Answer:
One method used by newborn mammals to generate heat is referred to as nonshivering thermogenesis. This method utilizes a protein channel called thermogenin which is present in high levels inside the mitochondria of brown adipose tissue.
Explanation:
Thermogenin forms a proton channel in the inner mitochondrial membrane in the brown adipose tissues. Most of the newborn mammals have these brown adipose tissues. In these tissues, the oxidation of nutrients do not produce ATP but generate heat. The heat is used by the newborn to maintain the warmer body temperature. The process is called nonshivering thermogenesis.
Use the following table to answer the question:
Element
Atomic Number
Oxygen
8
Lithium
3
Hydrogen
1
Carbon
6
Which element has the most protons in the nucleus?
Oxygen
Lithium
Hydrogen
Carbon
Answer:
I believe it's Oxygen because it's atomic number is 8 meaning it has 8 protons in the nucleus
Answer:
Oxygen
Explanation:
When we see oxygen in periodic table its atomic number is 8 which indicates that its 8th element which means that it has 8 protons in the nucleus.
Reference: Greenwood, Norman Neill, and Alan Earnshaw. Chemistry of the Elements. Elsevier, 2012.
How many FADH2 molecules are produced in the Krebs cycle?
A- 1
B- 2
C- 3
D- 4
Answer: B- 2
Explanation:
In aerobic respiration, the cell harvests energy from glucose molecules in a sequence of four major pathways: glycolysis, pyruvate oxidation, the Krebs cycle, and the electron transport chain. In the process of aerobic respiration, glucose is completely used. The 6-carbon glucose molecule is first cleaved into a pair of 3-carbon pyruvate molecules during glycolysis. One of the carbons of each pyruvate is then lost as CO2 in the conversion of pyruvate to acetyl-CoA; two other carbons are lost as CO2 during the oxidations of the Krebs cycle. All that is left to designate the passing of the glucose molecule into 6 CO2 molecules is its energy, some of which is preserved in 4 ATP molecules and in the reduced state of 12 electron carriers. 10 of these carriers are NADH molecules; the other 2 are FADH2.
40. Which of the following statements regarding substrate-level phosphorylation is correct?
a. Does not involve phosphate transfer
b. Does not produce ATP
c. Depends one distinct enzyme type
d. Occurs twice in glycolysis
Answer: c. Depends one distinct enzyme type
Explanation:
Glycolysis is the common metabolic pathway to all living things and consists of the incomplete oxidation of the glucose in pyruvate and occurs in the cytosol of eukaryotes and prokaryotes. During the substrate-level phosphorylation ATP is generated when a high energy P is directly transferred from a compound phosphorylated (substrate) to ADP. P got its energy during an initial reaction, in which the substrate itself is oxidized. During this process one distinct enzyme (Succinyl CoA) type is used.
Substrate-level phosphorylation involves the transfer of a phosphate group from a substrate molecule to ADP, producing ATP. It does not depend on one distinct enzyme type. It occurs twice in glycolysis.
Explanation:Substrate-level phosphorylation is a biological process where a phosphate group is transferred from a substrate molecule to adenosine diphosphate (ADP), producing adenosine triphosphate (ATP). Therefore, statement b is incorrect as ATP is indeed produced in this process. Statement c is incorrect because substrate-level phosphorylation does not depend on a single specific enzyme type, but multiple depending on the pathway (glycolysis or citric acid cycle). Referring to statement d, yes, in glycolysis, substrate-level phosphorylation occurs twice. Therefore, among the provided options, statement d is correct.
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What are Intramembranous and endochondral ossification?
Answer:
The direct conversion of mesenchymal tissue into bone is called intramembranous ossification .The process by which a cartilage intermediate is formed and replaced by bone cells is called endochondral osssification.
Explanation:
Intramembranous ossification is one of the two essential process during the fetal development of the gnathosome skeltal system by which rudimentary bone tissue is created. It is the process of bone development from fibrous membranes. It is involved in the formation of the flat bones of the skull, mandible and the clavicle. This type of ossification also helps in healing of bone fractures.
Endochondral Osssification: Method of forming a bone through cartilage intermediate. It is also involved in the formation of long bones.
1) how is cell division normally controlled
2) if DNA in G1 phase is damaged by UV or ionizing radiation, what will happen
Answer:
some specific safeguard protein are present
Explanation:
the specific protein which are present at G1 phase name cyclin dependent kinase they also initiate dna division
Which of the following populations would be considered r-selected? Black Widow Spiders - they produce 1000s of offspring and few survive. Opossums- They carry offspring in a pouch on the mothers body Mallard Ducks - They lay up to one dozen eggs and provide parental care of their chicks Mammals - They produce few offspring and provide parental care
Answer:
Option 1.
Explanation:
r-selected species may be defined as the species that has higher growth rate, shows less parental care and the rate of survival of the off spring is low as compared with k selected species.
Black widow spiders have the ability to produce 1000 offspring. Their chances of survival are extremely low and only few species of black widow spiders survive. Hence, black widow spider is r-selected species.
Thus, the correct answer is option (1).
Place a single word into each sentence to make it correct, then place each sentence into a logical paragraph order. Hemodialysis is the ____________ cleansing of wastes from the blood.
Answer:
Hemodialysis is the extracorporeal cleansing of wastes from the blood.
Explanation:
Hemodialysis is the process of kidney dialysis. This process is used for the purification of blood. Extracorporeal blood purification includes removal of waste product such as urea , water and toxic elements from patients body. This process is used when patient's kidney is not working properly.
Oh, no... you dropped your study flashcards, and now they're out of order! Put them back into the correct order for a eukaryotic cell. a. "Combine the mRNA strand with a ribosome and a tRNA carrying a methionine." b. "Unwind the DNA molecule near the promoter." c. "Exit the nucleus to the cytoplasm." d. "Transcribe the complementary RNA strand."
Answer:
The correct order would be:
b. "Unwind the DNA molecule near the promoter."
d. "Transcribe the complementary RNA strand."
c. "Exit the nucleus to the cytoplasm."
a. "Combine the mRNA strand with a ribosome and a tRNA carrying a methionine."
The given order is according to the central dogma of a cell. According to central dogma DNA is first transcribed into mRNA which is then moved out of nucleus (in eukaryotic cells).
For transcription, the DNA is unwind near the promoter region where transcription factors and RNA polymerase binds the template strand. It then make mRNA which is moved out of the nucleus.
The mRNA then is translated with the help of ribosome and tRNA. The start codon in mRNA usually codes for methionine due to which tRNA carrying methionine recognises the start codon and initiate the process of translation.
How do cells capture the energy released by cellular respiration?
Answer: by producing ATP in the cytoplasm and inside the mitochondria of the cell.
Explanation:
The complete oxidative degradation of glucose can be compartmentalized into four main biochemical steps: glycolysis, acetyl-CoA formation, Krebs (citric acid or tricarboxylic acids cycle) and the electron transport chain where oxidative phosphorylation occurs. During respiration, one Organic compound (usual sugar) is completely oxidized to form CO2 and H2O. In aerobic respiration, molecular oxygen, O2, serves as the ultimate acceptor of electrons. In anaerobic breathing, the acceptor end of electrons can be the NO3- (nitrate ion), SO42- (sulfate ion), CO2 or fumarate. If the oxidized substrate during respiration, is a protein then it also forms ammonia.
Cells utilize cellular respiration to convert the chemical energy stored in glucose into ATP, using glycolysis, the Krebs Cycle, and the electron transport chain. These stages culminate in the concentration and flow of hydrogen ions that power ATP synthase to generate ATP from ADP, securing energy for various cellular functions.
Explanation:How Cells Capture Energy Through Cellular RespirationCells capture the energy released by cellular respiration through a multi-step process involving the conversion of glucose into ATP (adenosine triphosphate), the primary energy currency of the cell. During cellular respiration, glucose is oxidized in the presence of oxygen, resulting in the production of carbon dioxide and water. The energy from glucose is transferred to ATP, which can then be used by the cell to perform various functions.
The process of cellular respiration includes several stages: glycolysis, the Krebs Cycle, and the electron transport chain. In the final stage, electron transport chains in the mitochondrial inner membrane capture high-energy electrons from reduced coenzymes NADH and FADH₂. These electrons are used to pump hydrogen ions across the mitochondrial membrane, creating an electrochemical gradient. When hydrogen ions flow back through ATP synthase, the energy is harnessed to convert ADP to ATP, effectively capturing the energy in a form that can be used by the cell for various metabolic activities.
Through this complex set of reactions, chemical energy from food is systematically transformed into an energy form readily available for the cell's use. In essence, cellular respiration allows cells to extract and utilize the energy stored in the chemical bonds of glucose, which is obtained from the organism's food intake and is originally derived from solar energy through photosynthesis.
Tubulovillous adenoma with high grade dysplasia
Explanation:
Adenomatous genes are most often found among neoplastic genes and are the target of 2/3 of all colon members. The factors of risks associated with its onset include advanced age, sedentary lifestyle, male gender and increased BMI / abdominal fat. Dysplasia gives it the potential for malignancy, constituting the precursors of most, however, only 5% of adenomas evolve to carcinoma by a process that runs from 7 to 10 years, with the greater risk of progression for advanced adenomas. Adenomatous polyps can be classified into 3 subtypes based on epithelial architecture:
Tubular adenoma: represent about 80% of all adenomas and are characterized by the presence of tubular glands in at least less than 75% of the architecture.
Villous adenoma: account for 5 to 15% of all adenomas and have glands with villous projections in at least 75% of its architecture.
Tubulovillous adenoma: correspond to 5 to 15% of adenomas and has mixed histology with less than 75% of both types of architectures.
High-grade dysplasia is characterized by a complex architecture where there are grouping and glandular irregularity as well as a cribriform pattern and cytological atypia, with loss of nuclear polarity, enlarged nuclei with nucleoli, atypical mitoses, and prominent apoptosis. The high-grade dysplasia has a higher risk of developing for carcinoma
Compare humoral immunity vs cell mediated immunity
Answer:
Humoral immunity:
Humoral immunity is mediated by the macromolecules present in the extracellular fluid. This immunity provides protection against the pathogens present in the extracellular fluid. B- cells are mainly produced by the humoral immunity. The plasma cell and memory cells are generated in the human body. Antibodies kill the pathogens in humoral immunity.
Cell mediated immunity:
Cell mediated immunity is activated by the infected cells. This immunity provides protection against intracellular pathogens. T- cells are mainly produced by cell mediated immunity. T helper cells and T killer cells are mainly produced in the cell mediated immunity. Cytokines kill the pathogens in case of cell mediated immunity.
Humoral immunity involves B cells producing antibodies to target extracellular pathogens and toxins, while cell-mediated immunity involves T cells targeting intracellular pathogens and some cancer cells.
Humoral immunity primarily involves B cells, which produce antibodies that circulate in the blood and lymph. These antibodies bind to pathogens and toxins in extracellular spaces, thereby preventing them from attaching to and invading host cells. On the other hand, cell-mediated immunity is concerned with T lymphocytes, or T cells, which target and eliminate intracellular pathogens such as viruses, some bacteria, and also some cancer cells. T cells can be subdivided into helper, cytotoxic, and regulatory T cells, each playing a unique role in orchestrating adaptive immune responses and maintaining immune system balance.
In pea plants, round peas (R) are dominant to wrinkled peas (r). You do a test cross between a pea plant with wrinkled peas (genotype rr) and a plant of unknown genotype that has round peas. You end up with three plants, all which have round peas. From this data, can you tell if the parent plant is homozygous dominant or heterozygous?
Answer:
The parent plant is a heterozygous.
Explanation:
If we have a recessive phenotype, the only way that the result ends up with a 75% of a dominant phenotype is doing a test cross with a dominant phenotype that have a heterozygous genotype. When both genotype are homozygous, the result ends up with a 100% of a dominant phenotype because if we do a test cross between them, the result is a combination of a dominant allele and a recessive one, therefore every time we have this combination (a heterozygous genotype) the phenotype that is showed is the dominant one, even tough there is a recessive allele.
Based on the data provided of three offspring all having round peas, it is not possible to conclusively determine if the parent of unknown genotype is homozygous dominant or heterozygous. Even if the parent is heterozygous, it could pass on the dominant gene each time leading to offspring with round peas. More data or a larger sample size of offspring would be needed for more certainty.
Explanation:The subject of your question relates to genetics, specifically referring to dominant and recessive traits in pea plants. As given, round peas (R) are dominant to wrinkled peas (r) and you've done a test cross between a pea plant with wrinkled peas (genotype rr) and a plant of unknown genotype, but has round peas. The three offspring plants also have round peas. This information alone, however, is not enough to definitively determine whether the parent plant of unknown genotype is homozygous dominant (RR) or heterozygous (Rr).
This is because even if the unknown parent is heterozygous (Rr), both dominant (R) and recessive (r) genes can be passed onto its offspring. If the recessive gene was passed on, the offspring would be rr (wrinkled peas), but if the dominant gene was passed on, the offspring would be Rr (round peas). By chance, it could be that the dominant gene was passed on all three times, leading to all three offspring having round peas. Hence, unless more data or offspring are available, it is not possible to conclusively determine the genotype of the unknown parent plant.
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Pepsin is a digestive enzyme. It is made in an inactive form, pepsinogen, which is activated by the acidity of the stomach to self-cleave. The cleaved molecule is active pepsin, which goes on to cleave more pepsinogen molecules, leading to a cascade of rapid pepsin activation. This pathway is an example of:
Answer: Positive feedback loop
Explanation:
Even before the food reaches the stomach (before ingesting it), the glands of the stomach mucosa begin to release its gastric secretion. The main characteristic of this secretion is acidity, as measured by pH (one of the physiological variables).
This acidity is a result of the presence of hydrochloric acid, which is part of the secretion composition. In addition to it, water, pepsinogen (which will give rise to the enzyme pepsin) and the intrinsic factor. The acid has a function of the protection of the entire system by eliminating microorganisms. It is also responsible for the activation of pepsin (which only occurs with acid pH), the enzyme that digests proteins containing the amino acids leucine or phenylalanine or tryptophan or tyrosine.
The concept of Positive feedback loop states that the body tries to increase the value of a variable (acidity, in the case of pepsinogen) when it is below its optimal value (called a point adjustment) and decreases this value when it is above optimal.
12. Which of the following statements is true regarding biological cell theory? a. All organisms are made of cells c. All cells come from mitosis b. All cells possess a nucleus d. All cells come from meiosis
Answer:
a
Explanation:
all organism is made of cell.they are the basic structursl unit of life.
Answer:
All Cells come from Meiosis
Action potential propagation in a skeletal muscle fiber ceases when acetylcholine is removed from the synaptic cleft. Which of the following mechanisms ensures a rapid and efficient removal of acetylcholine?
Answer:
a) Acetylcholine is degraded by acetylcholinesterase. :)
Suppose the cells of a tumor are idealized as spheres each with a radius of 5 mum (micrometers). The number of cells has a doubling time of 35 days. Approximately how long will it take a single cell to grow into a multi-celled spherical tumor with a volume of 0.2 cmcubed (1 cmequals10,000 mum)
Answer:
1044.3 days
Explanation:
Given,
Radius of sphere shaped cells of tumor [tex]= 5[/tex] micrometer
[tex]1[/tex] centimeter [tex]= 10,000[/tex] micrometers
Thus, radius of sphere in centimeters
[tex]= \frac{5}{10,000} \\[/tex]
Volume of a sphere
[tex]= \frac{4}{3} \pi r^{3}[/tex]
Volume of one cell of tumor
[tex]\frac{4}{3} *(3.14)*(\frac{5}{10,000})^{3}\\= {5.23 * 10^{-10}[/tex] centimeter cube
As we know ,
[tex]N(t) = N(0) e^{-kt}\\\\k = \frac{ln2}{35}\\ k = 0.0198 day^{-1}\\[/tex]
Substituting all the given values in above equation, we get -
[tex]0.2 = 5.23 * 10^{-10} * e^{-0.0198*t}\\t = 1044.3[/tex] days
Final answer:
It will take approximately 649.25 days for a single cell to grow into a multi-celled spherical tumor with a volume of 0.2 cm³, considering the cell has a doubling time of 35 days and originates from cells idealized as spheres with a 5 μm radius.
Explanation:
The question involves calculating the time it will take for a single cell to grow into a multi-celled spherical tumor of 0.2 cm³ volume. Since each cell is idealized as a sphere with a radius of 5 μm (μm stands for micrometers), and the cell number doubles every 35 days, we can use the formula for the volume of a sphere (V = (4/3)πr³) to find out the volume of a single cell and then determine how many such cells would fit into a tumor of 0.2 cm³.
First, we convert the volume of the tumor from cubic centimeters to cubic micrometers. 1 cm³ = 1,000,000,000 μm³ (since 1 cm = 10,000 μm). Therefore, 0.2 cm³ = 200,000,000 μm³. The volume of one cell with a radius of 5 μm is V = (4/3)π(5³) ≈ 523.6 μm³. To find the number of cells that fit into the tumor volume, we divide the tumor volume by the volume of one cell: 200,000,000 μm³ / 523.6 μm³ ≈ 382,239 cells.
Since the number of cells doubles every 35 days, we can find the number of doubling periods required to reach 382,239 cells by solving for n in the equation 2^n = 382,239, where n is the number of doublings. Solving this equation, we find n ≈ 18.55. Therefore, it will take approximately 649.25 days (18.55 × 35 days per doubling) for a single cell to grow into a spherical tumor with a volume of 0.2 cm³.
5. What is the critical distinction between allopolyploidy and autopolyploidy?
a. Allopolyploidy involves disruptive selection but autopolyploidy does not.
b. Autopolyploidy involves disruptive selection but autopolyploidy does not.
c. If speciation is allopatric rather than sympatric, the mode of speciation is allopolyploidy; if speciation is sympatric rather than allopatric, the mode of speciation is autopolyploidy.
d. In autopolyploidy, one parental species contributes to the polyploidy; in allopolyploidy, two parental species contribute to the polyploidy.
e. In allopolyploidy, one parental species contributes to the polyploidy; in autopolyploidy, two parental species contribute to the polyploidy.
Answer:
D. In autopolyploidy, one parental species contributes to the polyploidy; in allopolyploidy, two parental species contribute to the polyploidy.
Explanation:
Autopolyploidy is the polyploidy that arises when the organisms have more than two complete sets of the same genome. For example, if a diploid species is represented as "AA", its autopolyploid with four complete sets of the genome can be represented as "AAAA".
On the other hand, allopolyploidy occurs when the polyploid carries more than two complete sets of the genome from separate species. For example, if two diploid species are represented as "AA" and "BB", their allopolyploid with four complete sets of the genome can be represented as "AABB".
Which of the following animals has a heart in which oxygenated and deoxygenated blood mix?
a.human
b.fish
c.frog
d.bird
e.mammal
As you ascend from the medulla oblongata to the cerebrum, the functions of each successive level becomeA) more complex.B) simpler.C) better understood.D) more crucial to visceral functions.E) critical to reflexes
Answer:
The correct answer will be option A-more complex.
Explanation:
The brain is divided into three parts: cerebrum, cerebellum and Brainstem.
The medulla oblongata is a part of the brainstem which is involved in the response of the autonomic nervous system (ANS) like body temperature, breathing, digestion, heart rate, wake and sleep cycles, vomiting, sneezing, coughing and swallowing.
The cerebrum is the largest part of the brain which includes basal ganglia, hippocampus and olfactory bulb. This is involved in performing the higher functions of the central nervous system (CNS) like vision and hearing, speech, emotions, reasoning, learning and fine control of movement.
As we move from the medulla oblongata to cerebrum the response to stimuli changes from simple to complex as medulla oblongata is involved in simple processes of ANS whereas cerebrum is involved in the higher process of CNS like learning, speech and many more.
Thus, option A-more complex is the correct answer.
The function of coenzyme A in the citric acid cycle is most like: A. The function of coenzyme A in the citric acid cycle is most like a kid jumping up and down on a trampoline. B. a frog that turns into a prince. C. a limousine driver dropping off a couple at the school prom. D. throwing a baited hook into a lake and catching a fish.
Answer: A limousine driver dropping off a couple at the school prom.
Explanation: Coenzyme A is a co-factor that assists enzymes to perfom their functions of speeding up chemical reactions. They are non-proteins that can change form and be used by many different type of enzymes for assistance. Just like a driver, any other couple could have asked to be driven by the same driver for any other reasons other then school prom. Drivers are always there to assist but do not take part in the overall function of process or occasion just like coenzyme A.
Final answer:
Coenzyme A in the citric acid cycle is most akin to (option C) a limousine driver dropping off a couple at the school prom. It transfers an acetyl group to oxaloacetate to initiate the cycle, then leaves, mirroring the driver's role in facilitating arrival at an event.
Explanation:
The function of coenzyme A (CoA) in the citric acid cycle can be likened to option C: a limousine driver dropping off a couple at the school prom. Coenzyme A plays a crucial role in the citric acid cycle (also known as the TCA cycle or Kreb's cycle), which is a central part of cellular respiration taking place in the matrix of the mitochondria. In the first step of the cycle, Coenzyme A transfers its 2-carbon acetyl group to a 4-carbon compound, oxaloacetate, to form a 6-carbon molecule called citrate, thus initiating the citric acid cycle.
This transfer by Coenzyme A is comparable to a limousine driver who facilitates the arrival of a couple (the acetyl group) at the destination (the citric acid cycle), and then departs, allowing the couple to enjoy their event. The release of Coenzyme A after transferring the acetyl group mirrors the limousine driver departing after dropping off the prom attendees.
3. Suppose a person developed a mutation in a somatic cell that diminishes the performance of the body’s natural cell cycle control proteins. This mutation resulted in cancer, but was effectively treated with a cocktail of cancer-fighting techniques. Is it possible for this person’s future children to inherit this cancer-causing mutation? Be specific when you explain why or why not.
Answer: This cancer causing mutation will not be passed on to offspring.
Explanation: Cells are categorised into two, the germ cells and somatic cells. The germ cells are sex cells that produce or give rise to somatic cells called the body cells. Because somatic cells comes about secondarily from germ cells they only consist of random mutations (change in genetic make up) that maybe due to uv rays or a persons lifestyle thus these kind of somatic mutations cannot be passed down to offsprinds as they are not hereditary.
The double-stranded DNA molecule of the newly discovered Elradicus libanii has a length of 34 micrometers. How many base pairs are present in this molecule?
Answer:
100 000
Explanation:
The average length of a base pair (bp) is 340 pm [340 × 10^(-12) m]
Length of DNA = No. of bp × length of bp
34 × 10^-6 m = n × 340 × 10^-12 m
n = (34 × 10^-6)/(340 × 10^-12) = 100 000
The DNA molecule contains 100 000 base pairs.
To find the number of base pairs in a 34 micrometer DNA molecule of Elradicus libanii, convert the length to nanometers and divide by the length of one base pair (0.34 nm). The calculation results in 100,000 base pairs.
Explanation:To determine the number of base pairs in the DNA molecule of the newly discovered Elradicus libanii, we need to utilize the length of one base pair and the total length of the molecule. We know that each base pair in a DNA molecule is approximately 0.34 nm in length. First, we convert the given length of the DNA molecule from micrometers to nanometers; 1 micrometer = 1000 nanometers, so 34 micrometers is the same as 34,000 nanometers.
Now, to find the number of base pairs, we divide the total length of the DNA molecule by the length of one base pair:
Total length of DNA molecule = 34,000 nmLength of one base pair = 0.34 nmNumber of base pairs = Total length of DNA molecule ÷ Length of one base pairTherefore, the number of base pairs = 34,000 nm ÷ 0.34 nm/base pair = 100,000 base pairs.
If the number of photosynthetic organisms on the earth decreased drastically, what would happen?
Answer: There would be no life on earth.
Explanation:
Photosynthetic organism provides life to other life forms on earth by providing food. Basic is the basic need for all the living organism on earth.
Only photosynthetic organisms have the ability to synthesize food by the help of carbon dioxide and water in presence of sunlight.
This food( glucose) is used by the consumers to sustain life on earth. This energy is used by the organism to perform various tasks of the body. All the food chains starts by photosynthetic organism.
So, ultimately all the other organism would die without photosynthetic organisms.
How does stress affect your musculoskeletal system
Answer:
Stress causes tension in muscles that are very painful. This is how stress affects the musculoskeletal system.
Explanation:
Stress affects the musculoskeletal system causing muscle tension. Muscles can be so tight that they lead to a painful frame of muscle spasm. The back and neck muscles are particularly more sensitive to the effects of stress.
The slightest muscle strain can be the "drop of water" in a stressed person. For example, if the spinal nerves are restricted by scar tissue or calcium deposits, they can lead to minimal muscle tension that compresses the nerves and causes pain. Another example is sciatica, which can become much greater when the person feels stressed.
A client has a new tracheostomy. Which of the following interventions should the nurse include when performing tracheostomy care?Select one:a. Remove soiled dressing with sterile gloves.b. Suction the tracheostomy before beginning care. c. Change tracheostomy ties when soiled.d. Clean disposable inner cannula with hydrogen peroxide.
The interventions should the nurse include when performing tracheostomy care is to change tracheostomy ties when soiled. So, the correct option is C.
What do you mean by Tracheostomy?
It is a surgical procedure to create an opening through the neck into the trachea (windpipe). Its function is to relieve difficulties in breathing.
It is mandatory to change the tracheostomy ties whenever they are wet or dirty. If such conditions do not arise then they are changed every 24 hours. It is important to keep such ties clean and dry. During the process of changing, gloves should be properly washed to avoid the chances of contamination.
Therefore, the interventions should the nurse include when performing tracheostomy care is to change tracheostomy ties when soiled.
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When performing tracheostomy care, a nurse must adhere to hygiene and infection-prevention practices. This includes removing soiled dressings with sterile gloves, suctioning the tracheostomy before care, changing tracheostomy ties when soiled, and cleaning the disposable inner cannula with hydrogen peroxide to prevent the risk of infection.
Explanation:When performing tracheostomy care for a client with a new tracheostomy, the nurse must prioritize hygiene and infection-prevention practices. The nurse should start by adhering to strict handwashing protocol before and after patient interaction. It is crucial to reduce any potential bacterial flora from the nurse's skin to prevent infecting the tracheostomy site.
Based on the options provided, and taking into consideration standard medical procedures, all the interventions are necessary. The nurse should a. Remove soiled dressing with sterile gloves in order to prevent contamination. Similarly, c. Changing tracheostomy ties when soiled is important to reduce the risk of microbes spreading to the tracheostomy site. The step b. Suction the tracheostomy before beginning care is typically required to free the airway passage from any mucus or discharges. Lastly, the nurse should d. Clean disposable inner cannula with hydrogen peroxide. Hydrogen peroxide is a commonly used antiseptic that aids in eliminating bacteria, thus preventing the risk of infection.
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6. Which of the following has the lowest cost of production of recombinant proteins?
Bacteria / Mammalian cell culture / Transgenic plants / Yeast
Answer:
Bacteria
Explanation:
Bacteria constitutes large group of unicellular organisms that lacks membrane bound organelles and constitute cell wall. Bacteria are used for the production of recombinant proteins.
Bacteria can be selected easily for the recombinant protein formation. The plasmid of bacteria is helpful for the integration of desired genes and the production of protein. The bacteria culture can be maintained easily in the laboratory. The different strains of bacteria with different traits can be used for the production of different recombinant proteins at low cost.
Thus, the correct answer is option (1).
16. Mexican hairless dogs have little hair and few teeth. When a Mexican hairless is mated to another breed of dog, about half of the puppies are hairless. When two Mexican hairless dogs are mated to each other, about 1/3 of the surviving puppies have hair, and about 2/3 of the surviving puppies are hairless. However, about two out of eight puppies from this type of cross are born grossly deformed and do not survive. Explain this pattern of inheritance.
Here you can tell that it is heterozygous with a dominant allele to be a hairless dog, for example, Hh, where H is dominant.
When you cross one hairless (Hh) with a not hairless dog (hh), Punnett square will show chances are half hairless and half not hairless.
When you cross two hairless (Hh), Punnett square will show three cases: HH, Hh, and hh.
hh is not hairless.
Hh is hairless.
HH is a codominance in which both parents gave the hairless allele, and this may be the not surviving dogs.
The pattern of inheritance in Mexican hairless dogs follows classical genetics, with dominant and recessive alleles determining whether the dogs are hairless or not. When these dogs are mated, the mix of dominant and recessive alleles between the parents results in the different traits observed in the puppies. It is suggested that those puppies receiving two dominant alleles are born with severe deformities.
Explanation:The pattern of inheritance described regarding the Mexican hairless dogs can be explained through genetics. A Mexican hairless dog presumably carries a dominant allele (H) that causes hairlessness, and a recessive allele (h) responsible for hair formation. If a hairless dog (Hh) is mated to a dog with hair (hh), approximately half of the puppies will be hairless (Hh) and half will have hair (hh), as the dominant H allele is shared between them.
When two Mexican hairless dogs are mated together, both of them being Hh, the mix of genes in the puppies varies. Approximately 1/3 of the puppies receive a pair of recessive alleles (hh) and hence, have hair. About 2/3 of the puppies receive either a homozygous dominant pair (HH) or a heterozygous pair (Hh), making them hairless.
However, dogs inheriting the homozygous dominant pair (HH) are presumably born with severe deformities due to the harmful nature of two dominant alleles. This means that 1/4 of all puppies, or 2/8 are born deformed and do not survive.
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The purpose of DNA profiling is to ______. transfer genes from one species to another species determine whether two samples of DNA came from the same individual definitively convict a suspect assess the safety concerns associated with genetically modified organisms
Answer: Determine whether the two samples of DNA came from the same individual definetively convict a suspect.
Explanation:
DNA profiling is a technique for examining the DNA composition of the organism. The DNA can be extracted from hair, skin, bodily fluids and nails. The technique involves the following steps:
1. Extraction of the DNA from the cells.
2. The DNA can be cut into fragments by using enzyme.
3. Separation of DNA fragments using a gel.
4. Transfer of DNA over the paper.
5. Adding the radioactive probe over the paper. The radioactive probe forms a complementary pair with the DNA strand.
6. Observing the strands on the X-ray film.
It is a useful technique for identification of the living being from which it has come from. It is useful in comparing the two samples that is which obtain from the crime scene and the one which comes from the suspect. Thus will help in conviction of suspect.
The synthesis of nucleotides requires several amino acids. Sort each amino acid as a precursor in the synthesis of purines, pyrimidines, both purines and pyrimidines, or neither purines nor pyrimidines. Note: If you answer any part of this question incorrectly, a single red X will appear indicating that one or more of the phrases are sorted incorrectly.
Purines Pyrimidines Both NeitherA. leucine B. lysine C. methionine D. glutamine E. glycine F. histidine G. aspartate H. alanine I. tryptophan
Glycine and Glutamine are precursors in the synthesis of purines while Aspartate is a precursor in the synthesis of pyrimidines. Leucine, Lysine, Methionine, Histidine, Alanine, and Tryptophan are not directly involved in the synthesis of either purines or pyrimidines.
Explanation:The synthesis of nucleotides, the building blocks of DNA and RNA, requires several amino acids. Whether an amino acid serves as a precursor in the synthesis of purines, pyrimidines, both, or neither depends on the specific pathways involved in nucleotide biosynthesis.
Purines: Glycine (E) and Glutamine (D) Pyrimidines: Aspartate (G) Both Purines and Pyrimidines: None Neither Purines nor Pyrimidines: Leucine (A), Lysine (B), Methionine (C), Histidine (F), Alanine (H), Tryptophan (I)
It is important to note that while these amino acids are not directly involved in the synthesis of purines or pyrimidines, many of them do play indirect roles in nucleotide metabolism.
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The synthesis of nucleotides requires specific amino acids. Some amino acids are precursors for the synthesis of purines, some for pyrimidines, some for both, and some are not involved in the synthesis of these nucleotides.
Explanation:The amino acids that serve as precursors in the synthesis of purines include glycine, glutamine, and aspartate.
The amino acids that serve as precursors in the synthesis of pyrimidines include aspartate and glutamine.
The amino acids that serve as precursors for both purines and pyrimidines are glycine, glutamine, and aspartate.
The amino acids that are considered neither purines nor pyrimidines are leucine, lysine, methionine, alanine, histidine, and tryptophan.
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3. Which of the following is not an X-linked trait?
hemophilia
cystic fibrosis
Duchenne muscular dystrophy
color-blindness
Answer:
Cystic fibrosis
Explanation:
X- linked trait may be defined as the trait that is located on X chromosome or the inheritance of the trait is based on the X chromosome. The disorders associated with X-linked trait is known as sex linked disorder.
Cystic fibrosis is an autosomal recessive disorder that affects the digestive system, lungs and liver. The homozygous recessive condition is responsible for the disease cystic fibrosis. Hence, cystic fibrosis is not an X-linked trait.
Thus, the correct answer is option (2).