Answer:
availability of water
temperature
natural disaster
how is ethics and politics related to environmental policy
a glucose molecule enters aerobic respiration and all the six carbons are oxidized to co2. What happens to the six carbons in a glucose molecule as the molecules go through aerobic cellular respiration?
a. they are given off as water
b. they become carbons of ATP
c. they are given off as carbon dioxide
d. they are destroyed completely
When the six carbons in a glucose molecule as the molecules go through aerobic cellular respiration c. they are given off as carbon dioxide.
During aerobic cellular respiration, glucose undergoes a series of metabolic reactions, such as glycolysis, the citric acid cycle (also known as the Krebs cycle or TCA cycle), and the electron transport chain. These processes lead to the complete oxidation of the six carbons in glucose to carbon dioxide (CO₂).
In glycolysis, a glucose molecule is converted into two molecules of pyruvate, each containing three carbons. The pyruvate then enters the mitochondria, where it undergoes further oxidation in the citric acid cycle. In this cycle, the carbons in pyruvate are released as carbon dioxide molecules, generating energy-rich compounds such as NADH and FADH₂.
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Please help meeee!!!!
A student asks, "How does exercise affect an individual’s heart rate?" Which step can the student complete to best help her answer the question through scientific inquiry? She can keep the results to herself. She can conduct background research. She can ask friends for their opinions. She can observe people who do not exercise.
I will mark you
Answer:
b
Explanation:
plzzzz mark as brainliest!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
WHY DID THIS HAPPEN I WAS AT 86.5% SOMEONE TALK TO ME AND MAKE ME HAPPY AGAIN :(
Answer:
it seems maybe you didnt turn in an assignment or late ?
Explanation:
Answer:
ahhh it'll be ok *pats back* trust me ;D
Explanation:
Why do complex organisms need specialized cells?
Answer:
Because they are more complex therefore they need more specalized cells.
Explanation:
Answer:
Explanation:
"Complex organisms need specialized cells because they are capable of performing more tasks than simple organisms."
Nothing much I don’t got a question
Okay bud. have a nice day B)
Fluoroacetate occurs naturally in some plants and is highly toxic to mammals and insects. In cells, it is converted to fluorocitrate, which binds tightly and inhibits an enzyme crucial for cellular respiration, causing citrate buildup. To what enzyme does fluorocitrate bind?.
It is transformed into fluorocitrate in cells, where it binds firmly and prevents an enzyme necessary for cellular respiration, leading to citrate accumulation. Aconitase enzyme and fluorocitrate bind.
The primary pathway for the synthesis of energy features the enzyme aconitase. Citrate is changed into isocitrate as part of the citric acid cycle. Citrate-to-isocitrate conversion is carried out by the citric acid cycle enzyme mitochondrial aconitase. Similar action is taken by an isoform in the cytoplasm. Several TCA cycle intermediates are also biosynthetic precursors of amino acids and heme, in addition to being used as energy sources when carbohydrates are broken down. Additionally, mAcn performs a second crucial function in the upkeep of mitochondrial DNA (mtDNA). In the tricarboxylic acid cycle, aconitase is an enzyme that catalyzes the non-redox-active stereo-specific isomerization of citrate to isocitrate via cis-aconitate. Proteins are made up of amino acids that are connected by one or more polypeptide chains to form enzymes.
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Select all of the biomolecules that have a carbon-nitrogen bond.
lipid
protein
peptide
nucleic acid
carbonhydrate
please answer fast.
Answer:
All have carbon-nitrogen except carbohydrate
The biomolecules that do not have any carbon-nitrogen bond are proteins, peptides, and nucleic acids. The correct options are b, c, and d.
What are biomolecules?Biomolecules are organic molecules composed of carbohydrates, protein, lipids, and nucleic acids.
They are necessary for living cells to survive. Some valuable biomolecules have high demand that cannot be met by renewable resources.
Biomolecules are organic compounds that are required for life to exist. These molecules serve important functions such as a source of energy, materials for constructing new body mass such as muscles, and other molecules that serve important functions such as hemoglobin, which transports oxygen in the blood.
Proteins, peptides, and nucleic acids are examples of biomolecules that lack a carbon-nitrogen bond.
Thus, the correct options are b, c, and d.
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Chemical messengers derived from cholesterol fall under which of the following categories?
A. steroid messengers B. eicosanoid messengers
C. peptide/protein messengers
B). Chemical messengers derived from cholesterol fall under the category of steroid messengers. Chemical messengers or hormones are involved in the regulation of various body processes.
Chemical messengers derived from cholesterol are steroid messengers. Steroid messengers include hormones such as testosterone, estrogen, and progesterone. These hormones are involved in the regulation of reproductive functions, among other things.Chemical messengers derived from arachidonic acid are eicosanoid messengers. Hormones like prostaglandins and leukotrienes fall under this category.
Peptide/protein messengers are another class of chemical messengers. Hormones such as insulin, growth hormone, and thyroid hormone are examples of peptide/protein messengers. These hormones regulate various body functions such as metabolism, growth, and development. The following table summarizes the different classes of chemical messengers: Hormone Type Examples Steroid messengers Testosterone, estrogen, progesterone Eicosanoid messengers Prostaglandins, leukotrienes Peptide/protein messengers Insulin, growth hormone, thyroid hormone, oxytocin, cortisol, ACTH, etc.
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Oxygen enters the bloodstream through diffusion.
False
True
Answer:
True
Explanation:
By the process of diffusion, oxygen is driven by a gradient and crosses into the blood, taken up by the hemoglobin (in the red blood cells), and then transported by circulation to the the rest of the cells in the body.
Answer: True
Explanation:
assume that simple mendelian inheritance is at play. which of the following represent the possible genotypes resulting from a cross between two individuals that are heterozygous for black hair (bb; where b is black and b is blonde)?
Bb, bb and BB. A mating experiment between two heterozygous organisms that are equally hybrid for two allele traits is referred to as a dihybrid cross.
A heterozygous organism is one that possesses two distinct alleles at a certain genetic location, making it a hybrid. Only gametes with particular parental genetic combinations will result from a cross between two heterozygous parents with connected loci. Progeny with the paternal phenotype will result from the fusion of these parental gametes. An organism with a heterozygous genotype possesses both a dominant and a recessive allele. This genotype's designation in our case is Bb. Finally, homozygous recessive refers to an organism's genotype that consists of two recessive alleles.
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what air compressor type should i use for medical ventilation.
Positive Displacement or Dynamoc compressor? explain pls
When it comes to medical ventilation, positive displacement air compressors are the preferred choice because they offer a continuous flow of compressed air.
Dynamoc compressors, on the other hand, produce compressed air in pulses, which can be problematic for medical equipment. An air compressor is a device that compresses air and converts it into potential energy stored in compressed air. They are used to power various pneumatic tools, from airbrushes to jackhammers. They can be used for a wide range of applications, including powering tools and providing ventilation in hospitals. In medical settings, air compressors are used to power ventilators, oxygen concentrators, and other life-saving equipment.
Air compressors that are specifically designed for medical use must meet strict safety and quality standards, including ISO 13485:2003 and the EU Medical Device Directive. A positive displacement air compressor works by trapping air in a chamber and reducing the volume of that chamber. This causes the pressure of the air to increase, which is then released through an outlet. Because the flow of compressed air is continuous, positive displacement compressors are the preferred choice for medical ventilation.
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What are the R side chains in each
Answer:
Each of the 20 amino acids has a specific side chain, known as an R group, that is also attached to the α carbon. The R groups have a variety of shapes, sizes, charges, and reactivities. This allows amino acids to be grouped according to the chemical properties of their side chains.
What occurs in photosystem 1? a. Glucose is made b.water is split into oxygen and hydrogen c.ATP is produced and NADPH is produced d.NADP and ADP are burned for energy
PLEASE HELP!
Answer:
A
Explanation:
Who do you agree I
With the most?explain why you agree with person and not the other
Answer:
Mika is right because respiration happens during the d at and the night. and photosynthesis happens only when there is sunlight.
A body cell has been growing and synthesizing proteins. In the nucleus of this body cell, DNA replication is taking place, and a copy of the cell's genetic material is copied. Which of the following is the best conclusion you can make about the life cycle of this cell?
The best conclusion you can make about the life cycle of this cell is that the cell is in the S phase of interphase and will move next to the G2 phase.
S phase (Synthesis Phase) is the phase of the cell cycle in which all of the chromosomes (DNA) are replicated within the nucleus. During this phase, the DNA is effectively doubled as each chromosome contains two sister chromatids. After the S phase, the cell enters the G2 phase where various proteins (such as microtubules) are synthesized.
What is aerobic respiration?
A. The process cells use to produce glucose and oxygen
B. The process cells use to release energy from glucose in the
presence of oxygen
c. Physical activity that involves large muscle groups and increases
breathing rate
D. Physical activity that causes the heart and lungs to work harder to
meet the body's need for oxygen
SUBMIT
Help?
Write a testable question that could be used to further investigate the topic of using the fossil record to identify changes in life on earth.
What kind of organisms existed in the past?
How did organisms move about? are some testable questions hat could be used to further investigate the topic of using the fossil record to identify changes in life on earth.
Fossils are the remains of extinct plants and animals that have been encased in layers of rock.The existence of fossil records demonstrates how living things have altered or evolved over time. The specific order in which these species have lived has been identified by paleontologists. For instance, older rocks contained simpler species, while younger rocks contained complex life.Existence of species with transitional traits that are placed in sequential stratigraphic order is one pattern in the fossil record. This indicates that the fossil record demonstrates minute changes in the structures of creatures across time, supporting evolution.learn more about fossil here: https://brainly.com/question/11224681
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What are 5 factors that decrease populations (explain each or tell how that factor causes a decrease in a population).
Answer:
Morality and Emigration
Explanation:
Mortality and emigration decrease the population. Thus, the size of any population is the result of the relationships among these rates. Natality, mortality, immigration, and emigration rates apply to every population, including the human population. The sum of these rates makes up the growth rate of a population.
If a mother has blood type A and dad has type B, what are the chances of having a child that has blood type A?
Science help PLZZZ!!!
Answer: b
Explanation:
Answer: B.
Explanation: Light cannot pass through an opaque object.
Help me please with 1 and 2 you’ll get brainlest if ur right
Answer:
Question 1~ "particles that make up matter have kinetic energy" & "particles that make up matter are always in motion".
Question 2~ a- liquid
b- solid
c-gas
hope this helps!!!
Warm air creates a rising, low pressure zone.
A. True
B. False
Answer:
B. False
Explanation:
Can someone help me with this chromosome assignment?
this is the answer for only first page.
homologous, diploid,gametes,haploid
the same thing as what the other person said.
di=2
hap=1
remember that for diploid and haploid
Compare the general appearance of the DNA molecule with the mRNA molecule.
Answer:
DNA has thymine as one of the two pyrimidines while mRNA has uracil as its pyrimidines base. DNA is present in the nucleus while mRNA diffuses into the cytoplasm after synthesis. DNA is double-stranded while mRNA is single-stranded. mRNA is short-lived while DNA has a long lifespan.
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how does the Digestive system help maintain homeostasis?
In your answer, include the most important organ(s) (list at least 2) and include what they do to make the system work. It must include at least 3 full sentences
Answer:
The bacterial flora in the intestines are essential to homeostasis in the body, they not only break down food so the nutrients can be absorbed, they produce vitamins like biotin and vitamin K and guard against harmful bacteria that enter the system. While your heart is a vital organ, the brain (and the nervous system that attaches to the brain) make up the most critical organ system in the human body. The digestive system ordinarily gets 20% to 25% of the oxygenated blood pumped out by the heart and the receptors in muscles provide the brain with information about body position and movement, the brain controls the contraction of skeletal muscle the nervous system regulates the speed at which food moves through the digestive tract.
12.36Un balón lleno contiene 12 kg de oxígeno, O2, bajo una presión ma-nométrica de 40 atm. Determine la masa de oxígeno que se ha extraído del balón cuando la presión absoluta alcanza el valor de 25 atm. Su-ponga que la temperatura del balón permanece constante. R: 7.3 kg.
Answer:
Quedan 7.317 kilogramos de oxígeno en el balón y se extraen 4.683 kilogramos de oxígeno en el proceso.
Explanation:
Asúmase que el balón es rígido. Supongamos que el oxígeno se comporta como un gas ideal, entonces la ecuación de estado es la siguiente:
\(P\cdot V = \frac{m\cdot R_{u}\cdot T}{M}\) (1)
Donde:
\(P\) - Presión, medida en atmósferas.
\(V\) - Volumen, medido en litros.
\(m\) - Masa, medida en kilogramos.
\(M\) - Masa molar, medida en kilogramos por kilomol.
\(T\) - Temperatura, medida en Kelvin.
\(R_{u}\) - Constante de los gases ideales, medida en atmósfera-litros por kilomol-Kelvin.
El aire experimenta un proceso isocórico e isotérmico, entonces obtenemos la siguiente relación:
\(\frac{P_{1}}{m_{1}} = \frac{P_{2}}{m_{2}}\) (2)
Nótese que los subíndices 1 y 2 representan los estados inicial y final de proceso.
Si sabemos que \(P_{1} = 41\,atm\), \(m_{1} = 12\,kg\) y \(P_{2} = 25\,atm\), entonces la masa final del aire es:
\(m_{2} = \frac{P_{2}}{P_{1}} \cdot m_{1}\)
\(m_{2} = \left(\frac{25\,atm}{41\,atm}\right)\cdot (12\,kg)\)
\(m_{2} = 7.317\,kg\)
La masa que se ha extraído del balón es:
\(\Delta m = m_{1}-m_{2}\)
\(\Delta m = 12\,kg-7.317\,kg\)
\(\Delta m = 4.683\,kg\)
Quedan 7.317 kilogramos de oxígeno en el balón y se extraen 4.683 kilogramos de oxígeno en el proceso.
The sticky material between cells of areolar connective tissue is called the A) gel matro. B) ground substance. C) cytoplasm D) crista. E) cytosol.
The sticky material between cells of areolar connective tissue is called the ground substance, which corresponds to option B.
The ground substance is an amorphous, gel-like substance that fills the spaces between cells and fibers in connective tissue.
The ground substance consists of a combination of water, glycosaminoglycans (GAGs), proteoglycans, and glycoproteins.
This mixture gives the ground substance its viscous and adhesive properties, allowing it to act as a medium for nutrient and waste exchange, as well as providing support and hydration to the cells within the tissue.
Glycosaminoglycans (GAGs) are long, linear chains of repeating disaccharide units. They are highly negatively charged due to the presence of sulfate or carboxyl groups, which contributes to the gel-like consistency of the ground substance.
Proteoglycans are large molecules made up of GAGs attached to a protein core, while glycoproteins are proteins with attached carbohydrate chains.
The ground substance also plays a crucial role in facilitating cell migration during tissue repair and inflammation. It provides a pathway for cells to move through the connective tissue matrix.
Areolar connective tissue is a loose connective tissue found throughout the body, surrounding and supporting various organs and structures. It is characterized by its high cellularity and abundant ground substance.
Areolar connective tissue contains a variety of cell types, including fibroblasts, immune cells, and adipocytes.
In summary, the sticky material between cells of areolar connective tissue is called the ground substance. It is a gel-like substance composed of water, glycosaminoglycans, proteoglycans, and glycoproteins.
The ground substance fills the spaces between cells and fibers, providing hydration, support, and facilitating cell movement within the connective tissue.
So, option B is correct.
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Question: What type of mutation is seen in the mutated DNA?
Normal DNA
ATG TGT GTT TAT GGA
Mutated DNA
ATG TGT GTC TAT GGA
Answer:
substitution
Explanation:
as it is changed T base with the C base read your description of substitution it is good as well :)
1
(a)
You should attempt ALL questions
Identify the components temperature and approximate composition (of wt.% C) at which the peritectic transformation occurs in Fe-C alloys. Explain why delta-ferrite is not expected to be retained on equilibrium cooling a steel of this composition through the invariant reaction.
[4 marks] (b) When a Fe-0.4 wt% C steel is austenitised at 1000°C for 30 minutes and is slow cooled to room temperature we expect eutectoid alpha-iron, pro-eutectoid alpha- iron and cementite to form. If instead a Fe-1.3 wt% C steel is austenitised at 1000°C for 30 minutes and is slow cooled to room temperature we expect alpha- iron, eutectoid cementite and pro-eutectoid cementite to form.
(c)
(d)
Use the equilibrium phase diagram together with the lever rule to describe as quantitatively as possible the key difference in terms of phases, phase fractions and microstructure between these alloys. Include sketches which illustrate (i) the stages in the development of the microstructures and (ii) the mechanisms of the phase transformations.
[13 marks]
If the steel described in (b) is quenched rapidly from 1000°C to room temperature, describe the martensitic phase and microstructure which will be formed and explain why this is different from that formed in any of the cases above.
[3 marks]
If reflected light optical microscopy were used to confirm the formation of martensite (e.g., lath type or plate type or mixed) a sample of the steel in part (c) would be polished flat to a sub-micron surface finish, etched with sodium bisulfate solution and examined.
(i)
(ii)
(iii)
Explain the reason for the polishing and etching steps during sample preparation.
[2 marks]
Typically what type of objective lens would you need to use in terms both magnification and numerical aperture, in order to observe the fine martensitic microstructure?
[2 marks] Depending on the exact martensitic microstructure, a variety of defects could be observed such as dislocations and twins. Describe the general form of both these defects and indicate how they might be formed.
[3 marks] (iv) If not all the austenite had transformed to martensite, using an additional technique, how could you determine the crystallographic orientation relationship (i.e. the orientation of one crystal lattice relative to another) between the retained austenite and the martensite?
[2 marks]
(e)
The XRD pattern for the material produced in part (c) is shown in Figure Q1 below. The XRD pattern was collected using Cu Ka radiation.
3
Peak Phase 28 (°) (hkl)
1
1 42.94 (121)|
Intensity (arbitrary units)
23
2
1
43.83 (210)
2
44.67
4
1
44.99 (103)
5
1
45.94 (211)
7
67
1
48.59 (113)
1
49.16 (122)
12
الثر
(8)
8
2
65.02
42
44
46
48
50
52
54
56
60
62
64
66
20 (") Figure Q1
The XRD pattern has been partially analysed, with the identification of two constituent phases, labelled as Phase 1 and 2 detailed in the table above. Each peak, which is labelled, has the 20 value listed in the accompanying table, and is identified as being part of Phase 1 or 2. Phase 2 is a cubic structure with a = 2.87 Å, Phase 1 is not cubic in structure.
(i)
(ii)
Calculate the d-spacing to two decimal places for Peaks 1 and 2, the first two peaks present in the scanned region for Phase 1.
[2 marks]
According to the database, Phase 1 is not cubic and there should be additional peaks, the (002) and (202) peaks. Suggest two reasons why these peaks are not immediately apparent in the XRD plot and table.
(iii)
[2 marks] Phase 2 is cubic; however, it is not primitive cubic. Using peaks 3 and 8, determine which type of cubic unit cell Phase 2 is.
[3 marks]
(iv) Draw the unit cells for the Miller indices identified for peaks 3 and 8.
(v)
[1 mark] Determine the Miller indices for the next peak at higher 20 for Phase 2 and calculate the 20 value to the nearest degree this will be detected at.
[3 marks]
The peritectic transformation occurs in Fe-C alloys at a temperature of about 1148°C and at about 0.17 wt.% C. Delta-ferrite is not expected to be retained on equilibrium cooling steel of this composition through the invariant reaction because when the steel is cooled slowly, all delta-ferrite will transform into austenite at a temperature range of 1400-912°C, and the remaining austenite will then transform into pearlite at a temperature range of 727-427°C. Delta-ferrite cannot exist in equilibrium in iron-carbon alloys containing less than 0.09% C at room temperature.
b) If a Fe-0.4 wt% C steel is austenitised at 1000°C for 30 minutes and is slow-cooled to room temperature, eutectoid alpha-iron, pro-eutectoid alpha-iron and cementite will form. When instead a Fe-1.3 wt% C steel is austenitised at 1000°C for 30 minutes and is slow-cooled to room temperature, alpha-iron, eutectoid cementite and pro-eutectoid cementite will form.
c) When using the equilibrium phase diagram together with the lever rule to describe as quantitatively as possible the key difference in terms of phases, phase fractions and microstructure between these alloys, it can be observed that Fe-0.4 wt% C steel consists of about 98% pearlite and 2% proeutectoid ferrite while Fe-1.3 wt% C steel contains about 60% cementite, 20% proeutectoid cementite and 20% pearlite. The Fe-0.4 wt% C steel is composed of a fine network of pearlite while the Fe-1.3 wt% C steel is made up of coarse cementite and pearlite in a lamellar form with the ferrite and cementite phases separating into separate regions during cooling.
d) If the steel described in (b) is quenched rapidly from 1000°C to room temperature, martensite will be formed which will consist of a supersaturated body-centred cubic (BCC) lattice of carbon in iron. This is different from that formed in any of the cases above because martensite is formed when austenite is rapidly cooled, the carbon atoms do not have sufficient time to diffuse to form either ferrite or cementite.
e) The d-spacing to two decimal places for Peaks 1 and 2, the first two peaks present in the scanned region for Phase 1 can be calculated using Bragg's law, which is nλ = 2d sin θ. For Peak 1, d = nλ / 2 sin θ = (1 × 0.15406) / (2 × sin 28) = 0.270 nm, and for Peak 2, d = nλ / 2 sin θ = (1 × 0.15406) / (2 × sin 43.83) = 0.219 nm.
The (002) and (202) peaks are not immediately apparent in the XRD plot and table due to their weak intensities and their positions overlapping with the peaks of Phase 2.
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