Answer:
The correct answer is they show weak positive gram staining reaction.
Explanation:
Mycobacterium phlei is a rod-shaped species of the genus Mycobacterium. They are acid-fast bacteria with mycolic acids rich cell wall due to which gram stain cannot penetrate easily and show light purple color while staining.
When gram staining is performed on the M.phlei , they take light purple color which indicates that it is a gram-positive bacteria and is related to Actinobacteria phylum of the gram-positive bacteria.
Thus, M. phlei is a gram-positive bacteria.
Mycobacterium plei cannot be accurately classified as Gram-positive or Gram-negative due to its unique cell wall composition; instead, it is categorized as acid-fast. This differentiation is important for understanding the bacterium's structural and resistance properties. The acid-fast characteristic suggests a lipid-rich cell wall, distinct from typical Gram-positive or Gram-negative bacteria.
Explanation:The question about whether Mycobacterium plei is Gram-positive, Gram-negative, or neither is rooted in the practice of Gram staining, which is a vital technique used in microbiology to classify bacteria. Mycobacterium spp., pertaining to the question, are better classified as acid-fast rather than strictly Gram-positive or Gram-negative due to their unique cell wall composition. This classification reveals that their cell walls contain mycolic acids and are thicker than typical Gram-positive bacteria, which contributes to their resistance to the Gram stain procedure and necessitates a different staining method, known as the acid-fast stain, for proper visualization.
Understanding the distinction between Gram-positive and Gram-negative bacteria is crucial because it involves the bacteria's cell wall structure. Gram-positive bacteria have a thick peptidoglycan layer, which retains the crystal violet dye used in the Gram stain, turning them purple. Gram-negative bacteria, on the other hand, have a thin peptidoglycan layer and an outer membrane that does not retain the crystal violet dye but takes up the safranin counterstain, appearing pink. The unique lipid-rich cell wall of Mycobacterium spp. prevents both the uptake and the retention of these dyes, making the acid-fast staining necessary.
In summary, recognizing the staining characteristics of bacteria, such as Mycobacterium plei, aids in their identification and provides insight into their structure and potential resistance mechanisms. The acid-fast quality of Mycobacterium indicates a cell wall composition distinct from the general categories of Gram-positive or Gram-negative, demonstrating the diversity and complexity of bacterial cell wall structures.
How many ATP and NADPH molecules are required in the C3 pathway to make one six-carbon sugar? If the synthesis of a molecule of ATP were to require four protons, would you expect that these relative requirements for ATP and NADPH could be met by noncyclic photophosphorylation in the absence of cyclic photophosphorylation?
Answer:
The correct answer is 18 ATP and 12 NADPH and noncyclic photophosphorylation can produce ATP molecule in the absence of cyclic photophosphorylation.
Explanation:
ATP and NADPH molecules are synthesized during light reaction of the photosynthesis which is utilized during the reactions of the dark phases. Dark reaction or Calvin cycle (C3 cycle) is the cyclic pathway of producing glucose triose phosphates (3C) from carbon dioxide and water. This reaction proceeds into 3 phases: carboxylation, reduction and regeneration.
First ATP and NADPH are utilized during the reduction step in the reduction of 3-phosphoglycerate to glucose-3-phosphate by the transfer of phosphate group from 6 ATP to 3-phosphoglycerate and 6 NADPH reduction as it donates an electron. Regeneration step also uses 3 ATP in conversion of G3P to RUBP molecules.
A total of 9 ATP and 6 NADPH are utilized in producing 3C G3P molecule, So, to produce 6C sugar molecule 18 ATP AND 12 NADPH are used.
During chemiosmosis synthesis of ATP, 4 protons produce 1 molecule of ATP which could be able to generate 3 molecules of ATP for each pair of NADPH. so, noncyclic photophosphorylation or Z-scheme will be able to produce ATP in the absence of cyclic photophosphorylation.
Thus, 18 ATP and 12 NADPH and noncyclic photophosphorylation can produce ATP molecule in the absence of cyclic photophosphorylation are the correct answer.
Define Undifferentiated connective tissue disease
Answer:
is a disease in which the body mistakenly attacks its own tissues. It is diagnosed when there is evidence of an existing autoimmune condition which does not meet the criteria for any specific autoimmune disease]
Can the genotype for a gray-bodied fly be determined? Why or why not? Describe all of the possible genotypes for a fly with that phenotype
Answer:
The genotype of a fly can be determined gray bodied is dominant over black body flies. The possible genotype of a fly is either YY or Yy.
Explanation:
The genotype of a gray bodied fly can be determined with the help of punnett square. The gray bodied fly is crossed with black bodied fly, the original genotype of the fly can be determined. This type of cross is known as test cross.
If the cross results in the formation of all gray bodied progeny, the fly is homozygous dominant with the genotype YY. The cross results in the mixture of progeny with gray and black body , then the genotype of a fly is heterozygous dominant Yy.
Thus, the genotype of a fly can be determined by the test cross. The fly may have genotype either YY or Yy.
Copper in soil is usually toxic to plants, but some plants that grow near copper mines evolve copper tolerance. When copper-tolerant plants are grown in copper-free soil, they are outcompeted by the non-tolerant genotype. Which of the following principles does this scenario exemplify?
a.
The evolution of tolerance to copper resulted in a dependence on copper
b.
Evolution always maximizes the improvement or progress of a species
c.
Evolutionary changes can be advantageous under some environmental conditions but disadvantageous under others
d.
Resistant phenotypes always out-compete non-resistant phenotypes
Answer:
C. Evolutionary changes can be advantageous under some environmental conditions but disadvantageous under others
Explanation:
Evolution means to change something to survive, but that is under specific context. If we change the context, maybe that evolution is not advantegerous on the new scenario or context.
In this case, the plant evolve to survive copper in soli. If the soil has no more copper, that evolution is no longer needed.
A fellow student in the class is looking at a slide of onion root tip cells and is excited that he can see the chromosomes undergoing anaphase 1. How do you know that he is incorrect The chromosomes would not be coiled up during Anaphase 1 and therefore he would not be able to distinguish them Onion root tip cells do not undergo anaphase 1 He would need to use an electron microscope to see chromosomes Plants do not carry out mitosis
Answer:
The correct answer should be the second option, that is, Onion root tip cells do not undergo anaphase 1.
Anaphase I is a part of meiosis which occurs only in cells involved in reproduction. For example, gametes.
Onion root tip cells undergo mitosis.
The chromosome can be observed using compound microscope.
Thus, all other options are incorrect except second option.
Answer:
Explanation:
In the body of a multicellular organisms such as onion plant, there are two types of cells, namely somatic cells and reproductive cells. Somatic cells undergo mitosis and involved in vegetative growth whereas reproductive cells undergo meiosis that form gametes during sexual reproduction. Root tip of opinion consists of meristematic cells which are somatic cells which do not divide by meiosis. A fellow student recognises anaphase I in root tip cells is incorrect because anaphase I is a stage of meiosis I. Root tip cells only divide by mitosis. However, during anaphase I, a pair of chromosome separate which cannot be observed during mitosis in root tip cells.
Does anyone know how asymmetric liposome works? asymmetric liposome
Answer:
Asymmetric liposomes have different lipids in outer and inner leaflets, which would greatly increase the flexibility of vesicle in drug delivery systems. It has been well known that the phospholipid distribution in natural membranes is asymmetric. For example, phosphatidyl tcholine and sphingomyelin concentrate at the outer leaflet whereas phosphatidylethanolamine, phosphatidylinositol and phosphatidylserine are mainly localized in the inner leaflet. Typically, Lipids are self-assembled symmetrically in artificial liposomes regardless of the preparation methods. As drug delivery carriers, asymmetric liposomes with advanced functions are appealing candidates for targeted accumulation and controlled drug release. Their outer and inner leaflet could be manipulated depending on the nature of encapsulated drug molecules. For example, asymmetric liposomes help deliver negatively charged siRNA to target organs by having positively charged inner layer that encapsulates siRNA with high efficiency, and negatively charged outer surface prevents nonspecific uptake of the asymmetric liposomes. The unique tunability of asymmetric liposomes opens a wide door for multi-site functionalization, resulting in highly engineered liposomes as advanced drug delivery vesicles
Fred has chronic emphysema. Blood tests show that his pH is low but almost normal but his bicarbonate levels are elevated significantly. How can this be? What would urinalysis show?
Final answer:
Fred's chronic emphysema has led to respiratory acidosis with compensatory metabolic alkalosis, indicated by low but almost normal pH and high bicarbonate levels. His kidneys are attempting to balance the blood's pH, likely leading to alkaline urine shown in urinalysis.
Explanation:
Fred has chronic emphysema, a condition often associated with respiratory acidosis due to hypercapnia, which is an accumulation of CO₂ in the bloodstream. The elevation in bicarbonate levels suggests that his body is attempting to compensate for the acidosis. In a compensatory response, the kidneys retain bicarbonate to neutralize the excess acid. A urinalysis in Fred's case might show an elevated pH due to the body trying to excrete excess hydrogen ions (acidic ions) and retain more bicarbonate.
Chronic respiratory conditions like emphysema impair the lungs' ability to adequately expel CO₂, leading to an increase in the partial pressure of CO₂ (pCO₂) and a decrease in blood pH, initiating acidosis. However, over time, the kidneys attempt to regenerate and reabsorb bicarbonate ions as a compensatory mechanism, which is why Fred's bicarbonate levels are significantly elevated. Meanwhile, the urinalysis might further confirm this compensation by showing increased bicarbonate or alkaline urine, indicating the kidneys are working to correct the pH imbalance.
Which of the meninges is a delicate connective tissue membrane that clings tightly to the brain like cellophane wrap following its every convolution?
Answer:
Pia mater
Explanation:
Three meninges cover the brain and spinal cord. Pia mater refers to the deepest meninx of the brain. It is a thin delicate layer of connective tissue. It is a transparent connective tissue layer like that of cellophane. Being a transparent and thin tissue layer, it looks like a cellophane wrap adhered to the surface of the brain.
The delicate connective tissue membrane that clings tightly to the brain and follows its convolutions is the pia mater, which is the innermost layer of the meninges.
The meninges are a series of protective membranes that cover the brain and spinal cord. Among the layers of the meninges, the innermost layer that clings tightly to the brain and follows its every convolution is called the pia mater.
This delicate connective tissue membrane is directly adjacent to the surface of the brain, fitting into the grooves and indentations of the cerebral cortex. The pia mater ensures that the brain's surface is well-protected, adapting to its intricate structures.
The other two layers of the meninges include the dura mater, which is the tough, fibrous, outermost layer, and the arachnoid mater, which is situated between the dura and pia mater and resembles a spider web.
If a person’s blood cortisol concentration is 10 µg/mL, and 10 mg of cortisol can be excreted from the body in a 24-hour period, what volume of blood plasma in mL was “cleared” during this amount of time?
Answer:
[tex]1000 ml[/tex] of blood plasma was “cleared”
Explanation:
Given,
Concentration of a person’s blood cortisol [tex]= 10[/tex]µg/mL
The amount of cortisol excreted from the body in a period of 24 hour [tex]= 10[/tex]mg
The volume of blood plasma cleared during this time period is equal to
amount of cortisol excreted from the body divided by Concentration of a person’s blood cortisol
The above definition could be represented as following-
volume of blood plasma cleared [tex]= \frac{A}{B} \\[/tex]
Where A represents amount of cortisol excreted from the body
while B represents the concentration of a person’s blood cortisol
Substituting the given values in above equation, we get -
[tex]= \frac{10 * 10^{-3}g}{10*10^{-6}\frac{g}{ml} } \\= 1000 ml[/tex]
[tex]1000 ml[/tex] of blood plasma was “cleared”
The volume of blood plasma cleared of cortisol in 24 hours is calculated to be approximately 1000 mL. Clearance rate is found to be roughly 0.694 mL/min.
To find the volume of blood plasma that was "cleared" of cortisol in a 24-hour period, we must use the given values.
The blood cortisol concentration is 10 µg/mL, and 10 mg (which is equal to 10,000 µg) of cortisol is excreted in 24 hours.
Calculate the total amount of cortisol excreted in 24 hours: 10 mg = 10,000 µg.Use the formula for clearance, where clearance (C) is the volume of plasma cleared per minute (mL/min):C = (Amount of Substance Excreted in 24 Hours) / (Concentration in Blood Plasma * Time in Minutes)
Substitute the known values into the formula:
C = 10,000 µg / (10 µg/mL x 1440 minutes)
C = 10,000 µg / 14,400 µg/mL
C ≈ 0.694 mL/min
Total Volume Cleared in 24 hours = Clearance Rate x Time
Total Volume Cleared in 24 hours = 0.694 mL/min x 1440 minutes
Total Volume Cleared in 24 hours ≈ 1000 mL
Hence, approximately 1000 mL of blood plasma was cleared of cortisol in this 24-hour period.
Describe the parts of a neuron including the cell body, dendrites, axons, and terminals.
Answer:
A neuron has a cell body, an axon and dendrites. Apart from these structures, the white myelin sheath covers the axons. The gaps where myelin sheath is absent are called nodes of Ranvier. Axon terminals are the extreme ends of the axon from where nerve impulse is carried to the dendrites of the post synaptic neuron.
Explanation:
A neuron has a cell body, an axon, and dendrites.
Cell body: It is the metabolic center of the neuron and does not take part in the conduction of nerve impulse.
Axon and dendrites: These are the extensions arising from the cell body. Dendrites are smaller and more in number while axon is a single large extension.
The function of the axon is to carry the nerve impulse from the cell body to the axon terminals. Dendrites receive the nerve impulse from the axon terminal and pass it to the cell body. The ends of axons are called axon terminals. They synapse with dendrites of postsynaptic neurons.
The myelin sheath is the fatty layer present on long axons and serves to accelerate the rate of conduction of nerve impulse. The nodes of Ranvier are the gaps where the myelin sheath is not present.
How does the myelin sheath help the neuron?
Answer:
Myelin sheath increases the speed of electrical impulse along the nerve cells.
Explanation:
Myelin sheath is an insulting layer of proteins and fatty acids that are present around the nerves of brain and spinal cord.
Myelin sheath increases the speed of conduction of nerve impulse from one nerve cell to another cell. Myelin sheath is formed by the concentric wrapping of schwann cells around the axon. This sheath decreases axonal membrane capacitance and increases the conduction of nerve impulse.
3. Which of the following situations is least likely to lead to geographic isolation and allopatric speciation?
a. An eight-lane highway divides a population of snails with limited mobility.
b. A deep, narrow river divides a population of hummingbirds.
c. A deep, narrow river divides a population of white-footed mice (not known for their swimming ability).
d. A patch of land bisects a river and divides a population of minnows.
e. A mountain range divides a population of moths.
Answer:
b. A deep, narrow river divides a population of hummingbirds.
Explanation:
Hummingbirds can fly across the river. The deep and narrow river can not separate a population of hummingbirds into two patches since hummingbirds from one side of the river can fly and reach to the other side of the river.
Without any geographical isolation, hummingbirds can continue to interbreed among themselves. The continued interbreeding would not allow their reproductive isolation and thereby would prevent the allopatric speciation.
Which of the following is known to promote fat storage in adipocytes? a. Glucagonb. Lipoprotein lipasec. Cellulite synthetased. Lipoprotein synthetase e. Adipose lipase
Answer:
Lipoprotein lipasec- b.
The correct option that promotes fat storage in adipocytes is Lipoprotein lipase. Lipoprotein lipase is an enzyme found on the surface of adipocytes that promotes the uptake of triglycerides from circulating lipoproteins, such as chylomicrons and very low-density lipoproteins (VLDL), into adipose tissue for storage. This process increases the amount of body fat, contributing to fat storage in adipocytes.
Explanation:The correct option that promotes fat storage in adipocytes is Lipoprotein lipase.
Lipoprotein lipase is an enzyme found on the surface of adipocytes that promotes the uptake of triglycerides from circulating lipoproteins, such as chylomicrons and very low-density lipoproteins (VLDL), into adipose tissue for storage.
This process increases the amount of body fat, contributing to fat storage in adipocytes.
Learn more about fat storage in adipocytes here:https://brainly.com/question/33301753
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List the parts of the Central Nervous System and the parts of the Peripheral Nervous System.
Answer:
Central nervous system includes brain and spinal cord. Peripheral nervous system includes autonomic nervous system and somatic nervous system.
Explanation:
Nervous system consists of electrically excitable cell called neurons and collection of nerves that transmit information in the body.
Nervous system is mainly divided into two parts - PNS (peripheral nervous system) and Central nervous system (CNS) . The CNS of the body consists of spinal cord and brain that regulates and coordinates all the functions of the body.
PNS consist of cranial and spinal nerves. This system is divided into Autonomic nervous system ( controls involuntary actions of body) and somatic nervous system ( control the voluntary actions of the body).
The Central Nervous System, composed of the brain and spinal cord, controls most functions of the body. The Peripheral Nervous System, made up of nerves and ganglia beyond the CNS, serves as a communication line connecting all parts of the body to the CNS. The PNS has two subdivisions: the autonomic and somatic nervous systems.
Explanation:The Central Nervous System (CNS), is composed of the brain and spinal cord, serving as the main data center within the body. It is essentially contained within the cranial and vertebral cavities of the body. In contrast, the Peripheral Nervous System (PNS) encompasses all bodily structures beyond the brain and spinal cord. It consists of nerves and ganglia, carrying messages from the CNS to the muscles, organs, and senses in the periphery of the body. It can be further divided into two major subdivisions: the autonomic and somatic nervous systems.
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Binding Ca2+ to the troponin complex causes ____________ to move
a. troponin
b. actin
c. the myosin binding site
d. tropomyosin
Answer:
Tropomyosin
Explanation:
The calcium binding to troponin and induces the muscle contraction. The active binding site gets exposed due to the calcium binding.
The calcium ions then attaches with troponin molecule. This changes the shape of troponin and removes the tropomyosin protein from the binding site. No ATP is available during this process.
Thus, the correct answer is option (d).
Answer:
d. tropomyosin
Explanation:
Binding Ca2+ to the troponin complex causes the tropomyosin to move.
Tropomyosin is a protein that helps muscle contraction.
Suppose you were interested in the effect of breastfeeding versus formula feeding on the composition of gut flora in newborns. As part of your experiment, stool samples from healthy babies in the study get plated on various agar media. You notice that greater numbers of colonies are recovered when samples are plated on blood agar compared to Luria Bertani (LB) agar. After some diligent subculturing, you isolate several bacterial species capable of slow growth on the blood but not on the LB agar. You correctly surmise that these organisms
Choose one:
A. are opportunistic blood pathogens.
B. should be classified as beta-hemolytic.
C. have extensive and complex nutritional requirements not met by LB agar.
D. grow because the blood agar suppresses the growth of faster-growing bacteria.
E. are Gram-positive and, thus, inhibited by the ingredients in LB agar.
Answer:
Option (C).
Explanation:
Luria Bertani is a rich nutritional medium generally used for the growth of bacteria. This media was created by Giuseppe Bertani.
Bacteria shows slow growth on blood but can grow grow easily in LB agar because they have complex and extensive requirement that cannot be met by LB agar. Blood agar is more enriched media and can grow the bacteria that are dependent upon the host organism.
Thus, the correct answer is option (C).
Final answer:
The bacterial colonies are able to grow on blood agar and not LB agar because the organisms have extensive and complex nutritional requirements. Blood agar is a rich medium compared to LB agar, which accounts for their difference in growth pattern.
Explanation:
If you notice that bacterial colonies grow on blood agar but not on Luria Bertani (LB) agar, it suggests that these organisms have complex nutritional requirements that are not met by LB agar. Blood agar is a rich medium that provides many nutrients, which may explain the growth of organisms that are not growing on the simpler LB agar formulation. The correct answer to the experiment question is C: The organisms 'have extensive and complex nutritional requirements not met by LB agar.'
This does not necessarily mean the bacteria are opportunistic blood pathogens, beta-hemolytic, or that their growth is due to suppression of other bacteria. Additionally, while isolation on blood agar may commonly be associated with Gram-positive bacteria, growth on blood agar alone doesn't confirm that the bacteria are Gram-positive; this would require a Gram stain test. Therefore, option E is not the correct answer without further evidence.
Selective media are designed to suppress the growth of unwanted microorganisms and support the growth of the organism of interest, while differential media allow one to distinguish between different bacteria based on observable reactions occurring in the media, such as changes in color or clarity around colonies in the presence of specific reactions like hemolysis.
Why is it important not to misuse or overuse antibiotics?
A- Antibiotics will begin to rupture healthy cells.
B- Viral infections will become more prominent.
C- Bacteria will evolve, leading to the development of resistant bacteria
D- Normal flora will become pathogens.
Answer:
C- Bacteria will evolve, leading to the development of resistant bacteria
Explanation:
It is important not to misuse or overuse antibiotics because bacteria will evolve, leading to the development of resistant bacteria.
Answer:
Option C) Bacteria will evolve, leading to the development of resistant bacteria.
Explanation:
I took the test
Birds, reptiles, and insects dispose of excess nitrogen atoms by excreting uric acid, rather than urea, sparing these animals the need for large quantities of water to dissolve the urea. Humans and other primates also make and excrete uric acid as the catabolic degradation product of purines. Uric acid has a very low water solubility. High concentrations in the blood can form insoluble crystals of sodium urate, forming painful stones in the kidneys and ureters. High concentrations in the blood can also lead to painful arthritic joint inflammation, a condition known as gout. The final step in the synthesis of uric acid is the oxidation of xanthine, catalyzed by the enzyme xanthine oxidase. This enzyme is a target for some drugs used to treat gout and a more serious condition of uric acid overproduction called Lesch-Nyhan syndrome. Draw the structures of xanthine and uric acid, the substrate and product of xanthine oxidase.
Answer:
Xanthine oxidase enzyme converts xanthine (substrate) to uric acid (product). Xanthine is known by its chemical or IUPAC name 3,7-dihydropurine-2,6-dione and chemical formula C₅H₄N₄O₂. It is a purine base present in a maximum number of human tissues and fluids and in other living organisms. It is converted from hypoxanthine by the action of the xanthine oxidoreductase enzyme. (see structure in the figure)
Uric acid is the compound created by the action of xanthine oxidoreductase. It is a heterocyclic compound of C, N, and O and H. The chemical formula of uric acid C₅H₄N₄O₃. It produced by the metabolic breaking of purine.
C₅H₄N₄O₂ (xanthine) + O₂ + H₂O → C₅H₄N₄O₃ (uric acid) + H₂O
A cell in the testes of a male bald eagle is getting ready to begin meiosis. The diploid number for this bird species is 66. How many replicated chromosomes are there in this testicular cell at metaphase I? How many sister chromatids are there in metaphase I? In each daughter cell, how many sister chromatids are there in prophase II? How many chromosomes are there in each sperm cell?
Final answer:
At metaphase I in eagle testicular cell, there are 66 replicated chromosomes and 132 sister chromatids. During prophase II, each daughter cell will have 33 sister chromatids. Upon completion of meiosis II, each sperm cell will have 33 chromosomes.
Explanation:
When a cell in the testes of a male bald eagle, which has a diploid number of 66, is preparing to undergo meiosis, it will first replicate its DNA during interphase. At metaphase I, there would be 66 replicated chromosomes, because the chromosomes have paired up but not yet separated into daughter cells. Since each replicated chromosome consists of two sister chromatids joined at the centromere, there would be a total of 132 sister chromatids at metaphase I.
In meiosis II, specifically prophase II, each daughter cell from meiosis I will have 33 sister chromatids, since the cells are now haploid. Finally, when meiosis II is completed, each sperm cell, which is a gamete, will have 33 chromosomes (haploid).
SOME ONE PLS HELP ME THIS IS URGENT
According to the image below, which season is about to begin in the southern hemisphere?
Image shows Earth tilted 23.5 degrees away from the sun.
Summer
Autumn
Winter
Spring
Answer:
Pretty sure it's Winter.
Explanation:
Tilted away from the Sun.
Answer:
Autumn
Explanation:
Saddle joints have concave and convex surfaces. Name the two bones of the hand that articulate to form a saddle joint. A. The trapezium of the carpal bone and the thumb's metacarpal. B. The scaphoid of the index finger and the triquetrum of the middle finger. C. The trapezium of the ring finger and the capitate of the fourth finger. D. The scaphoid of the middle finger and lunate of the index finger.
Answer:
Option (A).
Explanation:
Saddle joint is a type of synovial joint and also known as sellar joint. These joints are present in thumb, shoulder and inner ear.
The bone has one part concave inward and other part is convex outward. This bone can articulate with thumb's metacarpal and trapezium of the carpal bone. Saddle joints show abduction and adduction movements.
Thus, the correct answer is option (A).
The saddle joint in the hand is the first carpometacarpal joint, located between the trapezium carpal bone and the first metacarpal bone of the thumb, allowing for the distinct opposability of the thumb.
Explanation:The saddle joints in the human body are a type of synovial joint where the articulating bone surfaces each resemble a saddle, which is concave in one direction and convex in the other. This design allows for a wide range of movements.
Specifically, in the hand, the saddle joint is found at the first carpometacarpal joint, which is between the trapezium, one of the carpal bones, and the first metacarpal bone that forms the base of the thumb. This versatile joint enables the thumb to move across two planes, allowing for the distinctive human ability to oppose the thumb against the other fingers.
Based on this description, the correct answer to which two bones articulate to form a saddle joint in the hand is: A. The trapezium of the carpal bone and the thumb's metacarpal.
Select the correct statement. a) Some drugs readily pass through the skin. b) Drugs taken by inhalation are usually slower acting and less potent than drugs taken orally. c) Usually more drug is needed when given intravenously than when administered orally to achieve an effect. d) The more times a drug is injected in the same vein, the easier the injection becomes.
Answer:
for 1 drugs taken by inhalation act just as fast as drugs taken intravenously
your answer is a some drugs readily pass throught the skin brainliest?
Explanation:
Answer:
a) Some drugs readily pass through the skin.
Explanation:
The skin has the ability to absorb various substances. Some penetrate the most superficial layers and others the deepest of the organ and may even enter the bloodstream. Penetration of mediations, for example, is not always possible because of natural barriers - such as the stratum corneum (the outermost layer of the skin, composed of keratin). However, some medications pass quickly through the skin.
What are the explanations for the emergence of bipedalism? How can it be identified in the fossil record?
Explanation:
The evolution of bipedalism (walking upright) signals the origin of hominid evolution, although no one is positively sure why bipedalism developed. The reason why bipedalism evolved in hominids remains a matter of discussion. No tools emerged until 2.5 million years ago, so toolmaking seems an improbable cause. Alternative ideas suggest that walking upright is more agile and uses less energy than walking on four legs; an upright posture allows hominids to pick fruit from trees and see over tall grass; upright reduces the body surface exposed to the sun’s rays; an upright stance aided the wading of aquatic hominids, and that bipedalism frees the forelimbs of males to carry food back to females, encouraging pair-bonding. All of these suggestions have their proponents, and none are universally accepted.
Fossils demonstrate that bipedalism extended back 4 million years ago; knee joints, pelvis, and leg bones all exhibit the signs of an upright stance. Remarkable evidence that early hominids were bipedal is a set of some 69 hominid footprints found at Laetoli, East Africa. Their footprints preserved in 3.7-million-year-old volcanic ash.
Men have XY (or YX) chromosomes and women have XX chromosomes. X-linked recessive genetic diseases (such as juvenile retinoschisis) occur when there is a defective X chromosome that occurs without a paired X chromosome that is not defective. Represent a defective X chromosome with lowercase x, so a child with the xY or Yx pair of chromosomes will have the disease and a child with XX or XY or YX or xX or Xx will not have the disease. Each parent contributes one of the chromosomes to the child. Complete parts a through d below. a. If a father has the defective x chromosome and the mother has good XX chromosomes, what is the probability that a son will inherit the disease? 0.333
Answer:
The probability that a son will inherit the disease is zero.
The human males are XY and females are XX. In addition, fathers pass their Y chromosome to their sons and X chromosome to their daughters. In contrast, mother pass one X to chromosome to son and another X chromosome to daughters.
In the given scenario, father is xY and mother is XX. Thus, father will pass x to daughters to Y to sons. The mother pass one X chromosome to son as well as daughters.
Thus, the genotype of the sons will be XY and that of daughters will be Xx. Hence, all the daughters will be carrier and all the sons will be unaffected.
The same is explained in the Punnette square in the figure below.
Final answer:
The probability that a son will inherit an X-linked recessive disease when the father has a defective x chromosome and the mother has normal XX chromosomes is 0%, as sons will inherit a normal X from the mother and a Y chromosome (not carrying the defective gene) from the father.
Explanation:
If a father has the defective x chromosome and the mother has normal XX chromosomes, the probability that a son will inherit the disease is 50%. This is because sons inherit their Y chromosome from their father and their X chromosome from their mother.
Since the father's X chromosome is defective (x), and he can only pass Y to his sons, all his sons will be Yx - the presence of the Y chromosome makes them male, but they do not receive the defective (diseased) x chromosome from the father. Instead, they receive a normal X chromosome from their mother making them XY, which means they will not have the disease. Here is the scenario:
Father: xY (carrier of the defective allele but not affected because males are hemizygous for the X chromosome)
Mother: XX (no defective allele)
Their children's possible genotypes could be:
Daughters: xX (carriers of the disease) or XX (not carriers)
Sons: XY (not affected and not carriers)
Therefore, a son cannot inherit the X-linked recessive disease in this case since he receives the Y chromosome from his father and a normal X from his mother.
The autonomic nervous system (ANS) plays a crucial role in the stress responses by regulating chiefly the ________ and ________ systems. Multiple Choice respiratory; integumentary lymphatic; circulatory Incorrect vestibular; lymphatic circulatory; respiratory
Answer: lymphatic circulatory; respiratory
Explanation:
The autonomic nervous system controls the internal body functions and responses which are not under the voluntary control. The functions control by the autonomous nervous system are heart rate, body fluid circulation, respiration, urination, sexual arousal and pupillary response.
The lymphatic system is one of the part of the circulatory system, it consists of a network of connected tubes called as lymphatic vessels. These vessels distribute the lymph fluid throughout the body. The lymph fluid consists of white blood cells which provide immunity against the pathogen.
The respiratory system is the system which controls the exchange of gases (Carbon dioxide and oxygen) between the body and the environment.
Both lymphatic circulation and respiration are the two functions controlled by the autonomic nervous system.
The autonomic nervous system (ANS) regulates primarily the respiratory and circulatory systems in response to stress, adjusting various physiological processes such as heart rate, blood pressure, and breathing.
Explanation:The autonomic nervous system (ANS) plays an essential role in co-coordinating various biological functions and in enabling our bodies to respond to changes in the environment, including stressful situations. Primarily, the ANS governs the respiratory and circulatory systems in response to stress. By controlling these two systems, the ANS adjusts physiological processes such as heart rate, blood pressure, and breathing, preparing the body to respond effectively to stress. For example, during a stressful event, messages are sent to the brain which the ANS then uses to increase heart rate and respirations. This is why the answer to your question is: The autonomic nervous system (ANS) plays a crucial role in the stress responses by regulating chiefly the respiratory and circulatory systems.
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3. During meiosis,
cells divide once.
cells double in size but do not divide.
cells do not divide.
cells divide twice.
cells do not divide but lose half their chromosomes.
Answer:
Cells divide twice
Explanation:
Meiosis may be defined as the process of cell division in which a single cell divides to give four daughter cells and each daughter cells have half number of chromosomes.
Meiosis is also known as reduction division as the chromosome number reduces upto half in the progeny cells as compared with the parent cell. The cells divide twice in meiosis and give rise to four daughter cells.
Thus, the correct answer is option (4).
Meiosis is a cellular division that results in four daughter cells with half the number of chromosomes of the parent cell. Specifically, cells in meiosis divide twice across two phases: Meiosis I and Meiosis II.
Explanation:During meiosis, cells divide twice. Meiosis is a type of cellular division that results in four daughter cells, each with half the number of chromosomes of the parent cell. It consists of two successive divisions: Meiosis I and Meiosis II. During Meiosis I, homologous chromosomes pair up and then separate, resulting in two cells. In Meiosis II, these cells divide again, resulting in four cells with half the number of chromosomes as the original cell.
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24. What is/are the major potential energy product(s) formed in the Krebs Cycle (not the same as end-products formed a the end of cellular respiration)?
a. NADH/FADH2
c. Carbon dioxide
b. Adenosine Tri-phosphate
d. Glucose
Answer: a. NADH/FADH2
Explanation:
Complete oxidative degradation of glucose can be compartmentalized into four main biochemical steps: glycolysis, the formation of acetyl-CoA, the Krebs (citric acid or tricarboxylic acid cycle) and the electron transport chain where oxidative phosphorylation is performed. During respiration, an organic compound (usually sugar) is completely oxidized to form CO2 and H20. The Krebs cycle is the set of reactions leading to complete oxidation of glucose. It occurs in the mitochondria of eukaryotes and the cytoplasm of prokaryotes. Per each degraded glucose molecule, resulting in the final of the Krebs cycle:
• 6 molecules of NADH.
• 2 molecules of FADH2
• 2 molecules of ATP
• 4 molecules of CO2
Which gives our answer What is/are the major potential energy product(s) formed in the Krebs Cycle? a. NADH/FADH2 because 8 molecules of them in total are made.
Secrete a hormone that controls the blood calcium level
Answer:
The parathyroid glands secretes PTH hormone that controls the blood calcium level.
Explanation:
The PTH is the parathyroid hormone which is secreted by parathyroid gland. It maintains the blood calcium level whenever required in the body.
The hormone increases the amount of calcium by preventing calcium excretion from the urine. The PTH also stimulates the release of calcitriol. The calcitriol allows absorption of calcium from the diet.
the thyroid and parathyroid glands belong to the _________ system.
Answer:
Endocrine
Explanation:
The thyroid and parathyroid glands belong to the Endocrine system.
The Endocrine system is a group of glands that usually gives off hormones to certain parts of your body.
ANSWER:
Thyroid and Parathyroid gland are related to the endocrine system. These are glands that secret different hormones in the body that regulate the several functions of the body.
EXPLANATION:
Endocrine system is a "network of glands" in the body. The endocrine glands secrete various hormones that regulate and control the body in different ways.
The hormones are released into the "bloodstream" by these glands. It is important for the endocrine system to be healthy and function in a proper manner else the body develops several problems.
What type of cells line the ventricles of the brain?
Answer:
Ependymal cells called Ependymocytes
Explanation: