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How to become s software engineer?

What is the role of software engineering in life?

What is Value of female labor (single vs married, temporary vs long term, within minority groups, within classes)?

• inhabitant
• makeshift
• abundance

What is a Watershed?

What is a phyton?

Triple some in array

1. We will discuss the multiple approaches to solve this problem and finally we will write its Java code. The easiest approach is to use three loops to solve the problem. The expected time complexity is 1 square without using any extra space. The brute force approach is the simplest approach or you can say the brute force approach. The time complexity is bigger of and cube so basically we are using three four loops right and it 's a space complexity is big o of 1 as we are not using any extra space right. After that, basically we return the value true. If the array is sorted we do n't have to make all possible sets right, we can do the sum of the three numbers and accordingly move the pointers so let me explain this so we have given this array right. First, let's sort this array and then use two pointers to solve the problem.

What is spring boot and how its use ?

it is July 30,2017. The cheapest-to-deliver bond in a September 2017 Treasury bond futures contract is a 13% coupon bond, and deliveryis expected to be made on September 30,2017. Coupon payments on the bond are made on February 4 and August 4 each year. The term structure is flat, and the rate of interest with semiannual compounding is 12% per annum. The conversion factor for the bond is 1.5. The current quoted bond price is $110. Calculate the quoted futures price for the contract.

4.The following table provides the quoted price and terms for three separate treasury bonds

1 $152 Time-to-maturity = 20 years

Coupon = 14% per annum (payable semi-annually)

2 $140 Time-to-maturity = 25 years

Coupon = 12% per annum (payable semi-annually)

3 $124 Time-to-maturity = 30 years

The quoted futures price is $78.

(a) What are the conversion factors for the above three bonds?

(b) What is the cheapest-to-deliver bond?

5. The cheapest-to-deliver bond in a Treasury bond futures contract is a 13% coupon bond, and the delivery is expected to be made 62 days from today. Coupon payments on the bond are made every six months and the last coupon was made 177 days back. The next coupon is due 5 days from today and the coupon after that will be due 188 days from today. The annual risk-free rate is 11.65% per annum. The conversion factor for the bond is 1.5. The current quoted bond price is $110. Calculate the quoted futures price for the contract.

6.Company A, a British manufacturer, wishes to borrow U.S. dollars at a fixed rate of interest. Company B, a U.S. multinational, wishes to borrow sterling at a fixed rate of interest. They have been quoted the following rates per annum (adjusted for differential tax effects):

Company A 11.0% 7.0%

Design a swap that will net a bank, acting as intermediary, 10 basis points per annum and that will produce a gain of 15 basis points per annum for each of the two companies.

Company A,  wishes to borrow U.S dollars at a fixed rate of interest. Company B, wishes to borrow Japanese yen at a fixed rate of interest. The amounts required by the two companies are roughly the same at the current exchange rate. The companies have been quoted the following interest rates: 

Company X- Yen 4.8%
Company Y-Yen 6.2%

Company X-Dollars 9.5%

Company Y-Dollars 10.0%

Design  a swap that will net a bank, acting as intermediary, 20 basis points per annum. Make a swap appear equally attractive to the two companies and ensure that all foreign exchange risk is assumed by the company X and Company Y.  Also draw the cash flow chart. 

What is imperialism

What about indian history?

An intern gives you the following table for a normal adult (65kg) that just finished ingesting an unknown quantity of a sodium chloride solution (750mOsm/L). you notice a few missing entries.

Assuming that the “before” condition means before the NaCl drink; the “after” condition is when equilibrium has been achieved; no water is lost from the body; and that there is no movement of sodium chloride into or out of the cells, fill in the chart above. Fill in the missing information, including the amount of fluid the painting ingested, and all missing osmolarities and volumes.

Two (2) liters of a hypotonic 0.5% NaCl solution is infused intothe extracellular fluid compartment of a 40 Kg dog with initialplasma osmolarity = 300 milliosmoles/liter. After equilibration,what will the extracellular fluid volume, intracellular fluidvolume, and plasma osmolarity be? Molecular weight of NaCl =58.5

Ed Rivers, a 3rd year medical student, was alone in the hospitalER one night. It was unusually quiet that night, and the residentwas getting some much needed sleep. A patient, Mrs. X, was broughtin showing signs of serious dehydration. Ed tried to give herewater, but she vomited this back up. Feeling he must try something,and not wanting to wake the resident, Ed administered 1 liter ofsterile distilled water IV. Assume for simplicity that the redblood cells contain only solutes to which the rbc membrane isimpermeable, and that the rbcs and plasma are in osmoticequilibrium when the patient is brought in. The osmolarity of therbc is 300 mOsm/L.

What is Mrs. X red blood cell volume after the infusionequilibrates with the blood and why? Also what is the osmolarity ofthe red blood cells and plasma AFTER equilibration?

Ed Rivers, a 3rd year medical student, was alone in the hospital ER one night. It was unusually quiet that night, and the resident was getting some much needed sleep. A patient, Mrs. X, was brought in showing signs of serious dehydration. Ed tried to give here water, but she vomited this back up. Feeling he must try something, and not wanting to wake the resident, Ed administered 1 liter of sterile distilled water IV. Assume for simplicity that the red blood cells contain only solutes to which the rbc membrane is impermeable, and that the rbcs and plasma are in osmotic equilibrium when the patient is brought in. The osmolarity of the rbc is 300 mOsm/L.a. Mrs. X

Ed Rivers, a 3^rd year medical student, was alone in the hospital ER one night. It was unusually quiet that night, and the resident was getting some much needed sleep. A patient, Mrs. X, was brought in showing signs of serious dehydration. Ed tried to give her water, but she vomited this back up. Feeling he must try something, and not wanting to wake the resident, Ed administered 1 liter of sterile distilled water IV. Assume for simplicity that the red blood cells contain only solutes to which the rbc membrane is impermeable, and that the rbcs and plasma are in osmotic equilibrium when the patient is brought in. The osmolarity of the rbc is 300 mOsm/L.The volume of Mrs. X’s plasma was 3 liters before Ed administered the IV.

Had Ed given Mrs. X a sucrose solution instead of water, he might have helped her. What concentration of sucrose solution should he have administered to leave her rbc volume unaffected, and why?

Ed Rivers, a third-year medical student, was alone in the hospital emergency room one night. It was unusually quiet in the ER, so the residents were getting some much-needed sleep. A patient, Ms. B., was brought in showing signs of serious dehydration. Ed tried to give her water but she vomited. Feeling that he must try something else and not wanting to wake the residents, he administered 1 liter (L) of sterile distilled water intravenously (IV). The questions that follow are aimed at determining the consequences of Ed�s action. Assume for simplicity that red blood cells (RBC) contain only solutes to which the RBC membrane is impermeable. The osmolarity of the RBC, the concentration of solute particles, is 300 mOsm/L.

Ed Rivers, a third year medical student, was alone in the hospital emergency room (ER) one night. It was unusually quiet in the ER, so the residents were getting some much- needed sleep. A patient Ms. M.B. was brought in showing signs of serious dehydration. Ed tried to give her water, but she vomited. Feeling that he must do something else and not wanting to wake the residents, he administered 1 liter (L) of sterile distilled water intravenously (IV).

Work through each section one at a time. Once you feel you have mastered each section of the problem use your answers to help you summarize everything into a one to two page written report. Work on this as a group, but each person in the group needs to hand in their own typewritten report.

1. Predict the direction of change (increase, decrease, no change) you would expect Ed’s infusion to have produced in the listed parameters. Explain

2. The Volume of Ms. M.B., plasma was 3 L before Ed administered the IV. Assume that there was complete mixing of the administered water with her plasma, but no mixing with her interstitial fluid. Calculate Ms. M.B.’s plasma osmolarity after the infusion mixed with her plasma but before any water entered her RBC.

3. Assuming that Ms. M.B.’s RBC volume was 2 L before the IV was given. Calculate her plasma and RBC osmolarity after the infused water equilibrated between her plasma and her RBC. How much would her RBC volume change? (Again, assuming that there was no mixing with the interstitial fluid).

Ms. M.B. got worse after Ed’s treatment so the resident was called. She drew some blood from the patient and centrifuged it. The denser RBCs were forced to the bottom of the tube. The supernatant plasma was pink. The resident then decided to infuse a sucrose solution into the patient (The RBC membrane is impermeable to sucrose).

4. Why was the patient’ s plasma pink?

5. Had Ed given Ms. M.B. a sucrose solution instead of water, he might have helped her. What concentration sucrose solution should he have administered to leave the RBC volume unaffected?

The resident, knowing what Ed had done, infused the patient with 1 L of a 600 mM (i.e., 600 mOsm / L) sucrose solution. Recall that, after Ed’s treatment, the patient’s plasma was x (what you figured out above) L, her RBC volume was x (what you figured out above) L, and her osmolarity was x (what you figured out above) mOsm / L. Assume that this infusion equilibrated with Ms. M.B.’s plasma and RBCs.

6.Predict (increase, decrease, no change) the affect of the infusion given by the resident on the parameters listed. Explain.

7. Assuming that none of the administered fluid was excreted, calculate Ms. M.B.’s plasma osmolarity and RBC volume after the resident’s infusion had equilibrated with her plasma and RBC. (Again assume that no mixing takes place with interstitial fluid).

The resident, knowing what Ed had done, infused the patient with 1 L of a 600 mM (i.e., 600 mOsm / L) sucrose solution. Knowing that, after Ed’s treatment, the patient’s plasma volume was 3.6L, her RBC volume was 2.4L, and her osmolarity was 250 mOs/ L. Assume that this infusion equilibrated with Ms. M.B.’s plasma and RBCs.

6.Predict (increase, decrease, no change) the effect of the infusion given by the resident on the parameters listed. Explain your predictions.

Assume we have a cell which contains nonpermeating molecules (p; not ionized) at a concentration of 300 mM/l cell water. Assume the initial cell volume is 1.0nanoliter. What will be the final (i.e. equilibrium) volume when the cell is immersed in a large volume of the following solutions?

A) 300 mOsm sucrose (sucrose cannot penetrate the membrane)

B) 150 mOsm sucrose

C) 600 mOsm sucrose

D) 300 mOsm glycerol (glycerol can penetrate the membrane)

E) pure water

F) 600 mOsm glycerol

G) 300 mOsm sucrose & 300 mOsm glycerol

Please explain work. Label each test solution in terms of its osmolarity and tonicity.

What would the time course of volume change curve (graph Volume vs. time) for solution G look like?

Now assume that the normal Na+ and K+ concentrations of ICF and ECF are as given below, and that the man suffers a severe heart attack, which can cause his cells to lose a little K+ and gain a little Na+:

Na+ (mM/l)

ICF: 10 ECF: 140

K+ (mM/l)

ICF: 145 ECF: 5

After heart attack:

Na+ ICF: 14

K+ ICF: 141

Specifically, assume that the 4mM/l K+ los from the cells is exactly balanced by a gain of Na+. Thus, no other significant fluxes occur. What will be the ECF concentrations of Na+ and K+?

Can you cite this quote for me properly in APA?

“Two things fill the mind with ever new and increasing admiration and awe, the more often and steadily we reflect upon them: the starry heavens above me and the moral law within me."

1. Your goal is to make an isotonic intravenous solution for an emergency room patient who

has lost a lot of blood. The available solutes include glucose (300 mM), NaCl (150 mM), and
urea (300 mM). What solutes should she use to make up the IV solution with an osmolarity
of 300 mOsm (isosmotic), making sure that it will also be isotonic?

Explain your answer.
Assume that the cell is permeable to urea, glucose, but not to NaCl and that the osmolarity of
the cell is 300 mOsm.

2. If a 10% sucrose solution is separated from a 20% sucrose solution by a membrane
impermeable to sucrose, in which direction will net movement of water occur?
A) from the 10% sucrose solution to the 20% sucrose solution only
B) from the 20% sucrose solution to the 10% sucrose solution only
C) There will be no net movement of water in this case.
D) None of the answers are correct.

The value for Ψ in root tissue was found to be - 0.15MPa. If you take the root tissue and place it in a 0.1M sucrose solution (Ψ = -0.23). Which on the following should occur?

Select one:

a. Sucrose will be taken up by the root.

b. The root will synthesize sucrose to match the outside solution.

. The root will lose water to the outside solution.

d. The root will take up both water and sucrose

The value for Ψ in root tissue was found to be -0.15 MPa. If you take the root tissue and place it in a 0.1 M solution of sucrose (Ψ = -0.23 MPa), the net water flow would _____

An open beaker solution of 0.1 M sucrose has a ΨS = -0.244 MPa and contains a plant cell at equilibrium with a ΨP = 0.1 MPa. When we double the concentration of sucrose in the beaker solution, what is the new ΨS of the cell when the cell becomes flaccid?

1. You measure xylem pressure in two plants using the pressurechamber. In one, you apply 10 bars of pressure before you noticewater coming from the xylem of your sample. In the other you apply15 bars of pressure. Which plant is more water stressed? What isthe pressure potential of the water stressed plant in MPa?
   
   
   

2. If (water potentional)_P = 0.3 MPa and (waterpotentional)_S = -0.45 MPa, the resulting (waterpotentional)_W is ...

3. Briefly describe what happens in the xylem when cavitationdue to water stress occurs. Mention one structural feature of xylemin conifers that protects against cavitation. (3 - 5 sentences).

4.Solute and pressure potential is measured in two differentplaces in the same plant. Fill in the table below with plausiblenumbers (based on physiological principles). (Note: the plant ishealthy and well-watered).

If a plant cell has a pressure potential (ΨP) of 2 MPa and its ΨS = -3.5 bars, what is its water potential (Ѱ)? The plant cell from question above is placed in a beaker of sugar water with ΨS = -4.0 MPa. Will the water flow into or out of the cell?

If a plant cell with a solute potential of -0.2 MPa is placed in a beaker of pure water, at equilibrium it will have a pressure potential of +0.2 MPa because the cell wall prevents cell expansion. If that cell is then placed in a solution with a solute potential of -0.3 MPa that is open to the atmosphere (draw a picture if it helps you),

1.
What are the three most abundant macronutrients in plant tissues, and where are they obtained from?

A.)
N, P, K; obtained from the soil
B.)
C, H, O, obtained from water and air
C.)
N, P, K; obtained from the air
D.)
C, H, O; obtained from mycorrhizae

2. What is the function of the pericycle?

A.
It is the meristem layer in the root that forms new roots.
B.
It is the meristem layer that undergoes secondary growth in stems.
C.
It is the layer of cells in the roots that has a Casparian strip.
D.
It forms from axillary buds after apical dominance is removed.

3.The endodermal layer forces water and minerals to take the __________ route from cell to cell because of a suberin layer called the ____________.

A.
apoplastic; cuticle
B.
transmembrane; cuticle
C.
apoplastic; Casparian strip
D.
transmembrane; Casparian strip
E.
symplastic; cuticle

4. A plant cell has a water potential (Ψw) of -1 MPa, and is flaccid, having a turgor potential (Ψp) of 0 MPa. What is the solute potential (Ψs) of this cell?

A.
-2 MPa
B.
-1 MPa
C.
+2 MPa
D.
+1 MPa
E.
0 MPa

consider a root epidermal cell that is in contact with soil. if the pressure potential of the soil is 0 MPa, the solute potential of the soil is -0.3 Mpa, and the matric potential of the soil is -2.5 Mpa, and if the pressure potential in the root epidermal cells is +0.1 MPa, the solute potential is -2.9 MPa and the matric potential is 0 Mpa, which of the following is true:
a) water will move out of the root epidermal cell into the soil, B) water will move out the soil into the root epidermal cell C) the root is in equillibrium with the soil and there will be no net flow in either direction

If the?sand ?pinthe soil adjacent to a root hair is -0.2 and -0.7 MPa respectively,and ?sin the root hair is -1.1 MPa, what is ?pin the root hair?

Q. Where are root hairs present in root.

I’m writing a reflection on Immanuel Kant about how his philosophy is better then Jeremy Bentham, St Thomas Aquinas, and Joseph Fletcher. I need you to provide me with a good catchy introduction i can use before i add my thesis

My thesis is: Immanuel Kants philosophy of the categorical imperative aligns with peoples beliefs the most because it emphasizes the importance of treating people as ends rather than means, promoting the idea of universal ethics, and stressing the significance of individual autonomy in decision making.

Please provide me with a good catchy introduction with a good use of advanced words at a university level

Please read the article below and answer these questions based on the article and USE YOUR OWND WORDS!

(1) Why does thermogenesis not occur during the transition into torpor?

(2) In the study by Toien et al., how was the metabolic rate of hibernating black bears measured?

(3) Bears typically come out of hibernation leaner (i.e., less fat) than when they entered. Why is this?

(4) Suggest a human application based on your answer to the previous question.

Both the AM and the NF symbioses are based on the shared activity of a set of plant genes, SYM genes (3–5). This indicates that bacte- ria hijacked the signal transduction pathway that fungi had used to gain entry into plant tissues and cells. Op den Camp et al. provide evidence that in Parasponia, the only nonle- gume partner of rhizobia, a single receptor can recognize both the fungal and bacterial signals and induce the common SYM path- way to promote the intracellular accommoda- tion of the partner microorganisms.

Op den Camp et al. used Rhizobium strains, some of which were able to form nodules, and some of which were unable to form nodules, to prove that both Parasponia and legumes use lipochito-oligosaccharides called Nod factors to induce nodule develop- ment. They also showed that Nod factors act similarly in both symbioses via a common signaling cascade; in Parasponia, the intro- duction of a dominant active form of calcium/ calmodulin-dependent kinase (CCaMK), a key element of this pathway, resulted in spon- taneous nodulation, as in legumes.

Op den Camp et al. also provide insight into how bacterial Nod factor receptors (NFRs) evolved from receptors involved in plant-fungi partnerships. The most-studied legumes recognize rhizobia—or, more accu- rately, the bacterial Nod factors—via a pair of LysM-type receptor-like kinases, NFR1/ LYK3 and NFR5/NFP (6, 7). Because these NFRs are specif ic to bacterial symbiosis, investigators had hypothesized that they evolved either by duplication of the mycor- rhiza-specific receptors, which then gained new functions, or by the recruitment of new

receptors that turned on the common signal- ing pathway. Op den Camp et al.’s analysis indicates that receptor duplication was not essential for plants to acquire the ability to form a symbiotic relationship with NF bac- teria. Instead, the presence of a single NFR5- like receptor in Parasponia, and its indispens- able role in both symbioses, strongly suggests that rhizobia entered symbiotic interactions with plants through the same entrance used by mycorrhizal fungi. It also means that the molecular “keycard” that opens the door to plant partnerships for both bacteria and fungi—the bacterial Nod factor and mycor- rhizal (Myc) factor—must be very similar. Indeed, Maillet et al. (8) recently described the Myc factors of AM fungi as lipochito-oli- gosaccharide molecules that are very similar to Nod factors.

These results raise several questions: Why is the appearance of nitrogen-f ixing nod- ules, especially rhizobial ones, restricted to a small fraction of mycorrhizal plants? How do plants discriminate between symbiotic fungi and bacteria? Was it necessary for host plants to distinguish between the microbes to create different niches? Studies of genes from related plants suggest that plant fami- lies establishing rhizobial or actinorhizal (Frankia) symbioses belong to the same large lineage. This raises the possibility that, dur- ing the evolution of flowering plants, a pre- disposition for symbiotic nodule formation originated only once (9). Did this predispo- sition occur by changing the activity of one or more component(s) of the common sym- biotic pathway, for example, by enabling it to provide different outputs?

Both bacteria and mycorrhizal fungi induce changes in intracellular calcium (Ca2+-) concentrations (termed calcium spik- ing). However, the frequency and duration of the oscillations, as well as the speed of Ca movement, are different in the two symbio- ses (10). Early elements of the common SYM pathway, such as the LysM-type receptors and another receptor protein, the symbiosis receptor kinase (SYMRK), are required for the induction of the calcium spiking, which is then deciphered by CCaMK. It will be inter- esting to compare calcium spiking upon rhi- zobial and fungal inoculations in species that possess dual-functioning receptors.

There is not yet enough systematic data from different plant lineages to determine exactly how molecules like SYMRK and CCaMK contributed to the evolution of a predisposition to nodule formation. The real challenge is to find out why lineages with predisposition for nodulation (for exam- ple, certain legumes) are unable to establish NF symbiosis.

Do bears really hibernate? Their high body temperature during winter dor- mancy has raised some doubt about this behavior, as it is unlike the pronounced decreases observed in small mammals that enter this nonactive state. On page 906 of this issue, Tøien et al. (1) show that bears do indeed hibernate. Through continuous mea- surement of oxygen consumption, body tem- perature, and heart, muscle, and brain activ- ities, the authors show that black bears display unusual patterns of metabolic and ther- mal regulation during hibernation as well as when they emerge from this resting state in the spring.

Hibernation is a powerful behavior that reduces energy costs in mammals. However, in small mammals, it is frequently interrupted by arousals (2, 3), thereby reducing its effec- tiveness. Generally, after entrance into torpor, deep torpor is maintained for 1 or 2 weeks with body temperature close to the freezing point of body fluids, and is terminated by an arousal for about 1 day. During arousal, body temperature rises to a normal 36°C by endogenous heat production. Collectively, the arousal episodes require about 80% of the entire energy cost of the animal during the hibernation season. The reasons for the repeated arousals are still a mystery, but they may allow for the repair of neuronal damage induced by prolonged hypometabolism and brain inactivity at low temperature (4, 5).

Spontaneous hibernation behavior is dif- ficult to observe in captive animals, so its study has mostly relied on field studies of subjects in their natural habitat, or on ani- mals kept in conditions similar to their nat- ural environment. Studying large mammals (at least 10 kg) is particularly difficult because of the challenges of continuous and long-term monitoring. Tøien et al. observed five Alaskan black bears (Ursus americanus) (two females and three males ranging in body mass from 34.3 to 103.9 kg) that were kept in outdoor enclosures in a forest near Fairbanks, Alaska. The bears hibernated in isolated wooden nest boxes, which allowed continu- ous observation and measurement of oxygen consumption and body temperature as well as monitoring of physiological activities from implanted transmitters. The authors observed that during the hibernation period (November to March), the bears did not display repeated arousals, but instead showed multiday oscil- lations of body temperature between 30° and 36°C. Such a lack of periodic arousals dur- ing hibernation has so far only been observed in one small mammal [fat-tailed lemur (6)]. However, Tøien et al. found that the hiber- nating bears reduced their metabolic rate to 75% below basal metabolic rate (BMR). The observed minimum metabolic rate in hiber- nating bears (0.056 ml O2 g−1 hour−1) is within the range of those observed in small hibernat- ing mammals (0.02 to 0.06 ml O g−1 hour−1) (2, 3). This implies that bears use the entire mammalian scope of metabolic inhibition in torpor and are true hibernators. This reduction of metabolic rate to 75% below BMR is sub- stantially less prominent than that for small mammals (98% below BMR). The difference is largely due to the allometric scaling of BMR, indicating that hibernation is more effective in small mammals below 1 kg body mass.

Tøien et al. also observed that when the bears emerged from their dens in mid- April, they had a normal body temperature of 36.6°C. Yet, they maintained a low meta- bolic rate that was 47% below their BMR, and it took several weeks for it to rise to that of the active season (2.76 ml O2 g−1 hour−1). It is generally assumed that BMR is a species- specific constant that is necessary to maintain the vital physiological functions of an endo- thermic mammal resting at thermoneutrality. The findings of Tøien et al. show that bears can maintain their vital functions with a met- abolic rate that is reduced to nearly half of that normally required in an active state, indi- cating that BMR is not a constant but a physi- ologically controlled variable.

Transition into the torpid state includes three processes. Thermoregulatory heat pro- duction (by shivering or nonshivering thermo- genesis) is inhibited because thermoregula- tion is adjusted to a lower body temperature. Metabolic rate is depressed below the BMR at normothermic body temperature (active metabolic inhibition). This inhibition can be assisted by temperature effects on metabolic rate. Tøien et al. make the surprising finding that in hibernating bears, metabolic depres-sion is largely achieved by active metabolic inhibition, whereas temperature effects play only a minor role. In small mammals that hibernate, active inhibition and temperature- related metabolic depression, on average, may each be responsible for about 50% of total metabolic depression (3, 7, 8).

The molecular mechanisms and biochemi- cal pathways that underlie metabolic adjust- ment in torpor are still unclear. In general, torpor metabolism involves inhibition of pro- cesses that generate adenosine 5′-triphosphate such as glycolysis (metabolism is rerouted to lipid utilization instead) and mitochondrial respiration, as well as energy-consuming pro- cesses such as transcription, translation, and protein degradation (9–11). This ultimately impairs cell proliferation and differentiation. However, entrance into torpor also requires increased expression of hibernation-specific genes to support lipid metabolism, gluconeo- genesis, cytoprotection, and other measures required to maintain cells (12, 13).

In most mammalian orders, one or several species use torpid metabolic depression. The greatest numbers are found among marsupi- als, rodents, and bats, but also in small num- bers in insectivores, primates, and elephant shrews; and it is likely that more such exam- ples will be discovered in large mammals (14). Although long considered an adaptation to cold, hibernation is also found in tropical animals and desert species, and, as in bears, can occur without substantial drops in body temperature. Perhaps we will find that a hypometabolic state is the primary means by which most, if not all mammals, can reduce their energy expenditures for prolonged peri- ods of time.

Hi, here is the full answer for these questions, can you help me to check if they are all correct or not, if there is any wrong, please point out and also let me know if the rest are correct or not, please.

1. Remember the label from Garden Soil Inoculant said:

"Helps fix nitrogen in soil. Boost pea and bean yields with Garden Soil Inoculant. Simply shake granules into furrows as you plant. It contains nitrogen-fixing bacteria to promote stronger root systems and larger harvests. Easy-to-use sprinkler top makes application a snap. 8.7-oz. container treats a 150' row."

This product is recommended for growing plants such as beans and peas, for all of the following reasons EXCEPT:

a. Plants such as beans and peas can use nitrogen gas as their nitrogen source.

b. Plants such as beans and peas produce root nodules that harbor bacteria.

c. Plants such as beans and peas form symbiotic relationships with the bacteria in the inoculant product.

d. Plants such as beans and peas are legumes.

2. Molecular nitrogen (nitrogen gas) makes up 78% of the Earth’s atmosphere, but is unavailable for use by all organisms except for some nitrogen-fixing ____. (Choose all correct answers.)

a. bacteria b. eukaryotes c. protists d. fungi e. cyanobacteria f. prokaryotes

3. Nitrogen gas consists of two nitrogen atoms bound to one another by (number) _______(three) covalent bonds. A great deal of energy is required to break these bonds, which is why nitrogen gas is a chemically inert molecule. The organisms that can fix-nitrogen produce the enzyme _______(Mo-nitrogenase). This enzyme breaks the bonds that hold the two nitrogen atoms together.

4. In biological nitrogen fixation, nitrogen gas (N2) becomes _______[oxidized or reduced] when it accepts _______[Hydrogen(H2)/nitrogen(N2)/oxygen(O2)] atoms, resulting in the production of _______[nitrate(NO3)/ammonia(NH3)/water(H2O)/nitrogen gas(N2)].

5. Legumes produce ________(oxygen/leghemoglobin/nitrogen/nitrogenase/bacteria), which regulates the amount of ______(oxygen/leghemoglobin/nitrogen/nitrogenase/bacteria) present in the root nodule so _________(oxygen/leghemoglobin/nitrogen/nitrogenase/bacteria) activity is not inhibited.

6. Humans and other animals acquire their nitrogen from which of the following? (Choose all correct answers.)

a. plant protein b. protein from legumes c. protein from animals that eat plants d. protein from animals that eat other animals

7. Humans obtain nitrogen from _________(inorganic or organic) sources. Plants can obtain nitrogen from ammonia, which is an _________(inorganic or organic) source.

1. At Thanksgiving dinner, your great aunt refuses to eat with plastic silverware and proudly exclaims that she has discarded all plastics because “they cause cancer.” You wonder if chemicals you commonly use are mutagenic. You recall your time in BY 331, Genetics, and realize that you can use the Ames test to test for reversion of a point mutation in the HisG gene in the bacteria S. typhimurium. You added the chemical you wish to test into the growth medium for the bacteria. You carefully plated an equal number of cells on each mutagenesis plate. You also calculated percent survival for chemical treated cells relative to untreated cells. Remember that the plate media for survival is not selective. A summary of the results is shown below.

A. What medium must be used in the selective plate as part of the Ames test? Explain how a mutation gives rise to a revertant in this experiment.

[Type your answer here, use as much space as necessary to completely answer the question]

B. You are initially surprised to see revertants in the absence of any chemical that you are testing, but you realize that this is normal. Explain a specific way in which a revertant can arise in the absence of an added mutagen.

[Type your answer here, use as much space as necessary to completely answer the question]

C. Which chemicals would you identify as containing a mutagen? Explain your reasoning.

[Type your answer here, use as much space as necessary to completely answer the question]

D. Which chemicals would you identify as possibly antimutagenic? Explain your reasoning.

[Type your answer here, use as much space as necessary to completely answer the question]

1. At Thanksgiving dinner, your great aunt refuses to eat with plastic silverware and proudly exclaims that she has discarded all plastics because “they cause cancer.” You wonder if chemicals you commonly use are mutagenic. You recall your time in BY 331, Genetics, and realize that you can use the Ames test to test for reversion of a point mutation in the HisG gene in the bacteria S. typhimurium. You added the chemical you wish to test into the growth medium for the bacteria. You carefully plated an equal number of cells on each mutagenesis plate. You also calculated percent survival for chemical treated cells relative to untreated cells. Remember that the plate media for survival is not selective. A summary of the results is shown below.

A. What medium must be used in the selective plate as part of the Ames test? Explain how a mutation gives rise to a revertant in this experiment.

[Type your answer here, use as much space as necessary to completely answer the question]

B. You are initially surprised to see revertants in the absence of any chemical that you are testing, but you realize that this is normal. Explain a specific way in which a revertant can arise in the absence of an added mutagen.

[Type your answer here, use as much space as necessary to completely answer the question]

C. Which chemicals would you identify as containing a mutagen? Explain your reasoning.

[Type your answer here, use as much space as necessary to completely answer the question]

D. Which chemicals would you identify as possibly antimutagenic? Explain your reasoning.

[Type your answer here, use as much space as necessary to completely answer the question]

In your own words.

3. Two new species of plants were introduced into an area that was low in available nitrogen compounds in the soil. One formed nodules and one did not. Which species would you reasonably expect to grow more robustly? Explain.

Answer:

4. How does the location of the nodules relate to the function of the nodule? Explain.

Answer:

5. This diagram shows two groups of plant cells, one with Rhizobium and one without the bacteria. Label which is which, and explain the obvious size difference, using observations from Part C of the lab in your answer.

Answer: