Answer:
MALT or mucosa-associated lymphoid tissue refers to a bundle of lymphatic cells, known as lymphatic nodules, situated inside the mucous membranes, which envelopes the respiratory, gastrointestinal, urinary, and reproductive tracts. These nodules comprise macrophages and lymphocytes that fight against the entering bacteria and other pathogens, which moves into these pathways along with air, food, or urine. These nodules can be grouped together in clusters or can be present solitary.
The major clusters of lymphatic nodules comprise adenoids, tonsils, and Peyer's patches.
If the ease of oxygen pickup depends on oxygen concentration, would respiration be easier in air or in water?
Answer:
Air
Explanation:
depending on the concentration of Oxygen in any of the medium- air or water, Oxygen pick up will be easier in air than water. This is because the energy barrier for the exchange of gases is higher in a liquid medium than in a gaseous medium.
Respiration would be easier in air than water
What is Respiration ?Respiration is the process of taking in oxygen and giving out CO₂ through the oxidation of organic substances in the body. The process of respiration leads to the production of energy in the body.
Respiration would be easier in a gaseous medium ( air ) than it would be in water because air molecules are more loosely packed and therefore allows the free movement of oxygen ( lesser energy barrier ) .
Hence we can conclude that Respiration would be easier in air than water.
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Describe the biological roots of Behavioral Neuroscience.
Answer:
Explanation:
The biological perspective or roots, a way of looking at neuroscience is by studying the physical basis for animal and human behavior. It involves such things as studying the brain, immune system, nervous system, and genetics.
The biological perspective tends to stress the importance of nature.
The study of physiology and biological processes has played a significant role in neuroscience since its earliest beginnings. Charles Darwin first introduced the idea that evolution and genetics has roles to play in human behavior. Natural selection influences if certain behavior patterns will be passed down to future generations. Behaviors that help survival skills most likely are passed down than those that prove dangerous.
The biological perspective is a way of looking at human problems and actions. For instance in aggression, someone using the psychoanalytic perspective might view aggression as the result of childhood experiences, another might take a behavioral perspective to show how the behavior was shaped by association and punishment. A neuroscientist with a social perspective might look at the group dynamics and pressures that contribute to such behavior. The biological viewpoint, on the other hand would look at the biological roots that lie behind aggressive behaviors by considering how certain types of brain injury might lead to aggressive actions. Or might consider genetic factors that can contribute to such displays of behavior.
The conversion of 1 mol of pyruvate to 3 mol of CO2 via pyruvate dehydrogenase and the citric acid cycle yields ___mol of NADH, ____mol of FADH2, and ___mol of ATP (or GTP).
A. 3; 2; 0
B. 4; 2; 1
C. 4; 1; 1
D. 3; 1; 1
E. 2; 2; 2
Answer: Option C.
4 mol of NADH, 1 mol FADH2 and 1 mol of ATP.
Explanation:
Pryruvate dehydrogenase is an enzyme that convert pyruvate to acetyl CoA by a process called private decatboxylation. This take place in the mitochondria. Pyruvate dehydrogenase convert 1 mol of pyruvate to 3 mol of Co2.
Critic acid cycle or kreb cycle is the series of reaction that produce energy through the oxidation of acetyl CoA produced from pyruvate dehydrogenase conversion in living organisms. Citric acid cycle produce ATP and reduced form of 1 mol of FADH2 and 4 mol of NADH.
Cindy is 63 years old and at risk for osteoporosis. Which nutrients would be the most important to consume in adequate amounts to preserve bone mass?
Answer:
Consumption of adequate calcium and vitamin D
Explanation:
Osteoporosis is a condition whereby there is significant amount of porous partition on the surface of the bone,this disease reduces the density and quality of bone. As bones become more porous and fragile, the risk of getting fractured is greatly increased.
In a bid to overcome or to be resistant to Osteoporosis, Consumption of adequate calcium and vitamin D is required by the body. This because Calcium performs various function which include;building strong bones, regulating heart beat and fluid balance within cells.
Also Vitamin D works in conjunction with calcium to slow down or even reverse osteoporosis. That is to say, the body cannot absorb calcium at all without some vitamin D. Hence, Vitamin D is vital in assisting the body absorb and use calcium.
The most important nutrients to preserve bone mass and prevent osteoporosis are calcium and vitamin D. Regular exercise, particularly resistance training, is also crucial for maintaining bone density. Other nutrients like vitamin K, magnesium, and omega-3 fatty acids can also support bone health.
Explanation:The most important nutrients to consume in adequate amounts to preserve bone mass for someone at risk for osteoporosis are calcium and vitamin D. Calcium is a critical component of bone and must be obtained from the diet, while vitamin D is necessary for the absorption of calcium. Both nutrients play crucial roles in bone health and can help prevent bone loss.
Regular exercise, especially resistance training, is also important for preserving bone mass and preventing osteoporosis. Exercise stimulates the deposition of bone tissue and helps improve bone density.
Additionally, other nutrients like vitamin K, magnesium, and omega-3 fatty acids also support bone health.
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Which primates are included among the prosimians? Why is this taxonomic group problematic?
Answer:
The primates that are included among the prosimians are all the existing or extinct species of strepsirrhines. The taxonomic groups are problematic because they fail to mark specific distinctions between the species.
Explanation:
It is from 85 to 55 million years that the primates have been existing on the Earth. They emerged as small terrestrial animals and developed into some large ones through the continued evolution of species. Daubentoniidae, Tarsiidae, Lemuridae are some examples of primate families.You decide to try Alex’s response to glucagon. This test consists of injecting a high dose of glucagonintravenously and then drawing samples of blood periodically and measuring the glucose content ofthe samples. After the glucagon injection, Alex’s blood sugar rises dramatically. Is this the responseyou would expect in a normal person? Explain
Answer:
YES!
Explanation:
Yes, because the liver is responsible for regulating blood sugar.
Consider a population of 425 diploid giant Sequoia trees. In this population, you observe the following genotypic counts: 100 homozygous dominant genotypes, 250 heterozygous genotypes, and 75 homozygous recessive genotypes. What is the allele frequency of the recessive allele in this population (use two decimal places and the usual rounding conventions)
Answer:
q = 0.42
Explanation:
This question is an example of Hardy-Weinberg question and there are two equations necessary to carry out this question;
p + q = 1
p² + 2pq + q² = 1
where;
p = the frequency of the dominant allele
q = the frequency of the recessive allele
p² = the frequency of individuals with homozygous dominant genotype
2pq = the frequency of individuals with heterozygous genotype
q² = frequency of individuals with the homozygous recessive genotype
Since the total population = 425
q² = [tex]\frac{individuals with recessive genotype}{Total Population}[/tex]
= [tex]\frac{75}{425}[/tex]
q² = 0.1765
To find q; we need to square root both side to eliminate the square from q².
∴ [tex]\sqrt{q^2}=\sqrt{0.1765}[/tex]
q = 0.4201
q = 0.42 (to two decimal places)
16.Globular proteins fold up into compact, spherical structures that have uneven surfaces. They tend to form
multisubunit complexes, which also have a rounded shape. Fibrous proteins, in contrast, span relatively large
distances within the cell and in the extracellular space. Which protein is not classified as a fibrous protein?
(a)elastase (b)collagen (c)keratin (d)elastin
Answer:(a)elastase
Explanation:
Elastase is a protease enzyme the function involves the cleavage of peptide bonds after amino acids with small side chains. This is responsible for cleaving the peptide bonds between the elastin fibers and aids in digestion of the elastic protein.
Thus the elastase cleaves the protein fibers this can be said that the elastase is a non-fibrous protein.
Answer:
The answer is A: elastase
Explanation:
Fibrous proteins are insoluble, long, rod-like structured proteins that have high α-helix or β-sheet content, and are made up of a repeating motif. They are mechanically strong, cross-linked and play structural roles in the extracellular matrix. Examples are: collagen, keratin and elastin.
Individuals with Huntington's disease possess an allele that causes severe symptoms by
a. replacing pairs of chromosomes.
b. halting the process of gene expression.
c. producing certain abnormal proteins.
d. regulating the sequences of nucleotides.
Answer:
C)Producing certain abnormal protiens
Explanation:
Mutations in HTT gene causes Huntington disease. The HTT gene provides instructions for making a protien called huntingin. This protien plays an important role in the neurons of brain. Thus as the gene coding for this protien is mutated ,abnormal protien is synthesized.
Clearly cellulose is very abundant on earth, and it is a long-lasting stable substance. Many animals cannot digest cellulose. Given this, what prevents the bodies of dead plants from filling the earth? Something must decompose cellulose. This is where fungal decomposition comes in--fungi digest cellulose, as do many prokaryotes. Name two organisms that consume cellulose and make an educated guess as to whether each breaks down cellulose or simply excretes it as fiber.
Answer: Termites and herbivores.
Explanation:
The two organism that can digests cellulose are termites and herbivores. The termite contains protists known as mastigophorans carry out digestion of cellulose in the body.
The cellulose in this case is digested and prevents it from getting deposited in the environment.
The other organism are animals like ruminants which can digest cellulose in their gut. They partially digest the cellulose and regurgitate it into the mouth an broken down further.
This process of digestion of cellulose in the gut is anaerobic so methane is released in the environment as a product of digestion.
The human beings cannot digest cellulose and a very less amount of it is considered as fiber and is simply excreted.
Final answer:
Ruminants like cows and fungi have the ability to digest cellulose due to cellulases, allowing them to break down this complex carbohydrate and utilize it as a food source, contributing to nutrient cycles in ecosystems.
Explanation:
Despite the fact that cellulose is a major structural component of plant cell walls and abundantly found on Earth, not all organisms can digest it due to its robust β(1,4) glycosidic bonds. However, certain organisms such as ruminants and termites have symbiotic relationships with microorganisms in their guts that produce enzymes capable of breaking down cellulose. For instance, cows have bacteria in their rumens that secrete cellulases, allowing them to break cellulose down into usable sugars.
Fungi also play a crucial role in cellulose degradation in ecosystems. They secrete exoenzymes into the environment to decompose dead plant material, preventing accumulation of undegraded biomass. These enzymes cleave the cellulose into glucose monomers, which they then utilize for energy and growth, contributing to nutrient cycling in the environment. Therefore, both ruminants and fungi do not simply excrete cellulose as fiber but effectively break it down for consumption.
A cultured cell line appears to be having trouble surviving. You find that the cells do not appear to have normal chromosome separation. You decide to check for the presence of a mutated protein. Which of the following would your best target for analyzing chromosome separation?
a. Actin
b. Melanin
c. Tubulin
d. Shugoshin
Answer:
c. Tubulin
Explanation:
Tubulin protein is polymerized to form the cylindrical structures of microtubules. Microtubules form the spindle apparatus during cell division. The spindle microtubules become attached to the kinetochores of chromosomes and mediate the alignment of chromosomes at the equator of cells during metaphase. The shortening of spindle microtubules is responsible for the movement of sister chromatids during anaphase. The same event also moves the homologous chromosomes during anaphase-I.
Any failure in the formation of the spindle apparatus would not allow the proper separation of chromosomes. Therefore, the cell with abnormal chromosome separation might have a faulty or no tubulin.
What symbiotic partnerships form between plant roots and fungi, and increase water and mineral absorption by the plant?
Answer: It is a Mutualistic partnership between fungi and roots of plant
Explanation:
Mutualism is a relationship between two organisms in which both of the benefits from the relationship. Fungi live in the roots of plant thereby getting carbohydrates from plants made by photosynthesis and the fungi in turn put mycelia that help plant to increase absorption of minerals and water.
Answer: Symbiosis
Explanation:
The root nodules of legumes contains organisms of the kingdom, fungi that helps the legumes fix atmospheric nitrogen thus meeting it's need for mineral Nitrogen. In turn, the legume plants provides shelter to the fungi.
This codependent partnership is known as Symbiosis
Maria is a 121-lb endurance athlete who is planning to employ carbohydrate loading before her next race. Which strategy correctly follows the guidelines for carbohydrate loading?
Choose the statement below that correctly describes a good carbohydrate loading strategy for Maria:
O Four to six days prior to the event, Maria should consume 550 g of carbohydrate daily and decrease to 220 to 275 g of carbohydrate daily 1 to 3 days prior to the event.O One to three days before the event, 550 g of carbohydrate daily is recommended.O One to three days before the event, 1210 g of carbohydrate daily is recommended.O Six days before the event, her carbohydrate intake should be 550 g per day.O Two days before the event, an intake of 33 g of protein daily is adequate.
Final answer:
The correct carbohydrate loading strategy for Maria, a 121-lb endurance athlete, is to consume 550 g of carbohydrate daily for 1 to 3 days before her event to maximize glycogen storage for energy.
Explanation:
The correct carbohydrate loading strategy for Maria—an endurance athlete who weighs 121 lbs—would be to consume 550 g of carbohydrate daily during the 1 to 3 days prior to her event. This is because carbohydrate loading is aimed at maximizing the storage of glycogen in the muscles, which is used for energy during prolonged periods of intense exercise. Ingesting 550 g of carbohydrates daily, following the guidance, should increase her glycogen stores, providing an energy reservoir that can be tapped into during her endurance event.
Concerning the options given, the second option correctly follows the guidelines for carbohydrate loading: "One to three days before the event, 550 g of carbohydrate daily is recommended." This is based on the principle that consuming high amounts of carbohydrates can boost glycogen stores within muscles, effectively increasing the athlete's endurance capacity during the race.
Protists are very diverse, and many different classification schemes have been used to define relationships between the protists. Traditionally, protists have been classified by their source of energy and nutrients, while newer classification schemes use mode of locomotion. How are amoeboids classified using both of these schemes?
A. photosynthetic, cilia
B. heterotrophic by absorption, nonmotile
C. heterotrophic by ingestion, nonmotile
D. heterotrophic by ingestion, pseudopods
Answer:
D. heterotrophic by ingestion, pseudopods
Explanation:
Protists are generally classified as all eukaryotic organisms that are not plants, animals or fungi. Example is amoeba, paramecium etc.They may be unicellular or multi cellular in nature.Most exist in colonies.
Their mode of nutrition can be photosynthetic or hetrotrophic. Hetrotrophic protists can be divided into phagotrophs and osmotrops/saprotrophs. The phagotrophs makes use of the cell body to engulf the food materials as in amoeba ,carry out extracellular digestion before swallowing it.
Osmotrops absorbed dissolved food from surrounding liquid environments directly. (Some photosynthetic protists can also be heterotrophic.
Amoeboid movement is the mode of locomotion of protists and some other eukaryotes. It involved the protrusion of cytoplasm, which exert pressure on the cell membrane to form pseudopodia and the posterioly evolved Uropods.
Sol-gel theory has been proposed to expalin this movements, The ectopalsm of amoeba is gelly-like , while the endiplams is less viscpus and said to be sol. The interchange of the cytoplasmic fluis between the endo-and ecto plasm gives the SOL-GEL propulsion of the protopalms for the amoebic moveemnts .
The false feet(psuedopodium) drags the amoeba along in the direction of the flow of the cytoplasm.
Therefore option D is the right option
Which of the following is NOTa sign that a survivor may need stabilization?Select one:a. Excessive talkingb. Glassy and vacant eyesc. Strong emotional responsesd. Uncontrollable physical reactionse. Frantic searching behavior
Answer:
Excessive talking is not a sign that a survivor may need stabilization.
Explanation:
In humans alkaptonuria is a metabolic disorder in whichaffected persons produce black urine. Alkapotonuria results from anallele(a) that is recessive to the allele for normal metabolism(A). Sally has a normal metabolism, but her brother hasalcaptonuria. Sally's father has alcoptonuria, and her mother hasnormal metabolism.
a) Give the genotype of Sally,her mother,father and herbrother.
b) If Sally's parents have another child what is theprobabilty that this child will have alkaptonuria?
c) If Sally marries a man with alkaptonuria, what is theprobability that their child will have alkaptonuria?
If Sally marries a man with alkaptonuria, there is a 50% chance that their child will have alkaptonuria.
The genotype of an individual refers to the sum total of genes that the individual received from its parents. Since Sally has a normal metabolism, Sally is Aa. Sally's mother is Aa while Sally's father and brother are aa.
If Sally parents have another child, using the Punnet square method, there is a 50% chance that the child will have alkaptonuria. If Sally marries a man with alkaptonuria, there is a 50% chance that their child will have alkaptonuria.
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a) Genotypes: : Aa (carrier of the alkaptonuria allele) ,Mother: Aa (carrier of the alkaptonuria allele)Father: aa (has alkaptonuria)Brother: aa (has alkaptonuria)
b) If Sally's parents have another child:The probability of the child having alkaptonuria (aa genotype) is 25%.The probability of the child being a carrier (Aa genotype) is 50%.The probability of the child having normal metabolism (AA genotype) is 25%.
c) If Sally marries a man with alkaptonuria (aa genotype):The probability of their child having alkaptonuria (aa genotype) is 50% (as Sally is a carrier).The probability of the child being a carrier (Aa genotype) is 50%.
The probability of the child having normal metabolism (AA genotype) is 0% (as the husband has alkaptonuria).
Sally: Aa (normal metabolism carrier)
Mother: Aa (normal metabolism carrier)
Father: aa (alkaptonuria)
Brother: aa (alkaptonuria)
b) If Sally's parents have another child:
Probability of the child having alkaptonuria (aa genotype) is 25%.
Probability of the child being a carrier (Aa genotype) is 50%.
Probability of the child having normal metabolism (AA genotype) is 25%.
c) If Sally marries a man with alkaptonuria (aa genotype):
Probability of their child having alkaptonuria (aa genotype) is 50% (as Sally is a carrier).
Probability of the child being a carrier (Aa genotype) is 50%.
Probability of the child having normal metabolism (AA genotype) is 0% (as the husband has alkaptonuria).
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Autoimmunities are relatively uncommon. What usually happens to autoimmune antibody-producing clones during development
Answer
There is not enough antibody-producing clones during development therefore the immune system suffers.
What unusual ability do sea cucumbers and polychaete worms have at this depth to help them find food
Answer:
Bioluminescence
Explanation:
In the deep oceans, the sea cucumbers and polychaete worms use the bioluminescence for defense against predators and to find or attract prey by glowing so the predator can easily see it's glowing skin.
Final answer:
Sea cucumbers and polychaete worms utilize bioluminescence to navigate, communicate, and find food in the deep-sea's perpetual darkness, using a chemical reaction involving luciferin, luciferase, and ATP.
Explanation:
Sea cucumbers and polychaete worms found in the deep sea have an unusual ability to emit light through a process known as bioluminescence. This remarkable feature is utilized to find food in the perpetual darkness of their environment. For example, the anglerfish, which lives between 1000 and 4000 meters below sea level where no sunlight penetrates, has a rod-like structure with a glow-in-the-dark tip to lure prey. In similar fashion, Deep-sea organisms like sea cucumbers and polychaete worms may use bioluminescence to attract or detect prey, or for other reasons that are still being researched by scientists.
Bioluminescence involves a chemical reaction that includes a light-emitting molecule called luciferin, an enzyme called luciferase, and ATP which provides the energy for the reaction. The ability to create light is especially advantageous in the deep sea, allowing these organisms to navigate, communicate, or deter predators in an environment devoid of natural light sources.
Which statement taken from a student’s logbook would beconsidered an experimental result?a.The speed of a snail isb.The average speed of the controlc.A mercury thermometerd.If temperatureaffected by temperature.group of snails is 2.4 cm/min.was used to measure temperature.increases, then the speed of the snail decreases
Answer:
A. The speed of a snail is
Explanation:
An experimental result, in this context, is the raw result (data) from the experiment. It has not undergone any modification, neither is it due to any other calculations. It is simply observed from the experiment. The speed a snail would be an experimental result.
The average speed is a calculation derived from several observations.
A mercury thermometer is an instrument used for measurement and not an experimental result .
Indicate whether below is True/False.
1. Mammary glands are specialized for milk and hormone production.
2. Breasts contain areolar connective tissue but little dispose tissue.
3. Alveolar glands occur in lobes of mammary glands.
Answer:
1. True
2. True
3. False
Explanation:
Mammary glands are made up of glandular tissue. These are meant for the secretion of milk after the newborn birth. The milk ejection is done by different hormones, especially by progesterone and prolactin.
Glandular tissue acts as both exocrine and endocrine glands. The mammary glands are a suitable example as it secretes milk which is an exocrine secretion. The hormone progesterone and prolactin released from the mammary gland are endocrine secretions.
The mammary glands consist of a sac-like structure called alveoli.
These are group alveoli form grape-like appearance called lobules.
The alveolar sacs contain sweat glands. These are the modified sweat glands that secrete milk. It is made up of fibrous connective tissue.
Assume that a cross is made between two organisms that are both heterozygous for a gene that shows incomplete dominance. What genotypic ratio is expected in the offspring?
Answer:
Genotypic ratio: 1 homozygous for parental trait : 2 heterozygous : 1 homozygous for other parental trait
Explanation:
Incomplete dominance is a non-mendelian type of inheritance in which one allele of a gene portrays Incomplete dominance over the other allele, hence, they combine to form a third phenotype that is a blending of both parental phenotypes. A very good example of incomplete dominance gene is that of flower colour in snapdragon plants.
One allele of the flower colour gene codes for RED while the other codes for WHITE. However, an heterozygous intermediate phenotype (PINK) is formed as F1 offspring when both parents are crossed.
If the F1 heterozygous offsprings are self-crossed i.e. two heterozygous offsprings, four possible offsprings will be produced with the genotypic ratio 1:2:1, which means that 1/4 of the offsprings will be homozygous for one of the parental traits, 2/4 will be heterozygous for both alleles, 1/4 will be homozygous for the other parental trait.
The branch of biology which deals with the study of genes and inheritance is called genetics. In genetics, there are two types of alleles present in an organism. These alleles are as follows:-
Dominant Recessive
Gregor John Mendel is the father of genetics and describes the functioning of the gene in an organism.
Let's assume the gene of the parents, The genes are as follows:-
Mother - TtFather - TtThese different type of genes is known as Heterozygous genes. If we cross both the parents together, the gamete formed is "T" and "t".
After crossing the genes, the genotype of the offspring will be "TT", "Tt", "tt".
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A point restricted rabbit was mated with a chinchilla rabbit. The two offspring were albino and chinchilla.
What were the genotypes of the parents?
Answer:
The result shows that the parents innately is able to bread an albino,
Let's assume they both have a Can the allele or albinism.
One is plain color that is CCa
and the other is light gray so it has to be CchCa.
Outlined below are the possibilities of the offspring of the cross
Plain coloured offspring: CCch, CCa
Light gray: CchCa
Albino: CaCa
This explains that the offspring are possible and we are likely to have the outcomes linking to the parental genotypes.
Explanation:
The genotypes of the parents can be determined by analyzing the phenotypes of the albino and chinchilla offspring. One parent must have the genotype cc, while the other parent must have the genotype cchech or chcch.
Explanation:The genotypes of the parents can be determined by analyzing the phenotypes of the offspring. In this case, the offspring were albino and chinchilla.
Since one of the offspring was albino, which is expressed as white fur, this means that one of the parents must have had the genotype cc, as the albino phenotype is only expressed in individuals with two recessive alleles.
Since the other offspring was chinchilla, which is expressed as black-tipped white fur, this means that the other parent must have had the genotype cchech or chcch, as the chinchilla phenotype is dominant over albino and Himalayan, and incompletely dominant over Himalayan. The chinchilla phenotype requires at least one dominant allele for its expression.
The thoracic cavity contains the ________. It is found ________ to the vertebral cavity.
A) stomach and liver: superficial
B) heart and lungs: anterior
C) digestive viscera: inferior
D) kidneys and spleen: deep
Answer:
B Thoracic cavity contains the heart and lungs and is anterior/in front of the vertebral cavity
Explanation:
The thoracic cavity contains the heart and lungs, and it is found anterior to the vertebral cavity.
The thoracic cavity, a crucial anatomical region, houses the heart and lungs, playing a pivotal role in circulatory and respiratory functions. Positioned anteriorly to the vertebral cavity, it lies in the upper part of the trunk and is bound by the rib cage. This anatomical arrangement protects and supports vital organs within the thoracic cavity. The heart, a muscular organ responsible for pumping blood throughout the body, is centrally located, while the lungs flank it on either side, facilitating the exchange of oxygen and carbon dioxide. The anterior positioning of the thoracic cavity implies its presence towards the front of the body, emphasizing its strategic role in bodily functions. This anatomical knowledge is foundational for understanding physiological processes and aids in clinical contexts, guiding medical professionals in diagnosis and treatment.
Therefore, the correct answer is B) heart and lungs: anterior.
Which set of details correctly identifies a series of events in a sympathetic pathway?
a. thoracolumbar origin, long preganglionic fiber, NE release at ganglion, short postganglionic fiber, NE release at effector
b. craniosacral origin, short preganglionic fiber, ACh release at ganglion, long postganglionic fiber, ACh release at effector
c. thoracolumbar origin, short preganglionic fiber, ACh release at ganglion, long postganglionic fiber, NE release at effector
d.craniosacral origin, long preganglionic fiber, ACh release at ganglion, short postganglionic fiber, ACh release at effector
Answer:
Answer is C.
Explanation:
The sympathetic pathway is the pathway of the sympathetic nervous system. Sympathetic nervous system is a part of the autonomic nervous system , that comprises of neurons regulating the body's involuntary actions. It is also regarded as the pathway by which organisms respond to stress, because it prepares the body for stress.
The sympathetic nervous system help the body to respond to stress by increasing the rate of the heart beat, activating the release of adrenaline among others.
The other part of the autonomic nervous system is the parasympathetic nervous system, which brings the body to the state of calmness and relaxed feeling after a stressful event. This is done by slowing down the heart rate and increasing the rate of digestion.
1. Which of the following processes will most likely occur when you transfer a bacterial culture from 37 °C to 10 °C?
a) the bacteria will start synthesizing cholesterol
b) the bacteria will remove cholesterol from their membranes
c) the bacteria will reduce the number of unsaturated bonds in the fatty acid tails of their phospholipids
d) the bacteria will increase the synthesis of phospholipids with very long fatty acid tails and insert these into their membranes
e) the bacteria will increase the synthesis of phospholipids with unsaturated fatty acid tails and insert these into their membranes
2. Following up on question 1, which of the following best describes what would happen to a culture of yeast cells that was shifted from 37 °C to 10 °C?
a) The cells will die because their plasma membrane will be leaky
b) The cells will increase the synthesis of phospholipids with long, saturated fatty acid chains
c) The cells will increase the synthesis of phospholipids with short saturated fatty acid tails
d) The cells will insert more cholesterol into their ER membrane
e) The cells will insert more ergosterol into their plasma membrane
f) The cells will remove phospholipids with unsaturated fatty acid tails from their plasma membrane
Answers with Explanations:
1. Which of the following processes will most likely occur when you transfer a bacterial culture from 37 °C to 10 °C?
The answer is letter e, the bacteria will increase the synthesis of phospholipids with unsaturated fatty acid tails and insert these into their membranes.
Explanation: Bacteria are capable of surviving different types of environments with varying temperatures. When it is exposed to a low temperature, as in the situation above (from 37 °C to 10 °C), it regulates its transition phase. Th cytoplasmic membrane of organisms plays a vital role when it comes to physiological environments.
When it comes to bacteria, their membrane is primarily composed of "phospholipids." It is said that the phospholipids with unsaturated fatty acids have a transition that is lower to that of phospholipids with saturated fatty acids when it comes to temperature. Thus, the bacteria will have to increase the synthesis of phospholipids with unsaturated fatty acid tails and insert these into their membranes. This will help increase the membrane's fluidity, thus allowing it to survive in low temperatures.
2. . Following up on question 1, which of the following best describes what would happen to a culture of yeast cells that was shifted from 37 °C to 10 °C?
The answer is letter b, The cells will increase the synthesis of phospholipids with long, saturated fatty acids chains.
Explanation: The growth of yeast is affected by several factors, naming temperature as one. They are sensitive to temperature, such that they can be killed in water if it reaches a peak of 60°C.
At 10 °C, the yeast will continue its metabolic activity but with low growth rates. Its biological membrane now comes to play its part when it comes to regulation. The growth of temperature is directly related to the degree of the cell's lipid unsaturation. This means that when temperatures are lower, the saturation of lipids is low. So, the cells would most likely increase the synthesis of phospholipids with long, saturated fatty acid chains when the temperature is low.
You are looking at chromosome 1 in a human being. Assuming there is no crossing over, what is the source of all the genes on this chromosome?
Final answer:
The source of genes on chromosome 1 in a human being, assuming no crossing over, is one of the individual's parents. Each parent contributes one set of 23 chromosomes, and the combination of genes from these chromosomes determines the individual's characteristics and traits.
Explanation:
If you are looking at chromosome 1 in a human being and assuming there is no crossing over, then the source of all the genes on this chromosome is either the individual's mother or father. Each parent contributes one set of 23 chromosomes at conception, when the sperm (from the father) and the oocyte (from the mother) combine. Chromosome 1 is one of those 23 chromosomes. The genes located on chromosome 1 determine a variety of characteristics and traits, with alleles for these genes possibly varying between the two parental chromosomes.
Each copy of the homologous pair of chromosomes originates from a different parent, leading to variation in individuals within a species. The specific combination of the genes inherited from both parents causes this variation. Beyond this, certain traits, such as blood type, are determined by which specific versions of a gene are inherited from each parent.
Furthermore, genes on the same chromosome are linked, and their alleles usually segregate together during meiosis, unless separated by crossing over. Therefore, in the absence of crossing over, all the genes on chromosome 1 come from the same parent and are likely to be inherited together.
Bird feathers are modified scales.
a. True
b. False
Answer:
A. True.
Explanation:
True. b is wrong
The statement Bird feathers are modified scales is a. True.
Bird feathers and reptile scales share a common evolutionary origin. Feathers are considered to be highly modified scales that have evolved over millions of years.
This evolutionary relationship is supported by both developmental and genetic evidence.
Feathers and scales are made of the same protein, beta-keratin, which is a protein unique to reptiles and birds.
The fundamental structure of feathers and scales is quite similar, with both composed of layers of beta-keratinized cells.
This similarity in composition and structure strongly suggests a common ancestry.
During embryonic development, both feathers and scales start as simple epithelial cells.
As development progresses, these cells differentiate and form either feathers in birds or scales in reptiles, depending on the genetic instructions within the organism.
The evolutionary transition from scales to feathers is thought to have provided advantages such as improved insulation, enhanced aerodynamics for flight, and eventually elaborate displays for courtship. Feathers underwent various modifications and specialized into diverse forms serving different functions, from flight feathers for flying to down feathers for insulation and display feathers for attracting mates.
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The GART, a gene which is involved with Down syndrome, is found toward the bottom of chromosome 21. This is an example of a ________.
Answer: It is an example of TRISOMY 21.
Explanation:
Trisomy 21 is a genetic condition in which 21 chromosome fail to separate during the growth of egg cell and sperm which make either of the two to an extra copy of chromosome 21. This is also known as down syndrome. This lead to slow and in appropriate mental development in affected individual.
Phosphoriboglycinamide transformylase(GART) is agene found on chromosome 21, which lead to down syndrome is an example of trisomy 21.
Glucose moves from the plasma into a skeletal muscle cell, where it is used for energy. Through which fluid compartment does glucose move between the plasma and the skeletal muscle cell?
(A) Intracellular fluid
(B) Interstitial fluid
(C) Extracellular fluid inside of blood vessels
(D) Cytosol
Answer:
(B) Interstitial fluid
Explanation:
The interstitial fluid and blood plasma together make the extracellular fluid. The extracellular fluids are present outside the cells. The extracellular fluid that is present in the narrow spaces between cells of tissues is known as interstitial fluid. When a substance moves from blood plasma into the cells of a tissue, it crosses the interstitial fluid present between its cells. Therefore, when a skeletal muscle cell picks glucose molecules from blood plasma, it moves from plasma to the interstitial fluid to enter the cell.
Glucose moves from the plasma, through the interstitial fluid, which surrounds cells, to reach the skeletal muscle cells where it's used for energy. The correct answer is (B).
Explanation:The movement of glucose from the plasma into a skeletal muscle cell follows a specific path through different fluid compartments in the body. After digesting carbohydrates, glucose is absorbed into the bloodstream and circulates in the blood plasma. From here, glucose must then pass through the interstitial fluid that surrounds the cells. This movement is facilitated by a concentration gradient, where glucose levels are higher in the blood compared to inside the cells, and by glucose transport proteins present in the cell membrane. Thus, the correct answer is (B) Interstitial fluid, as the glucose moves from the plasma, through the interstitial space, and finally into the skeletal muscle cells where it can be utilized for energy production.
It is important to note that insulin plays a significant role in this process by stimulating the uptake of glucose into cells, particularly into liver and muscle cells for storage and energy use. Additionally, the sodium-potassium pump and facilitated diffusion mechanisms are involved in maintaining proper electrolyte balance and glucose transport, respectively.
Which of the following answers describes the most direct consequence of tropomyosin binding to F-actin in muscle cells? a. The (-) end of F-actin is stabilizedb. Loss of actin-ADP subunits from the (+) end is preventedc. Myosin binding to F-actin is blockedd. ATP hydrolysis by myosin is blockede. The movement of myosin towards the (+) end is blocked
Answer: A
Explanation:
The "-" end of F actin is stabilized. The myosin head bind to actin and makes the actin filament to slide