp QQ individual are p Q and P q.
When these gametes combine in the F2 generation, the possible genotypes and phenotypes will be as follows:
PP QQ: Phenotype P QPP Qq: Phenotype P QPp QQ: Phenotype P QPp Qq: Phenotype P Qpp QQ: Phenotype p Qpp Qq: Phenotype p QTherefore, the phenotypic ratio in the F2 generation would be 3 P Q : 3 p Q or 1:1.
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Water molecules have one oxygen atom that is more electronegative than the two hydrogen atoms bound to it. As a result, which type of bonding holds water molecules together?
A.) non polar covalent bonding
B.)polar covalent bonding
C.)ionic bonding
Answer: Option B.
Polar covalent bonding.
Explanation:
Hydrogen bonding or polar covalent bonding exist between water molecules. Water have one oxygen molecules that is electronegative and two hydrogen molecules that is slightly electropositive. Two hydrogen atoms are covalently bonded to one oxygen atom. In covalent bonding, there is sharing of electrons between atoms. In water, there is unequal sharing of electron between oxygen atom and hydrogen atoms. Oxygen atom tend to attract more electrons that hydrogen atoms, which make water a polar molecule.
A specialized protein in saliva breaks up starch molecules in food into smaller chains of simple sugars. In this reaction, which molecule is the enzyme, which is the substrate, and which is the product? One of the four answer choices will not be used.a. Saliva
b. Starch molecules
c. Smaller chains of simple sugars
d. specialized protein
Answer:
option d, b, c
Explanation:
Starch molecules taken into the mouth from food substances are processed to an extent of 30% of its digestion. this is carried out by a specialized protein/ an enzyme that is present in the saliva; called amylase or ptyalin. this enzyme acts on the substrate molecule which in this case is starch molecules and convert it into smaller chains of simple sugars that includes maltose and dextrin which is digested in the small intestine.
Action potentials are generated at the axon hillock of a neuron and travel in one direction towards the synaptic terminals. Why are action potentials only transmitted in one direction under normal conditions
Answer:
The sodium channel have refractory period.
Explanation:
Action potential are generated at axon hillock of neuron and travel across the neuron in response to stimuli. For the the generation of action potential the steps are stimulus, depolarization, repolarization, hyperpolarization and resting state.
So the first step for action potential after stimulus is depolarization. Depolarization requires the opening of sodium channels to make the inside of neuronal membrane positive. These sodium channels have refractory period. They cannot open immediately after activation. So the sodium channels of one direction in neuron (upstream) become refractory and it is convenent to open the sodium channels of other direction (downstream) of neuron. That why action potential is transmitted in one direction.
It is like the gunpowder placed at a line the fire cannot go back in burnt gun powder direction.
Gram staining It is a differential staining method to differentiate between gram positive and gram-negative bacteria based on the differences in their cell wall composition. Gram positive bacteria has a thick meshwork of peptidoglycan layer due to which crystal violet iodine complexes get entrapped in between them and it appears purple. Gram negative bacteria does not have a thick meshwork of peptidoglycan, instead it is thin and therefore unable to trap crystal violet iodine complexes and loses purple color on alcohol wash. Rather, it gets a reddish oranges color on staining with safranine. Procedure - 1. On a slide, make bacterial smear. 2. Heat fix the cells. 3. Add a drop of crystal violet on the slide and wait for 1 minute. Wash off excess stain with water. 4. Add a drop of iodine on the slide and wait for 1 minute. Wash off the excess stain with water. 5. Wash with alcohol and wait for 30 seconds. Wash with water. 6. Wash with safranine and wait for 30 seconds. Wash with water. Gram positive will appear purple and gram negative will appear orange. Flu is causes by virus and therefore, cannot be detected by gram staining. Spore stain is primary staining method which uses malachite green dye to stain endospores. Simple stain is used to visualize bacteria by using only one stain. If it stains the bacteria, then it is positive stain and if it stains the background and bacteria is colorless then it is negative stain
Gram staining serves as a key technique in distinguishing between gram-positive and gram-negative bacteria, based on the differences in their cell wall composition.
Explanation:Gram staining is a critical technique in microbiology used to differentiate between two major types of bacteria: gram-positive and gram-negative. The process, which is based on the variance in the cell wall composition of bacteria, involves using a crystal violet iodine complex and a counterstain, safranine. After the staining process, gram-positive bacteria present a purple color due to a thick peptidoglycan layer that traps the purple-colored crystal violet iodine complex, while gram-negative bacteria portray an orange color after being stained with safranine because of their thinner peptidoglycan layer. The Gram staining procedure includes making a bacterial smear on a slide, heat fixing the cells, applying crystal violet, then iodine, and rinsing each with water, applying alcohol, rinsing with water again and finally applying safranine and giving a final water rinse.
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Explain how the tendinous intersections of the rectus abdominis of the cat differ from those found in humans?
Answer:
The tendinous intersections refer to the three fibrous bands that cross the rectus abdominis muscle. The upper band is located at the xiphoid process, the lower one at the umbilicus level, and the middle one is situated between the two bands. The tendinous intersections in humans are formed of tendinous fibers, which differentiates the rectus abdominis into four components.
On the other hand, the rectus abdominis muscle is spontaneous in the case of cats, that is, it arises from the pubis and merges with the costal cartilages numbering five to seven and the sternum. The alternate overlapping of the internal oblique and rectus abdominis muscles produces the intersections, however, no true intersections are witnessed in it.
Sugar loaded into the source end of the phloem draws water into the sieve tubes by osmosis, raising the pressure. What happens to the water at the sink end?
Answer:
it is pushed out harder and faster
Explanation:
The roots of plants are located underground where light does not penetrate; therefore, photosynthesis cannot occur. Where do plant cells located in the roots obtain energy for their metabolic needs? [2 pts] aerobic respiration – the root cells metabolize the sugar produced during photosynthesis oxidative phosphorylation – the root cells directly convert phosphate in the soil to ATP fermentation – the root cells ferment the sugar produced during photosynthesis anaerobic respiration – the root cells metabolize sugars absorbed from the soil
Answer:
The correct answer is: aerobic respiration -the root cells metabolize the sugar produced during photosynthesis
Explanation:
From the germination of seeds, roots will depend exclusively on energy in the form of photosynthates supplied from the aerial parts of the plant. Photosynthesis takes place in the leaves. Sugar is produced which is transported to all the other plants via phloem tube.Roots absorbs air from the air spaces present between the soil particles and is able to carry out aerobic respiration. The sugar is metabolized and energy is produced.
Plant roots obtain energy for their metabolic needs primarily through aerobic respiration, a process where the glucose produced from photosynthesis is metabolized using oxygen to produce ATP.
Explanation:The roots of plants obtain energy for their metabolic needs primarily through aerobic respiration. While photosynthesis primarily occurs in leaf cells where sunlight is accessible, the energy stored in the form of glucose travels to all parts of the plant, including roots. During aerobic respiration in the root cells, this glucose is metabolized, or broken down, to generate ATP, the energy currency of cells. This process requires oxygen, which is delivered to the roots from air spaces in the soil and from aboveground parts of the plant.
Beyond aerobic respiration, plants do not typically use oxidative phosphorylation to directly convert soil phosphate to ATP. Anaerobic respiration and fermentation processes can occur in the absence of oxygen, but these are not the main source of energy for root cells. Instead, these processes tend to be more common in microorganisms and in plant or animal cells under conditions of oxygen deprivation.
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Estimate the reduction in bacteria during the passage of wastewater that initially contains 106 organisms per milliliter through three stabilization ponds that are arranged in series. The volumes of the three ponds are 10,000, 20,000 and 6,000 m3, respectively. The flow rate is 1,000 m3/d. Assume that steady-state conditions apply, that ponds are mixed completely because of wind action, that first-order decay kinetics apply
Answer:
an improvement in efficiency is observed with η = 99.99%
Explanation:
the voltage can be calculated using the formula: the detection time can be calculated using the equation:
t1 = V/Q = 10000/1000 = 10 days
t2 = 20000/1000 = 20 days
t3 = 6000/1000 = 6 days
the exit of the organisms will be:
1rd pond = 10^6/(1+(10*10)) = 90909.1 org/mL
2nd pond = 90909.1/(1+(1*20)) = 4329 org/mL
3rd pond = 4329/(1+(1*6)) = 618.4 org/mL
η = ((Cin-Cout)/Cin)*100 = ((10^6-618.4)/(10^6))*100 = 99.93%
detection time t = 12000/1000 = 12 days
Cout1 = 10^6/(1+(1*12)) = 76923.1
Cout2 = 76923.1/(1+(1*12)) = 5917.2
Cout3 = 5917.2/(1+(1*12)) = 455.2
η = ((10^6-455.2)/(10^6))*100 = 99.99%
an improvement in efficiency is observed
17. Which of the following happens first as a nephron processes blood?
A. reabsorption
B. excretion
C. osmosis
D. filtration
E. secretion
Answer: Option D) Filtration
Explanation:
Blood brought to the kidney by renal artery. As it circulates through the capillaries of glomerulus of each bowman's capsule, water, urea, nitrogenous compounds, glucose etc are filtered into the capsule. This process of filtering materials from the glomerulus into the bowman's capsule is called ultra filtration
Answer: filtration
Explanation:
this process allows water and small molecules to filter out of the plasma of the blood.
It has been proposed that loss of telomeric DNA as a result of replication might be responsible for cellular (and organismal) senescence. Since telomerase should be able to solve the problem, why is this proposal a reasonable explanation?
Answer:
Cells during division process cells reduce its size gradually and form an end association that leads to the Damage in DNA. Due to this disruption in DNA the cell signal to cause the cell to replicative senescence.
It is termed as the cell arrest, it this cell checkpoints are not present the telomeres keep on reducing its size that leads to M2 stage. It is shown by the researcghes the telomerase dysfunction is leads to cellular and organ cancer progression.
Telomeric DNA loss during replication can lead to cell aging due to genomic instability and cessation of cell division. While the telomerase enzyme can counteract this, it's not strongly expressed in most cells, creating a form of 'cellular clock'. Meanwhile, cancer cells often upregulate telomerase, becoming 'immortal', which shows telomere length maintenance is not unilaterally beneficial.
Explanation:The concept of telomeric DNA loss and its potential role in cellular and organismal senescence is a central topic in the biology of aging. Telomeres are the ends of chromosomes, and they gradually shorten as a cell divides and replicates its DNA. This shortening of telomeres can lead to genomic instability, cessation of cell division and eventual cell death, which can contribute to the process of aging.
Indeed, telomerase, an enzyme that can add DNA sequence repeats ('TTAGGG' in all vertebrates) to the 3' end of DNA strands in the telomere regions, should be able to prevent or even reverse this telomere shortening. However, its activities are usually not sufficient in most somatic cells, partly because it is not strongly expressed in them. This situation can create a so-called 'cellular clock', by which the cell's age and its passage towards senescence can be measured.
On the other hand, several types of cancer cells are known to upregulate the expression and activity of telomerase to maintain the length of their telomeres, essentially making them 'immortal'. This indicates that maintaining telomere length might not be an unambiguously positive factor for an organism's health, even if it might prevent aging.
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How much waste alone comes from toxic and hazardous wastes that are released into the environment?
A 40 Million Metric Tons
B 265 Million Metric Tons
C 11 Million Metric Tons
D 100 Million Metric Tons
Answer:
A 40 Million Metric Tons
Explanation:
40 million metric tons of toxic and hazardous wastes are produced globally each year