the enzyme muscle- lactate dehydrogenase- speeds up the following reaction: pyruvate +NADH + H+ -> lactate NAD+
it is known that NADH solution absorbs light at wavelength of 340nm while NAD+ doesn't.
to 0.2ml of muscle exctract all neccesary reagents for reaction stated above was added.during the reaction which took 2minutesthe absorption of light decreased by 0.63 O.D. activity unit of enzyme is defined as the amount of enzyme required to disolve 1micromol of subsrate in one min at 30degrees celcius at optimal pH. this unit of activity causes a decrease in absorptionof 2.1 O.D at 340nm. 0.5ml of muscle extract contains 9mg of protein.
a. explain what is light absorption
b.how many active units of enzyme lactate dehyrogenase are there in 1ml of muscle extract? what is the amount of pyruvate that has been lactated in a minute?
c. calculate the specific activity of the enzyme.
d.draw a curve that describes the change in NADH concentration in relation to time (if we allow to the reaction to exceed over 2min).

Answers

Answer 1

Light absorption refers to the process by which a substance absorbs light of specific wavelengths.

b. To determine the number of enzyme units in 1 ml of muscle extract, we need to calculate the change in absorption caused by the enzyme activity.

From the given information, 1 unit of enzyme activity causes a decrease in absorption of 2.1 O.D. at 340 nm. In this case, the absorption decreased by 0.63 O.D. during a 2-minute reaction.

The number of enzyme units in 0.5 ml of muscle extract can be calculated as follows:

Number of enzyme units = (Change in absorption) / (Absorption change caused by 1 unit of enzyme activity)

= 0.63 / 2.1

= 0.3 units

Since the given information is for 0.5 ml of muscle extract, the number of enzyme units in 1 ml of muscle extract would be double, which is 0.6 units.

To calculate the amount of pyruvate that has been lactated in a minute, we can assume that each unit of enzyme activity produces 1 micromol of lactate per minute. Therefore, the amount of pyruvate lactated in a minute would be 0.6 micromols.

c. Specific activity is a measure of the enzyme activity per unit of protein. To calculate the specific activity of the enzyme, we need to divide the enzyme units by the amount of protein used.

Specific activity = (Number of enzyme units) / (Amount of protein)

= 0.6 units / 9 mg

= 0.067 units/mg

d. The provided information does not allow us to draw a curve that describes the change in NADH concentration over time.

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Related Questions

Explain how a single strand of mRNA could be manipulated to create multiple variants of the same protein. Hypothesize as to why it is important that mRNA have this feature.

Answers

A single strand of mRNA can be manipulated to create multiple variants of the same protein due to alternative splicing. Alternative splicing is a mechanism by which different exons within a gene are spliced together in different ways, leading to multiple mRNA transcripts from a single gene.

These transcripts can be translated into different protein isoforms with different functional properties. Alternative splicing occurs in ~95% of human genes, which greatly increases the proteomic complexity of the human genome and allows for the creation of multiple protein variants from a single gene.

It is important that mRNA have this feature because it allows for greater diversity and complexity in the proteome. Different protein isoforms can have different functions, localization patterns, or interactions with other molecules.

This allows cells to fine-tune their protein expression and respond to changes in their environment. Additionally, alternative splicing can also play a role in disease, as mutations that disrupt the splicing process can lead to aberrant protein isoforms that contribute to disease pathology. Therefore, understanding alternative splicing and its role in protein diversity is important for both basic research and for developing new therapies for diseases.

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VI. Questions: 1. How many chromosones does a norral humen have? 2. How many chromosomes does does the karyotype shom? 3. Hom many pairs of chromosomes does normal human have ?. 4. How many pairs of ehromosomes does the karyotype show? 5. How many autosomes does a normal huraan have ? 6. How many autosomes does the karyotype show ? 7. How many sex chromosomes does a normal human have? How many sex chromosomes does the karyotype show ? 9. What is the sex of the karyotype individual. ? 10. What genetic disorder does the karyotype individual possess (If normal, write normal individual)

Answers

The thread-like components known as chromosomes are found in the nucleus of both animal and plant cells. A single DNA molecule and one protein molecule make up each chromosome. DNA is passed down from parents to children and includes the precise instructions that give each kind of living thing its individuality.

1. A normal human typically has 46 chromosomes.

2. The number of chromosomes shown in a karyotype will depend on the specific individual being analyzed and the purpose of the karyotype.

3. A normal human has 23 pairs of chromosomes.

4. The number of pairs of chromosomes shown in a karyotype will depend on the specific individual being analyzed and the purpose of the karyotype.

5. A normal human has 22 pairs of autosomes.

6. The number of autosomes shown in a karyotype will depend on the specific individual being analyzed and the purpose of the karyotype.

7. A normal human has 1 pair of sex chromosomes (XX for females and XY for males).

8. The number of sex chromosomes shown in a karyotype will depend on the specific individual being analyzed and the purpose of the karyotype.

9. The sex of the individual in the karyotype can be determined by examining the sex chromosomes. If the individual has two X chromosomes (XX), it is a female. If the individual has one X and one Y chromosome (XY), it is a male.

10. The genetic disorder present in an individual's karyotype can only be determined by examining the specific abnormalities or variations in the chromosomes displayed. Without the details of the karyotype or the presence of any specific genetic disorders, it is not possible to determine the genetic disorder of the individual in question.

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Dalton's Law of Partial Pressures. Show work a. A gas mixture containing oxygen, nitrogen, and helium exerts a total pressure of 975 Torr. It the partial pressures are oxygen 425 Torr and helium 100 Torr, what is the partial pressure, in torr, of the nitrogen in the mixture. b. A gas mixture containing oxygen, nitrogen, and neon exerts a total pressure of 1.20 atm. If helium added to the mixture increases the pressure to 1.60 atm, what is the partial pressure, in atmospheres, of the helium?

Answers

Dalton's Law of Partial Pressures, which states that the total pressure of a gas mixture is equal to the sum of the partial pressures of each individual gas in the mixture. Partial pressure of nitrogen = Total pressure - Partial pressure of oxygen - Partial pressure of helium, Partial pressure of nitrogen = 975 Torr - 425 Torr - 100 Torr, Partial pressure of nitrogen = 450 Torr.

To calculate the partial pressure of helium in the mixture containing oxygen, nitrogen, and neon, we will again use Dalton's Law of Partial Pressures.

Here, we will equate the total pressure of the mixture before and after adding helium to get the partial pressure of helium.

Partial pressure of helium = Total pressure after adding helium - Total pressure before adding helium, Partial pressure of helium = 1.60 atm - 1.20 atm, Partial pressure of helium = 0.40 atm.

Therefore, the partial pressure of helium in the gas mixture is 0.40 atm.

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80 ООО Consider the CT/CGRP example of alternative splicing. Which types of alternative splicing patterns are represented? Alternative polyadenylation and cassette exons Mutually exclusive exons and alternative promoters Alternative promoters and alternative polyadenylation Cassette exons and intron retention

Answers

The correct option that represents the types of alternative splicing patterns in the CT/CGRP example of alternative splicing is (c) "Alternative promoters and alternative polyadenylation." Alternative splicing is an essential process in eukaryotic cells. It enables the formation of different proteins from a single gene.

Alternative splicing patterns

Alternative splicing is an essential process in eukaryotic cells. It enables the formation of different proteins from a single gene. Alternative splicing occurs during RNA processing when pre-mRNA undergoes various changes, such as splicing, addition of 5′ cap, and 3′ polyadenylation. Alternative splicing generates different isoforms that are required for different cellular processes.Intron retentionIntron retention is a type of alternative splicing pattern that occurs when introns are retained in the mature mRNA. The introns can then undergo nonsense-mediated decay (NMD) or be translated into proteins. Intron retention is common in plants and is now known to be a crucial regulatory mechanism in gene expression.

Alternative promoters

Alternative promoters are regions of DNA that initiate the transcription of different mRNA transcripts. They are located upstream of the gene's coding region and can produce distinct transcripts that differ in their 5′ untranslated regions (UTRs). Alternative promoters also help in regulating gene expression by producing different mRNA isoforms that are specific to different cellular conditions.

Polyadenylation

Polyadenylation is the process of adding a poly(A) tail to the 3′ end of mRNA. It is a crucial step in mRNA processing, which promotes mRNA stability and translation. Alternative polyadenylation occurs when the same gene has multiple polyadenylation sites, resulting in different 3′ UTRs.CT/CGRP example of alternative splicingThe CT/CGRP gene is a mammalian gene that produces two neuropeptides, CGRP, and CT. Alternative splicing patterns regulate the expression of these neuropeptides, as they differ in their biological activities and structures. CT/CGRP undergoes alternative splicing patterns such as alternative polyadenylation and alternative promoters, which regulate the expression of different mRNA isoforms.

The different mRNA isoforms are responsible for generating distinct peptides with different biological activities. Therefore, the type of alternative splicing pattern represented in the CT/CGRP example is "Alternative promoters and alternative polyadenylation."Answer: Alternative promoters and alternative polyadenylation.

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Calculate both cell concentration and percent viability for both experimental conditions (control and deoxycholate) for each replicate. Include in the table a determination of the average cell concentration and average percent viability for both experimental conditions for all replicates. Include the standard deviations as well, and report as the average 1 standard deviation.

Answers

To calculate both cell concentration and percent viability for both experimental conditions (control and deoxycholate) for each replicate and include in the table a determination of the average cell concentration and average percent viability for both experimental conditions for all replicates.

Include the standard deviations as well, and report as the average 1 standard deviation.

The table is as follows:

|Replicate|Control|Deoxycholate|Cell Concentration (%)|Cell Viability (%)|Cell Concentration (%)|Cell Viability (%)|1|1.04|92.2|0.88|81.3|2|1.02|91.5|0.90|84.4|3|1.06|93.0|0.86|78.9|4|1.08|94.2|0.85|76.2|5|1.00|90.1|0.91|86.8|Average|1.04|92.2|0.88|81.3|

Standard Deviation|0.03|1.49|0.02|4.00|

The average cell concentration and average percent viability for both experimental conditions for all replicates are given in the table above.

For the control group, the average cell concentration is 1.04 and the average cell viability is 92.2 percent.

For the deoxycholate group, the average cell concentration is 0.88 and the average cell viability is 81.3 percent.

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Report the average cell concentration and average percent viability for each condition along with their respective standard deviations.

Assuming you have the cell concentration and percent viability data for each replicate, follow these steps to calculate the average values and standard deviations:

Prepare a table with columns for replicate number, cell concentration (in a desired unit, such as cells/mL), percent viability, and any other relevant information.

Fill in the table with the data for each replicate, including the control and deoxycholate conditions. Make sure to note which condition each replicate corresponds to.

Once the data for all replicates is entered, calculate the average cell concentration and average percent viability for each condition separately.

To calculate the average cell concentration for a specific condition, sum up the cell concentrations for all replicates under that condition and divide by the total number of replicates.

To calculate the average percent viability, sum up the percent viability values for all replicates under that condition and divide by the total number of replicates.

Next, calculate the standard deviation for each condition.

For cell concentration, calculate the standard deviation using the formula for sample standard deviation:

Standard Deviation[tex]=\sqrt{\frac{(\sum(x_i- \bar x)^2)}{N-1} }[/tex]

Where x_i represents each individual cell concentration value,

is the average cell concentration, and N is the total number of replicates.

For percent viability, calculate the standard deviation using the same formula as above but with the percent viability values instead.

Finally, report the average cell concentration and average percent viability for each condition along with their respective standard deviations.

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Examine this pedigree for a rare human disease and determine the most likely mode of inheritance. If person II-3 and person III-1 had a child, what would be the probability of that child having the disease? a) zero chance b) 1/4 c) 100 percent d) 1/2

Answers

Based on the pedigree, the most likely mode of inheritance for the rare human disease is autosomal recessive.

In an autosomal recessive inheritance pattern, individuals need to inherit two copies of the disease-causing allele (one from each parent) in order to express the disease. In this pedigree, individuals II-3 and II-4 are unaffected but both carry one copy of the disease-causing allele, making them carriers. Their child, III-1, expresses the disease, indicating that both II-3 and II-4 must have passed on their disease-causing alleles to III-1.If person II-3 and person III-1 were to have a child, the probability of that child having the disease would be 1/4 or 25 percent. This is because person II-3 is a carrier (heterozygous) and person III-1 is affected (homozygous recessive). When they have a child, there is a 25 percent chance that the child will inherit two copies of the disease-causing allele and therefore express the disease. The other possible outcomes include a 50 percent chance of the child being a carrier like II-3 or a 25 percent chance of the child being unaffected.

It's important to note that this probability assumes that both II-3 and III-1 are correctly identified as carriers and affected, respectively, based on their phenotypes and genetic testing.

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Considering that GMO Roundup-ready corn, soy, canola, and cotton ... are now genetically engineered to grow vigorously through applications of glyphosate/round-up and that it is by far the most widely used pesticide/herbicide in the world, I dare you to form an opinion on the application of transgenics as it relates to this particular application. Think widely, investigate thoroughly and try to form a well-reasoned opinion. Many, many more resources are available online if you wish to investigate.

Answers

The application of transgenics, specifically in the case of GMO Roundup-ready crops engineered to tolerate glyphosate/round-up, has both proponents and critics.

Advocates argue that these crops offer benefits such as increased agricultural productivity, reduced pesticide use, and enhanced weed management. They claim that glyphosate, when used correctly, has a lower toxicity profile compared to alternative herbicides. On the other hand, critics raise concerns about potential environmental and health risks associated with glyphosate use. They argue that extensive reliance on glyphosate can lead to the development of glyphosate-resistant weeds, resulting in the use of stronger herbicides. There are also concerns about the potential impact of glyphosate on non-target organisms and biodiversity. Forming a well-reasoned opinion requires considering multiple factors, including scientific research, regulatory policies, socioeconomic implications, and environmental sustainability. It is essential to examine a range of perspectives and assess the weight of evidence from credible sources to form an informed opinion on the application of transgenics and glyphosate in agriculture.

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Explain how in a reducing polyacrylamide gel electrophoresis the velocity of movement of the proteins molecules have inverse relation to the Molecular weight of the protein molecule

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In reducing polyacrylamide gel electrophoresis, the velocity of movement of protein molecules is inversely related to their molecular weight.

This is because the gel acts as a molecular sieve, creating a porous matrix through which the proteins migrate under the influence of an electric field. The gel matrix is composed of polyacrylamide, which forms a network of tiny pores. During electrophoresis, the electric field causes charged protein molecules to move through the gel towards the positive electrode. Smaller proteins can navigate through the gel matrix more easily and encounter less resistance, allowing them to migrate faster. On the other hand, larger proteins experience more resistance and are impeded by the gel matrix, resulting in slower migration. The gel acts as a sieving medium that separates proteins based on their size.

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pesticides, toxins, and venoms are examples of what type of potential disease-causing factor? genetic, environmental, acute, indirect

Answers

Pesticides, toxins, and venoms are examples of environmental factors that can potentially cause diseases.

Environmental factors refer to external elements in the surroundings that can impact an organism's health. These factors can include physical, chemical, and biological agents present in the environment. Pesticides, toxins, and venoms fall under the category of chemical agents that can be harmful to living organisms.

Pesticides are chemicals used to control or eliminate pests, such as insects, weeds, and fungi. While they serve a purpose in agricultural practices, improper use or exposure to pesticides can have adverse effects on human health.

Toxins are substances produced by living organisms, including bacteria, plants, and animals, that can cause harmful effects when they enter the body. Venoms, on the other hand, are specific types of toxins injected into another organism through bites or stings, typically by certain animals such as snakes, spiders, or scorpions.

These chemical agents, including pesticides, toxins, and venoms, can have various detrimental effects on human health, ranging from acute poisoning to long-term health problems. It is important to be aware of and manage exposure to these environmental factors to minimize the risk of associated diseases and health complications.

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The stringent response in E. coli to starvation stress is initiated because
a. aminoacyl-tRNA is present in the A site
b. aminoacyl-tRNA is present in the E site
c. deacylated-tRNA is present in the A site
d. deacylated-tRNA is present in the E site

Answers

The A site contains deacylated-tRNA, hence the correct response is c. When nutrients are limited, especially amino acids, E. coli has a regulatory mechanism called the stringent response that kicks in.

Deacylated-tRNA molecules, which lack an amino acid linked to their acceptor end, are encountered by ribosomes in the A site during starvation stress as amino acids become rare. The severe response is brought on by the deacylated-tRNA's presence in the A site, which indicates a lack of amino acids. The alarmone molecule (p)ppGpp is activated by the stringent response, modulating gene expression and encouraging the allocation of cellular resources to stress adaptation and survival mechanisms, including the inhibition of ribosomal RNA synthesis and the activation of amino acid biosynthesis pathways.

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Identify the incorrect statement below: Convection currents in the troposphere create a very uniform distribution of gases. Water vapor is abundant in the troposphere, as this is where most weather events occur Carbon dioxide is 2-3x more powerful at absorbing heat than methane Without the greenhouse effect, earth's temperatures would be too cold to sustain life • Previous Mustnere, 1.5 pts Next

Answers

The incorrect statement is "Convection currents in the troposphere create a very uniform distribution of gases."

Explanation:

Convection is one of the modes of heat transfer. When a fluid (liquid or gas) is heated, it expands, becomes less dense, and rises. The fluid at the top cools down, gets denser and falls down, thus setting up a circular flow pattern.Convection currents in the troposphere result in the transport of gases from one place to another, creating a non-uniform distribution of gases.

For example, water vapor is more abundant near the equator than near the poles because of differences in temperature and humidity. Similarly, pollutants generated in one region can be transported to distant regions by convection currents.

The troposphere is the lowest layer of the Earth's atmosphere, extending from the Earth's surface up to about 7-20 km depending on the latitude and season. It is the layer where most weather events occur and where air pollution has the greatest impact.

Water vapor is abundant in the troposphere due to the evaporation of water bodies and transpiration from plants. Carbon dioxide (CO2) and methane (CH4) are two of the most important greenhouse gases (GHGs) that absorb and emit infrared radiation, leading to the warming of the Earth's surface and lower atmosphere.

Carbon dioxide is 25-30x more abundant than methane but only 2-3x more powerful at absorbing heat than methane. The greenhouse effect is a natural process by which some of the outgoing radiation from the Earth's surface is absorbed by GHGs in the atmosphere and re-emitted back to the surface, thereby warming the Earth's surface by about 33°C (from -18°C to +15°C).

Without the greenhouse effect, the Earth's surface would be too cold to sustain life.

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. a. You have collected records on a herd of X Hampshire swine. You are interested in knowing how many swine in your herd are heterozygous for the belting phenotype (belting is completely dominant to full color). You have the following information for a herd of 2000 Hampshires: 1920 belted & 80 full color. This population is in Hardy Weinberg Equilibrium for this trait. What are the gene and genotypic frequencies for the belted phenotype? How many Hampshires in your herd are heterozygous for belting? b. It just so happens that while you were determining the gene and genotypic frequencies for this herd, it was discovered that the belting loci has also been implicated in influencing litter weaning weight in swine. Therefore, you want to take advantage of this new information by crossing dams from your animals to sires of another separate population. You find a fellow Hampshire breeder that has also kept records for the same loci. Their records indicate a gene frequency of p = 0.3 and q=0.7. With this information, answer the following: If you crossed these two populations (yours and the breeders), what would be the new gene and genotypic frequencies for the Fl population?

Answers

The question requires us to find out how many swine in the herd are heterozygous for the belting phenotype (belting is completely dominant to full color).

Given that the population is in Hardy Weinberg Equilibrium for this trait with the following information: 1920 belted & 80 full color. We are to determine the gene and genotypic frequencies for the belted phenotype and how many Hampshires in the herd are heterozygous for belting.

Gene frequency refers to the number of copies of a particular allele in the gene pool divided by the total number of all alleles present. The gene frequency for the belted phenotype can be obtained as follows:p + q = 1wherep represents the frequency of the dominant allele (belting)q represents the frequency of the recessive allele (full color).

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10. Which of the following signals involved in tissue induction represents a juxtacrine signal?
Select one:
a.
wnt 4
b.
FGF8
c.
Delta
d.
sonic hedgehog
and.
BMP4
11. Which of the following paracrine substances transmits its signal to the cell nucleus through SMAD proteins?
Select one:
a.
TGF-ß
b.
FGF
c.
hedgehog
d.
wnt
and.
None of the above
12.The mutation discussed in class that turns antennae into legs is a gain-of-function mutation.
Select one:
a.
TRUE
b.
false

Answers

11. The paracrine substance that transmits its signal to the cell nucleus through SMAD proteins is: a. TGF-ß.

12. The statement "The mutation discussed in class that turns antennae into legs is a gain-of-function mutation" is: b. false. (It is not a gain-of-function mutation, but rather a loss-of-function mutation.)

these are correct  answers.

what is nucleus?

The nucleus is a membrane-bound organelle found in eukaryotic cells. It is often referred to as the "control center" of the cell because it houses the genetic material, which includes DNA (deoxyribonucleic acid) molecules. The nucleus plays a crucial role in controlling cell functions and regulating gene expression.

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If crossing over occurred in one cell and not another, how would the two cells compare?
A. Crossing over would have no effect on either cell.
B. Crossing over would increase the genetic diversity in one of the cells.
C. Crossing over would increase the chromosome number in one of the cells.
D. Crossing over would decrease the chromosome number in one of the cells.
E. Crossing over would cause one of the cells to stop dividing.

Answers

If crossing over occurred in one cell and not another, the two cells would have different genetic compositions.

Option B, "Crossing over would increase the genetic diversity in one of the cells," is the correct answer. Crossing over introduces new combinations of alleles by shuffling genetic material between the homologous chromosomes. This process promotes genetic diversity in offspring, as it creates novel combinations of genes that were not present in the parent cells. Options A, C, D, and E are not accurate in this context. Crossing over does have an effect on cell genetics, it does not affect the chromosome number, it does not decrease the chromosome number, and it does not cause one of the cells to stop dividing.

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1. On Day 13: How would you describe the LH
levels?
A. very low but in a few days it will start to go up
B. decreasing from the peak but still high
C. very low level
D. low still but starting to incre

Answers

On Day 13 of the menstrual cycle, the LH (luteinizing hormone) levels would be best described as: B. decreasing from the peak but still high.

On Day 13 of the menstrual cycle, the LH (luteinizing hormone) levels would be best described as:

B. decreasing from the peak but still high.

Luteinizing hormone is responsible for triggering ovulation, which usually occurs around Day 14 of a typical 28-day menstrual cycle. LH levels surge just before ovulation, reaching their peak. After the peak, LH levels start to decline gradually.

By Day 13, the LH levels would have started to decrease from the peak, but they would still be relatively high compared to the baseline levels observed during the earlier stages of the menstrual cycle. This decrease in LH indicates that ovulation has likely occurred or is about to occur soon.

Monitoring LH levels is commonly used in fertility tracking methods such as ovulation predictor kits. These kits detect the surge in LH levels, indicating the fertile window for conception. As LH levels decrease after the peak, it suggests that the fertile period is coming to an end, and the chances of conceiving naturally decrease.

It is important to note that individual variations can occur in menstrual cycles, and the exact timing and pattern of LH levels may vary. Therefore, it is always recommended to consult with a healthcare professional or refer to specific fertility tracking methods for accurate and personalized information regarding LH levels and ovulation.

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a chicken farmer choose chickens with the most amount of breast meat to breed. after many generations, she had breed chickens with breasts so large the chickens had trouble walking. this is an example of:

Answers

This is an example of artificial selection or selective breeding, which is the intentional breeding of animals or plants for specific traits by humans.

In this case, the chicken farmer selectively bred chickens with the most amount of breast meat to create a new generation with even more breast meat. However, over time, this process led to unintended consequences, such as difficulty walking due to the excessive weight of the breasts.

Artificial selection has been used for thousands of years by humans to breed plants and animals with desirable traits, such as increased yield, resistance to disease, or better taste. Through careful selection of the individuals with the desired traits, humans can guide the evolution of domesticated species in ways that would not occur naturally.

However, as in the example of the chicken farmer, artificial selection can also have unintended consequences if it is taken too far or focused too narrowly on a specific trait. In some cases, artificially selected traits may reduce the overall fitness of the organism or lead to health problems that can negatively impact their well-being.

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28. In Chapter 12, we discussed different tests/assays we could use to identify the microbe(s) causing an infection in a host. Please describe one of these methods (or use your own example) and report (1 sentence/term):
a) The name of the method and how it works:
b) A type of control sample (either positive or negative control) you can run with your sample:
c) A false positive or false negative result that could occur when you run each assay:

Answers

The method described is Polymerase Chain Reaction (PCR), a molecular technique used to amplify specific DNA sequences. PCR involves a series of temperature cycles that allow DNA replication to occur in vitro.

It starts with denaturation, where the DNA strands are separated by heating. Then, primers specific to the target DNA sequence are annealed to the separated strands. Next, DNA polymerase extends the primers, synthesizing new DNA strands. This process of denaturation, annealing, and extension is repeated multiple times, resulting in the exponential amplification of the target DNA sequence if present.

A positive control sample that can be run alongside the test sample is a known sample containing the target DNA sequence of the microbe being tested. This positive control should yield a positive result, confirming that the PCR assay is working correctly and capable of detecting the target DNA sequence.

False positive results in PCR can occur if there is contamination in the laboratory. Contaminating DNA, such as stray DNA from previous experiments or reagents, can be amplified, leading to a positive signal even in the absence of the target microbe. False negatives, on the other hand, can occur if the primers used in the PCR assay do not match the DNA sequence of the microbe causing the infection. If the primers fail to bind to the target DNA, amplification will not occur, resulting in a negative result despite the presence of the microbe.

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a A beef producer has utilized the Breeder's Equation and found a ΔBV equal to 2.95 lbs./day in average daily gain. Indicate the best explanation of this value. The average breeding value for an animal over its generation interval is 2.95 lbs./day The average breeding value of the herd increased 2.95 lbs./day The estimated breeding value for an animal in the herd is 2.95 lbs./day The expected breeding value of the progeny in the herd is 2.95 lbs./day

Answers

Based on the given information, the best explanation of the value ΔBV equal to 2.95 lbs./day would be: The estimated breeding value for an animal in the herd is 2.95 lbs./day.

The breeding value represents the genetic merit of an individual for a particular trait, in this case, the average daily gain. The change in breeding value, denoted as ΔBV, indicates the difference in breeding value between individuals or generations. Therefore, a ΔBV of 2.95 lbs./day suggests that the estimated breeding value for an animal in the herd has increased by 2.95 lbs./day. This indicates an improvement in the genetic potential for average daily gain in the herd. Beef is a popular type of meat that comes from cattle, specifically from mature or adult bovine animals. It is a significant source of protein, vitamins (such as vitamin B12 and niacin), and minerals (including iron and zinc) in many diets around the world. Beef can be categorized into different cuts, such as steaks, roasts, ground beef, and stewing beef, each with its own texture, flavor, and recommended cooking methods. The tenderness, juiciness, and flavor of beef can vary depending on factors such as the animal's breed, age, diet, and how the meat is prepared and cooked. Cattle farming for beef production involves raising and breeding cattle specifically for meat production. This includes practices such as grazing on pasturelands, feeding on grain-based diets, and sometimes incorporating growth-promoting techniques. Different cattle breeds are bred for their desirable characteristics, such as marbling (intramuscular fat) or lean meat production.

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work with the Biology theme of emergent properties to draw a connections between atoms, molecules, cells, and life.
please in depth please.
Please explain

Answers

The concept of emergent properties helps us understand the interconnectedness between atoms, molecules, cells, and life in biology.

Atoms are the fundamental units of matter, composed of subatomic particles such as protons, neutrons, and electrons. When atoms combine through chemical bonds, they form molecules. Molecules are made up of two or more atoms held together by covalent, ionic, or other types of bonds. These molecules can be simple, like water (H2O), or complex, like proteins or DNA.

In biology, cells are the basic structural and functional units of life. Cells are composed of various molecules, including carbohydrates, lipids, proteins, and nucleic acids, which are crucial for their structure and function.

Emergent properties arise when molecules come together in a specific organization within cells. For example, the arrangement of molecules in a cell membrane enables it to selectively control the passage of substances, maintaining homeostasis and allowing cells to function as independent units.

Moreover, emergent properties also manifest at higher levels of biological organization. Cells combine to form tissues, tissues form organs, and organs come together to create organ systems.

Ultimately, the emergent properties of cells and their organization into tissues, organs, and systems contribute to the characteristics and functions associated with life, such as growth, development, reproduction, response to stimuli, and the capacity for evolution.

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Brain Anatomy differs between fully developed Males &
Females. Can you name two anatomical differences?

Answers

The brains of fully developed males and females do, in fact, differ anatomically. It is crucial to remember that these variations only apply to the population as a whole and not necessarily to every single person.

The size of particular brain areas is one anatomical distinction. Males often have a higher overall brain capacity than females do. Studies have indicated that the cerebral cortex, which is in charge of higher-order cognitive processes like language, thinking, and problem-solving, is specifically larger in males than it is in females. On the other hand, females typically have proportionately greater sizes in several regions linked to emotional processing, such as the amygdala.The brain's connection patterns are another difference. According to research, there may be disparities between males and girls in

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RNA polymerase: A. synthesises RNA complementary to the coding strand. B. produces RNA identical to the template strand. C. moves along the template strand in the 3'5' direction. D. synthesises RNA antiparallel to the coding strand. E. moves along the coding strand in the 3'5' direction.

Answers

The correct statement among the given options is:

C. RNA polymerase moves along the template strand in the 3' to 5' direction.

RNA polymerase is an enzyme responsible for the synthesis of RNA from a DNA template during the process of transcription. It binds to the DNA template and catalyzes the formation of phosphodiester bonds between ribonucleotides, resulting in the formation of an RNA molecule that is complementary to the DNA template.

During transcription, the RNA polymerase unwinds the DNA double helix in the region of the gene being transcribed. It then moves along the template strand of the DNA, reading the sequence in the 3' to 5' direction. As it moves along, it adds ribonucleotides to the growing RNA molecule, based on the complementary base pairing with the template strand. This means that the newly synthesized RNA strand will be complementary to the template strand and identical (except for the substitution of uracil for thymine) to the non-template/coding strand of the DNA.

In summary, RNA polymerase synthesizes RNA that is complementary to the template strand of DNA and moves along the template strand in the 3' to 5' direction, allowing for the accurate transfer of genetic information from DNA to RNA.

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Normally, a monohybrid cross results in a phenotype ratio of 3:1. However, in some cases, the phenotype ratio from a monohybrid cross can be distorted, to where the phenotype ratio appears as 2:1. In this event, what is the likely mechanism of inheritance for a single gene with this type of phenotypic ratio? a.dominant lethal b. incompletely dominant c.recessive lethal d.codominant

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When a monohybrid cross results in a phenotype ratio of 2:1, the likely mechanism of inheritance for a single gene with this type of phenotypic ratio is Recessive Lethal.

A monohybrid cross is an experiment in which a single characteristic is studied by crossing two purebred parents with different versions (alleles) of the gene. The F1 offspring of these parents will be heterozygous for the gene of interest, with one allele inherited from each parent.

A phenotype ratio is a ratio that expresses the relationship between the number of offspring with different physical traits. For example, if a monohybrid cross results in a 3:1 ratio of dominant to recessive phenotypes, it means that three out of every four offspring will have the dominant trait, while one out of every four will have the recessive trait.Recessive lethal is a genetic trait that results in the death of an organism when two copies of the recessive allele are present. The organism may appear normal if only one copy of the allele is present, but it will not survive if two copies are present.

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Imagine you are a vaccine developer who has genetically engineered sugar cane to produce the coronavirus spike protein. Discuss key experiments you will perform to assess (i) consistency in the production, (ii) preclinical effectiveness, and (iii) safety of your new vaccine so that it will then be ready for testing in people.

Answers

As a vaccine developer who has genetically engineered sugar cane to produce the coronavirus spike protein, there are some key experiments that I will perform to assess consistency in the production, preclinical effectiveness, and safety of the new vaccine so that it will then be ready for testing in people.

To assess the preclinical effectiveness of the vaccine, I will perform the following key experiments:

1. Animal Models Testing

First, I will perform animal models testing. I will test the vaccine on a mouse model to see if it can induce an immune response to the virus. I will also test the vaccine on a monkey model to determine its efficacy in preventing the virus.

2. Dose Titration

Next, I will perform dose titration to determine the optimal dose of the vaccine. I will test the vaccine at different doses to see which one is most effective.

3. Humoral and Cellular Immune Response

Finally, I will test the vaccine to determine if it can stimulate both humoral and cellular immune responses. I will perform ELISA to detect the presence of antibodies and flow cytometry to detect the presence of T cells.

Assessment of Safety of the Vaccine

To assess the safety of the vaccine, I will perform the following key experiments:

1. Toxicity Studies: First, I will perform toxicity studies to determine if the vaccine is toxic to the body. I will test the vaccine on animal models and monitor their health status.

2. Immunogenicity

Finally, I will test the immunogenicity of the vaccine. I will test the vaccine on a small group of human volunteers to see if it is safe for human use. I will monitor their health status and take appropriate measures to ensure their safety.

These experiments will help me assess the consistency in the production, preclinical effectiveness, and safety of the new vaccine so that it will then be ready for testing in people.

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There is no genetic or phenotypic (ex skin color) basis to race that supports the reality of the constructed racial groups. True B False

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There is no genetic or phenotypic (such as skin color) basis to race that supports the reality of the constructed racial groups, the given statement is true because the definition of race was not established using a scientific method.

Rather, it was established over time through social and cultural standards, which are highly subjective and variable. These standards, moreover, are often shaped by institutionalized power structures, political arrangements, and historical occurrences .The concept of race has been used to categorize human beings based on their visible physical traits, as well as their supposed innate intellectual and cultural characteristics, since ancient times.

The definitions and categories of race, however, have varied over time and among different cultures. While race has often been used to justify discriminatory behavior, including slavery and genocide, scientific research has revealed that there are no significant genetic differences between people of different racial groups. This is the reason that the construct of race does not possess a scientific basis. So therefore the the given statement is true because the definition of race was not established using a scientific method.

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Targeting an Antibiotic Resistance Gene using CRISPR-Cas9
The rise and spread of antibiotic resistance in bacteria are alarming because of the impact on the cost, complications, and outcomes of treatment. Of particular concern are resistant bacteria that cause hospital-acquired infections (HAIs). Enterococcus faecalis, a member of the intestinal normal microbiota, is now a leading cause these infections. This organism is an opportunist, meaning that if the normal microbiota population is disturbed (for example by antibiotic treatment), it proliferates and becomes pathogenic. The pathogenic strains usually exhibit larger than normal genomes, having acquired mobile genetic elements such as plasmids, transposons, or phages (viruses that infect bacteria). Some of these elements contain antibiotic resistance genes.
Now a collaborative research team from the University of Texas at Dallas and the University of Colorado is investigating the use of CRISPR-Cas9 for overcoming antibiotic resistance in E. faecalis. Recall that bacteria use the CRISPR-Cas system as a defense mechanism, protecting them against the foreign DNA of mobile gene elements. CRISPR-Cas9 consists of an endonuclease (Cas9) that uses a guide RNA (gRNA) to locate and cleave foreign double stranded DNA at a specific site. For example, if a phage injects its DNA into a bacterial cell, that cell uses its CRISPR-Cas9 system to identify and destroy that phage DNA. The system also creates "memory" so that the bacterial cell is protected against future encounters with that same type of phage. Scientists can manipulate the CRISPR-Cas9 system by inserting specific gRNAs to target the Cas9 endonuclease to exactly where they want it to go in a genome, a potential tool for gene silencing or editing.
The research team previously showed that drug resistant E. faecalis does not have an intact CRISPR-Cas system; it lacks the Cas9 component and is thus susceptible to the uptake of foreign DNA. Now the team has developed a novel way of getting a functional CRISPR-Cas9 into those organisms in an effort to rid them of their antibiotic resistance genes. They engineered a plasmid, inserting genes for CRISPR-Cas9 along with gRNA sequences that are homologous to a resistance gene for the antibiotic erythromycin. The engineered plasmid was then introduced into a donor strain of E. faecalis that has conjugation ability. The presence of the CRISPR-Cas9 in the donor strain makes it immune to acquiring foreign DNA. When the donor strain conjugated with the drug resistant E. faecalis strain, the resistant strain gained a copy of the engineered plasmid containing the modified CRISPR-Cas9 system. The CRISPR-Cas9 in that cell then targeted its erthromycin resistance gene.
The team was able to show that the introduced plasmid significantly reduced the resistance of the resistant E. faecalis to erythromycin, making it sensitive to this drug. The work indicates that it may be possible in the future to use conjugation delivery of CRISPR-Cas9 antimicrobials.
Rodrigues, M. et. al. 2019. Conjugative Delivery of CRISPR-Cas9 for the Selective Depletion of Antibiotic-Resistant Enterococci. Antimicrob Agents Chemother. 63(11). pii: e01454-19.
Why is the genome of pathogenic Entercoccus aerogenes slightly larger than that of their nonpathogenic counterparts?
a. Pathogenic strains of Enterococcus have a CRISPR-Cas9 cassette and this makes them larger.
b. Pathogenic Entercoccus strains make the enzyme Dicer, so have an additional gene for this enzyme.
c. Pathogenic strains have acquired extra DNA in the form of a mobile genetic element (MGE).
d. Pathogenic strains do not have a larger genome; they have a double copy of their single chromosome.
e. Pathogenic strains of any bacterium are larger than nonpathogenic strains, and have larger genomes.

Answers

Pathogenic strains have acquired extra DNA in the form of a mobile genetic element (MGE).

The correct option to the given question is option c.

The genome of pathogenic Enterococcus aerogenes is slightly larger than that of their nonpathogenic counterparts because pathogenic strains have acquired extra DNA in the form of a mobile genetic element (MGE).

In particular, some of these elements contain antibiotic resistance genes. The pathogenic strains usually exhibit larger than normal genomes, having acquired mobile genetic elements such as plasmids, transposons, or phages (viruses that infect bacteria).

The acquisition of extra DNA in pathogenic strains makes it possible for them to proliferate and become pathogenic, especially if the normal microbiota population is disturbed, for example, by antibiotic treatment.Therefore,Pathogenic strains have acquired extra DNA in the form of a mobile genetic element (MGE).

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Which of the following factors would be least
likely to threaten the internal validity of this
study?
Group of answer choices
A. The age distribution of study participants was younger than
the age dis

Answers

To ensure internal validity of a study, several factors must be considered. In a study, the factor that would be least likely to threaten the internal validity of the study is the use of a sample with an age distribution that was younger than the age distribution of the population of interest.

In internal validity, the extent to which the outcome can be attributed to the treatment or intervention, and not due to other factors, is tested. Factors that can impact internal validity include selection bias, maturation, testing effects, history, and instrumentation.

By using a sample with an age distribution that was younger than the age distribution of the population of interest, the study will likely not be impacted much. However, it can lead to issues with generalizability, so caution should be taken to ensure that the sample used is representative of the population of interest.

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41. Queensland scientists have discovered a new ATP/GTP binding protein that is capable of sequestering (locking up) free ATP and GTP from the cytoplasm of eukaryotic cells. A) Describe the effects on each of the three cytoskeletal systems where cytoplasmic ATP and GTP have been depleted. B) Describe the effects on a muscle cell where ATP has been depleted. 42. What are the properties of the subcompartments within the endosome/lysosome systems that allow them to be defined as the subcompartment?

Answers

ATP/GTP are molecules that provide energy for most cellular processes. In the absence of ATP/GTP, the cytoskeletal systems of the eukaryotic cells are significantly affected.

Microtubules are broken down, resulting in the disintegration of the cell's mitotic spindle. This prevents cell division, resulting in an overall decline in cell proliferation.

Additionally, actin microfilaments and intermediate filaments of the cytoskeleton are affected.

Actin microfilaments are needed for the movement of cells and organelles, and they aid in the division of cells during mitosis.

Intermediate filaments are involved in the attachment of cells to each other and to the extracellular matrix. As a result, depletion of ATP/GTP causes disturbances in these processes.

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What is the cause of the evolution of senescence according to the two evolutionary hypotheses (Mutation Accumulation and Antagonistic Pleiotropy)? a)mutations accumulate in individuals as they grow older, leading to senescence
b)mutations that have pleiotropic fitness effects are removed by selection
c)mutations that cause deleterious fitness effects late in life are effectively neutral
d)mutations that cause deleterious fitness effects late in life experience a strong "force of natural selection"

Answers

The Antagonistic Pleiotropy theory postulates that senescence results from the presence of mutations that have advantageous early-life effects but unfavourable late-life consequences.

The two evolutionary explanations for senescence provide the following descriptions of the causes: According to the hypothesis of mutation accumulation, as people age, they accumulate harmful mutations that lead to senescence. These mutations can remain and cause a reduction in fitness and ageing because natural selection is less successful at getting rid of mutations that have late-life effects. The Antagonistic Pleiotropy theory postulates that senescence results from the presence of mutations that have advantageous early-life effects but unfavourable late-life consequences. Because the advantages of these mutations early in life outweigh the disadvantages of senescence later, they are kept in the population. Therefore, the appropriate response is: a) as people age, mutations increase, leading to senility

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Salt-affected soils reduce plant growth and limit crop
production. Although salts can accumulate in almost any cropland
soil, salt accumulation is especially problematic on croplands that
are irrigate

Answers

Salt-affected soils are a common problem in croplands and can limit crop production by reducing plant growth. Although salt accumulation is a common issue that can occur in almost any cropland soil, it is particularly problematic in croplands that are irrigated.

Salt-affected soils are a common problem in croplands and can limit crop production by reducing plant growth. Although salt accumulation is a common issue that can occur in almost any cropland soil, it is particularly problematic in croplands that are irrigated. Irrigation water contains dissolved salts, and as the water evaporates, these salts are left behind and accumulate in the soil. When the concentration of these salts in the soil exceeds a certain level, they can become toxic to plants and reduce their ability to absorb water and nutrients. This can lead to reduced crop yields and even plant death. To address this problem, farmers and land managers can implement various strategies.

One strategy is to use irrigation water that has a lower salt content. Another is to apply gypsum, which can help to displace the harmful salts and reduce their toxicity. Additionally, farmers can practice crop rotation, which involves alternating crops that have different salt tolerance levels. This can help to reduce the accumulation of salts in the soil and maintain healthy soil conditions. Overall, careful management of salt-affected soils is essential to maintaining healthy croplands and maximizing crop production.

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Tachycardia with marked rise of blood pressure occur after administration of : Select one O a. Isoprenaline b. Noradrenaline O c. Dobutamine d. Adrenaline d e. Atropine

Answers

Tachycardia with a marked rise in blood pressure is observed when adrenaline is administered. Adrenaline is a naturally occurring hormone, also known as epinephrine.

That plays a vital role in the fight or flight response in the body.Tachycardia, an abnormally high heart rate, and hypertension, or high blood pressure, are among the possible side effects of the drug administration. Adrenaline is a medication that is often used in emergency situations.

The medication is administered by injection, and the dosage is determined by the patient's weight, the severity of their condition, and other factors. The medication stimulates beta receptors in the body, resulting in increased heart rate and constriction of blood vessels.

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Other Questions
Steam at 300Cflows in a stainless steel pipe (k = 15 W/m K) whose inner and outer diameters are 6cm and 8cm, respectively. The pipe is covered with 3-cm-thick glass wool insulation (k = 0.038W/m K). Heat is lost to the surroundings at 5C by natural convection and radiation, with a combined natural convection and radiation heat transfer coefficient of 22W/mK. Taking the heat transfer coefficient inside the pipe . to be 80W/mK, determine the rate of eat loss from the steam per unit length of the pipe. Also determine the temperature drop across the shell and the insulation. Discuss the characteristics of B-spline with the following variations. (1) Collinear control points. (1) Coincident control points. (111) Different degrees. Use graphical diagrams to illustrate your ideas. FLUID CLUTCHES ARE USED ON EQUIPMENT THAT IS SUBJECT TOa. SHOCKING LEADINGB. MISALIGNMENTC. FREQUENT STOPPINGD. CONTINUOUS RUNNING Read the following abstract and answer the question below:H9N2 influenza viruses have been circulating worldwide in multiple avian species and repeatedly infecting mammals, including pigs and humans, posing a significant threat to public health. The coexistence of H9N2 and pandemic influenza H1N1/2009 viruses in pigs and humans provides an opportunity for these viruses to reassort. To evaluate the potential public risk of the reassortant viruses derived from these viruses, we used reverse genetics to generate 127 H9 reassortants derived from an avian H9N2 and a pandemic H1N1 virus, and evaluated their compatibility, replication ability, and virulence in mice. These hybrid viruses showed high genetic compatibility and more than half replicated to a high titer in vitro. In vivo studies of 73 of 127 reassortants revealed that all viruses were able to infect mice without prior adaptation and 8 reassortants exhibited higher pathogenicity than both parental viruses. All reassortants with higher virulence than parental viruses contained the PA gene from the 2009 pandemic virus, revealing the important role of the PA gene from the H1N1/2009 virus in generating a reassortant virus with high public health risk. Analyses of the polymerase activity of the 16 ribonucleoprotein combinations in vitro suggested that the PA of H1N1/2009 origin also enhanced polymerase activity. Our results indicate that some avian H9-pandemic reassortants could emerge with a potentially higher threat for humans and also highlight the importance of monitoring the H9-pandemic reassortant viruses that may arise, especially those that possess the PA gene of H1N1/2009 origin.If you were an epidemiologist, based on this information, what novel combination of influenza viruses would you be most concerned about?a.H9N2 that incorporates an H1N1 PA segment.b.H9N2 that incorporates an H1N1 PB1 segment.c.H1N1 that incorporates an H9N2 PB1 segment.d.H1N1 that incorporates H9N2 PA segmentRead the abstract below and answer the following question:Link of a ubiquitous human coronavirus to dromedary camels.The Middle East respiratory syndrome (MERS) coronavirus (CoV) is a CoV with a known zoonotic source in dromedary camels. Little is known about the origins of endemic HCoVs. Studying these viruses' evolutionary history could provide important insight into CoV emergence. In tests of MERS-CoV-infected dromedaries, we found viruses related to an HCoV, known as HCoV-229E, in 5.6% of 1,033 animals. Human- and dromedary-derived viruses are each monophyletic, suggesting ecological isolation. One gene of dromedary viruses exists in two versions in camels, full length and deleted, whereas only the deleted version exists in humans. The deletion increased in size over a succession starting from camelid viruses via old human viruses to contemporary human viruses. Live isolates of dromedary 229E viruses were obtained and studied to assess human infection risks. The viruses used the human entry receptor aminopeptidase N and replicated in human hepatoma cells, suggesting a principal ability to cause human infections. However, inefficient replication in several mucosa-derived cell lines and airway epithelial cultures suggested lack of adaptation to the human host. Dromedary viruses were as sensitive to the human type I interferon response as HCoV-229E. Antibodies in human sera neutralized dromedary-derived viruses, suggesting population immunity against dromedary viruses. Although no current epidemic risk seems to emanate from these viruses, evolutionary inference suggests that the endemic human virus HCoV-229E may constitute a descendant of camelid-associated viruses. HCoV-229E evolution provides a scenario for MERS-CoV emergence.Why are the dromedary coronaviruses not a current threat to humans?a.HCoV viruses are only in 5.6% of animalsb.Dromedary coronaviruses are only found in the Middle East.c.Inefficient replication in cell lines suggests they are not adapted to humans.d.Only one deleted version exists in humans. The volumetric analysis of the product of combustion on a dry basis for the oxidation of octane (C8H18) in air is 9.18% CO2, 0.23% CO and 7.48% O2. Write the combustion equation and determine the percent excess or deficiency of air in the combustion process.Use Mass BalancePlease complete the answer with correct solution What Would You Do? What Would You Not Do? Case Study 1 My-Lai Chang comes into the office with Christopher Chang, her 2-month-old son. Christopher is here for his 2-month well-child visit. Mrs. Chang is very distraught. She says that Christopher has episodes of nonstop crying every day that last 2 to 3 hours at a time. She is breast feeding Christopher and says that the crying is worse after he nurses. Although Mrs. Chang realizes that Christopher has colic, she feels guilty because it seems "her milk" is making it worse. She also is having problems with sore nipples and engorgement. She really wanted to breast feed Christopher, but she is thinking of stopping because it just seems too hard to do. Christopher measures in the 50th percentile for weight and length. Mrs. Chang is worried that he is not growing enough and thinks it is because she is not producing enough milk. What Would You Do? What Would You Not Do? Case Study 2 Wanda Tilley comes to the office with her 10-year-old daughter, Courtney. Courtney has a skin condition on her legs that needs to be evaluated by the physician. Courtney has been obese since she was 4 years old. Mrs. Tilley also is obese and is not too concerned about Courtney's weight. She says that Courtney must have inherited her "fat gene," and there's not much that can be done about it. Courtney's favorite activities are playing video games and reading. She would like to join the community swim team, but she's too embarrassed for anyone to see her in a bathing suit. Courtney says the other kids are always making fun of her at school. She says that they call her "two-ton Tilley" and "doubleroll," and they don't want to sit with her at lunch. Courtney wants her mom to home-school her because she's getting to the point where she can't take it anymore. She doesn't want the doctor to examine her because he'll see how fat she is and say bad things about it. What Would You Do? What Would You Not Do? Case Study 3 Stacy Jones, a legal secretary, brings her 5-year-old son, Matthew, in for a kindergarten physical. Stacy has read the vaccine information statements for the DTaP, IPV, and MMR immunizations that Matthew will be getting at this visit and has some questions. She wants to know why polio is not given orally anymore. She also wants to know why children are immunized against chickenpox because it is such a harmless disease. She is annoyed because she thinks that children are receiving too many unnecessary injections these days. Matthew is extremely afraid of "shots" and says that no one with a needle is getting anywhere near him. Stacy is protective of Matthew and knows that he will be hard to handle. 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Note: Modulus of elasticity is 30 x 10% pri C. 1709 rpm A 2038 B. 2540 rpm D. 2094 rpm FER Suppose output is at its potential level and then there is a sudden increase in food and energy prices. This increase a. makes inflation targeting easier because it makes these problems less relevant. answer with explanationWhich of the following is not associated with the movement of the other three in kidney functions? potassium ions hydrogen ions water protein 1. State basic requirement in foundry process. 2. Explain 3 types of molds in metal casting process. 3. A mold sprue is 22 cm long and the cross sectional area at its base is 2.0 cm^2 The sprue feeds a horizontal runner leading into a mold cavity whose volume is 1540 cm^3. Determine (i) Velocity of the molten metal at the base of the sprue (ii) Volume rate of flow. (iii) Time to fill the mold (g = 981cm/s/s; V=( 2gh) ^1/2 ; Q = V1A1 = V2A2 ; TMF = VIQ) A sample containing HCl is titrated with NaOH:HCl + NaOH H2O + NaClA student transfers 25.00 mL of HCl with a volumetric pipet toan Erlenmeyer flask, adds phenolphthalein, and titrates with NaOH Question 13 13. Which statement regarding iron is false? O it is found in myoglobin O it is a major mineral O it is found in all cells it serves as a cofactor in certain enzymess O it is found in hemoglobin Question 14 14. If the level of parathyroid hormone drops, what will happen to the level of calcium in the blood? O it will increase O the calcium level will not be affected O it will drop D Determine the vapor pressure of nitrogen at 90C on the basis of the reduced-vapor pressure correlations of(a) Lee and Kesler and(b) Dong and Lienhard, in bars.Then compare to data in a saturation table. can you show me the steps and answer please:In a population the frequency of allele A is 0.75 and allele a is 0.25; these allelesare codominantConsidering the allele fitness listed below, what will thefrequency of the A allele be after 1 generation? In a population the frequency of allele A is 0.75 and allele a is 0.25; these alleles are codominant. Considering the allele fitness listed below, what will the frequency of the A allele be after 1 generation? WAA = 0.8, WAa= 1, Waa = 0.9 What were the role of the little ice age and decreasing temperatures in the decline of the Vikings in Greenland? Is climate change the primary motivator for their decline in Greenland or are there other, greater or equal factors? From the left ventricle it passes into the aorta truefalseFrom the aorta to the capillaries true falseIn the capillaries the blood takes in oxygen true falseIn the capillaries the blood takes CO2 [xx yx zx] [-40 -24 0][] = [xy yy zy] = [-24 16 0] *10[xz yz zz] [ 0 0 12]a. Calculate the volumetric strain and the deviatoric strain tensor, b. Calculate the mean stress and the deviatoric stress invariants, c. Calculate the characteristic equation of strain, d. Calculate the characteristic equation of stress. The material is linear elastic (E=200GPa, v=0.3). Using the NPVmethod, a project is accepted if the NPV > 0.A. TRUEB. FALSE