The mean for hemoglobin is 14.0 and the standard deviation is 0.20. The acceptable control range is ± 2 standard deviations. What are the allowable limits for the control?
A. 13.8-14.2
B. 13.6-14.4
C. 13.4-14.6
D. 13.0-14.0

Answers

Answer 1

Option B is right. The acceptable control range for hemoglobin is ± 2 standard deviations from the mean, which would be 14.0 ± 0.40 (2 x 0.20). This means the allowable limits for the control are 13.6-14.4, so the correct answer is B.

To find the allowable limits for the control of hemoglobin, we need to calculate the range within ± 2 standard deviations from the mean.
The mean for hemoglobin is 14.0, and the standard deviation is 0.20.
Step 1: Multiply the standard deviation by 2.
0.20 * 2 = 0.40
Step 2: Add and subtract the result from the mean.
14.0 + 0.40 = 14.4
14.0 - 0.40 = 13.6
So, the allowable limits for the control are 13.6 to 14.4.
Your answer: B. 13.6-14.4

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Related Questions

what may happen when a mutation changes the amino acid sequence of a protein? the protein may become less effective, more effective, or altogether nonfunctional

Answers

When a mutation changes the amino acid sequence of a protein, various outcomes can occur. The protein may become less effective, meaning its function is impaired or weakened. This could lead to a decrease in the efficiency of the biological processes it's involved in.

Alternatively, the mutation could make the protein more effective, enhancing its function and potentially benefiting the organism. However, in some cases, the mutation may render the protein altogether nonfunctional. This means the protein loses its ability to perform its designated role, which can have significant consequences for the organism, depending on the importance of the protein's function. Overall, the effect of a mutation on the amino acid sequence of a protein can range from negative to positive or even neutral, depending on the specific change and its impact on protein function.

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.When preparing a smear from a bacterial colony (as opposed to a broth culture)... None of these are correct Water is not required as the solid medium has enough water Add a drop of water to the glass slide prior to transferring the bacterial sample Add the stain before preparing the bacterial smear

Answers

When preparing a smear from a bacterial colony, a drop of water should be added to the glass slide prior to transferring the bacterial sample.

This helps to disperse the cells and create an even layer for better visualization under the microscope. It is important to use a sterile loop or needle to transfer the bacterial sample to avoid contamination. The bacterial smear should then be air-dried and heat-fixed to ensure the cells adhere to the slide and are not washed away during staining. Only after the smear has been properly prepared, the stain can be added to visualize the cells and their structures.

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QUESTION 1:
How is delayed-onset muscle soreness (DOMS) different from muscle fatigue?
a. Delayed-onset muscle soreness occurs a day or more after the physical exertion
b. Muscle performance is impaired
c. There may be damage to the sarcolemma
e. Delayed-onset muscle soreness involves pain

Answers

Delayed-onset muscle soreness (DOMS) differs from muscle fatigue in that DOMS involves pain that occurs a day or more after physical exertion, while muscle fatigue refers to a temporary decline in muscle performance during or immediately after exercise.

DOMS typically manifests as muscle tenderness, stiffness, and soreness, often accompanied by reduced range of motion. It is commonly experienced after engaging in unfamiliar or intense physical activities or after performing eccentric exercises (such as downhill running or weightlifting with eccentric contractions).

The pain associated with DOMS is thought to be caused by microscopic damage to muscle fibers and connective tissues, as well as inflammation and the release of pain-inducing substances.

On the other hand, muscle fatigue is a transient decline in muscle force-generating capacity during or immediately after exercise. It is characterized by feelings of exhaustion, weakness, and reduced ability to generate force. Muscle fatigue can result from various factors, including depletion of energy stores, accumulation of metabolic byproducts, impaired neuromuscular signaling, and diminished muscle fiber excitability.

While muscle fatigue affects immediate performance, DOMS develops over time and is primarily characterized by the presence of pain and discomfort after the exertion has ended.

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Sunflowers have 34 chromosomes in diploid cells. How many chromosomes would be found in a haploid pollen grain?a. 17b. 34c. 51d. 68

Answers

a. 17 Explanation( Haploid cells have half the number of chromosomes as diploid cells)

An example of a nutrient that is important for primary productivity is:
a. calcium.
b. silica.
c. oxygen.
d. nitrogen.

Answers

The correct answer to the question is d. nitrogen. Nitrogen is a key nutrient for primary productivity as it is essential for the growth and development of plants and other photosynthetic organisms.

Nitrogen is a component of many important biomolecules such as chlorophyll, DNA, and amino acids, which are necessary for the processes of photosynthesis and growth. Without sufficient nitrogen, primary productivity would be limited, and ecosystems would be less productive. In fact, nitrogen is often a limiting factor in many ecosystems, meaning that the availability of nitrogen controls the rate of primary productivity.

Therefore, the management of nitrogen inputs and cycling is important for maintaining healthy ecosystems and ensuring sustainable food production. As a result, nitrogen plays a vital role in supporting primary productivity and the overall health of an ecosystem.

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discuss the anatomical differences between strepsirhine and haplorhine primates. how do these anatomical differences influence behavior and socialization among the respective primate groups?

Answers

Strepsirhine and haplorhine primates are two suborders of primates that have several anatomical differences that influence their behavior and socialization.

Strepsirhines are considered the more primitive of the two suborders, and they have several distinct anatomical features that set them apart from haplorhines. One of the most significant differences is the presence of a moist rhinarium, which is a moist, hairless area around the nostrils that is used for olfactory communication.

Strepsirhines also have a dental comb, a specialized set of lower incisors and canines that are used for grooming and obtaining food. Additionally, they have a tapetum lucidum, a layer of reflective cells in the eye that enhances night vision.

Haplorhines, on the other hand, lack a moist rhinarium and dental comb, and they have a more forward-facing placement of their eyes. They also lack a tapetum lucidum, which suggests that they are more diurnal than strepsirhines.

These anatomical differences have significant implications for the behavior and socialization of these primates. Strepsirhines tend to be more solitary and less social than haplorhines, and they rely heavily on olfactory communication for social interactions. The presence of the dental comb suggests that grooming is a crucial part of their social behavior. The tapetum lucidum enhances their night vision, allowing them to be more active at night, which may influence their behavior and socialization patterns.

Haplorhines, on the other hand, tend to be more social and less reliant on olfactory communication. They have a more forward-facing placement of their eyes, which enhances their depth perception and color vision, allowing them to be more diurnal and visually oriented. The lack of a dental comb suggests that grooming is less important in their social behavior, and they may rely more on other forms of social bonding, such as vocalizations and physical contact.

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The pentose phosphate pathway provides a number of critical functions including production of ribose-5-phost v that is vital for synthesis of nucleotides and Y. This pathway is divided into oxidative and non-oxidative steps. The latter steps are directly involved in NADF v If NAD+ v levels are high, . flux through the pentose phosphate pathway is promoted by allosteric regulation of glucose-6-phosphate dehydrogenase. If reducing equivalents derived form the pentose phosphate pathway are high, glucose-6-phosphate is directed toward glycolysis

Answers

The pentose phosphate pathway is a highly regulated metabolic pathway that can be adjusted based on the cellular needs for energy and reducing equivalents.

The pentose phosphate pathway plays a crucial role in the production of ribose-5-phosphate which is essential for the synthesis of nucleotides and NADPH. The pathway is divided into two parts, oxidative and non-oxidative steps. The non-oxidative steps are directly involved in NADPH production and do not involve the generation of ATP, unlike glycolysis. When NAD+ levels are high, flux through the pentose phosphate pathway is increased due to allosteric regulation of glucose-6-phosphate dehydrogenase. However, if reducing equivalents derived from the pathway are high, glucose-6-phosphate is directed towards glycolysis.

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I am playing hide-and-seek with my daughter. I hide for 20 minutes in a box and the carbon dioxide levels are increasing. Which of the following is TRUE? A. higher CO2 levels lead to bronchodilation to decrease airway resistance B. I'm panicking and my adrenaline causes my airways to bronchoconstrict C. higher CO2 levels lead to bronchoconstriction to decrease airway resistance D. higher CO2 levels lead to bronchodilation to increase alfway resistance

Answers

Higher CO2 levels lead to bronchoconstriction to decrease airway resistance.

The correct answer is C

In general ,  if ventilation is inadequate, as can happen when one is in a small enclosed space like a box, CO2 levels can continue to rise and cause a variety of physiological responses, including bronchoconstriction. Bronchoconstriction is the narrowing of the airways in the lungs, which can make it more difficult to breathe.

So , If you are playing a game that involves hiding in a confined space, it is important to ensure that the space is well-ventilated and that you can easily breathe.

Hence , C is the correct option

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All of the following are essential components in the transcription of DNA into mRNA EXCEPTA. Bene in the DNAB. transcription factors.C. terminator sequence.D. DNA polymerase. promoter

Answers

The correct answer is D. DNA polymerase is not involved in the transcription of DNA into mRNA, but rather in the replication of DNA. The essential components in transcription include the promoter,

which signals the start of transcription and is recognized by RNA polymerase; transcription factors, which bind to the promoter and help recruit RNA polymerase to the correct site; the template DNA strand, which serves as the blueprint for mRNA synthesis; and the terminator sequence, which signals the end of transcription. During transcription, RNA polymerase reads the DNA template strand and synthesizes a complementary RNA strand, which is then processed and exported to the cytoplasm for translation into protein.

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Endotherms are foregut fermenters while ectotherms are hindgut fermenters.
True
False
False, except for endothermic and ectothermic birds
False, except for lamprey

Answers

Endotherms are foregut fermenters while ectotherms are hindgut fermenters. The statement is False.


What are endotherms and ectotherms?
Endotherms are warm-blooded animals that generate heat internally to maintain a stable body temperature, while ectotherms are cold-blooded animals that rely on external sources of heat. Fermentation, respiration, and metabolism are processes involved in energy production within organisms. However, the classification of animals as foregut or hindgut fermenters is not solely based on them being endothermic or ectothermic. Foregut and hindgut fermentation refer to the location of fermentation in the digestive system of animals, which is independent of their endothermic or ectothermic nature.

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Endotherms are foregut fermenters while ectotherms are hindgut fermenters. The statement is False.


What are endotherms and ectotherms?
Endotherms are warm-blooded animals that generate heat internally to maintain a stable body temperature, while ectotherms are cold-blooded animals that rely on external sources of heat. Fermentation, respiration, and metabolism are processes involved in energy production within organisms. However, the classification of animals as foregut or hindgut fermenters is not solely based on them being endothermic or ectothermic. Foregut and hindgut fermentation refer to the location of fermentation in the digestive system of animals, which is independent of their endothermic or ectothermic nature.

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This question refers to freshwater stickleback losing their pelvic spines:
The mutation to the pelvic switch region of the Pitx1 gene affected which stage of the gene expression process?
Is it transcription or mRNA processing, or perhaps neither?
The mutation to the pelvic switch region also meant that the Pitx1 gene was only primarily functional in the rest of the body.
Is this true or false?

Answers

Hi! I'd be happy to help with your question. The mutation in the pelvic switch region of the Pitx1 gene affected the transcription stage of the gene expression process. Transcription is the process by which the DNA sequence is copied into a complementary mRNA molecule.

In the case of the freshwater stickleback, the mutation in the Pitx1 gene's pelvic switch region led to a reduced expression of the gene specifically in the pelvic spine. As for the second part of your question, it is true that the mutation in the pelvic switch region meant that the Pitx1 gene was only primarily functional in the rest of the body. The mutation affected the gene expression in the pelvic region, but the gene remained functional in other areas where it is typically expressed.

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When Manny Johnson received his stem cell therapy, the goal was to a. enhance the production of myoglobin to replace the ineffective hemoglobin. b. destroy the abnormal sickle cell proteins in his blood cells. c. stimulate the production of fetal hemoglobin instead of adult hemoglobin.
d. replace the incorrect nucleotide with the correct nucleotide so the new blood cells would form properly. e. stimulate the production of adult hemoglobin instead of fetal hemoglobin.

Answers

Manny Johnson received stem cell therapy with the goal of stimulating the production of fetal hemoglobin instead of adult hemoglobin. Fetal hemoglobin has a higher affinity for oxygen than adult hemoglobin, which is beneficial for individuals with sickle cell anemia.

Sickle cell anemia is a genetic disorder that causes abnormal hemoglobin to be produced, resulting in misshapen red blood cells that can clog blood vessels and cause pain, organ damage, and other complications. By stimulating the production of fetal hemoglobin, the hope is that Manny's red blood cells will function more effectively and reduce the severity and frequency of sickle cell-related symptoms. This is achieved by using stem cells from a donor with a similar genetic makeup to Manny, and stimulating them to produce healthy red blood cells with the appropriate fetal hemoglobin levels.

In summary, while stem cell therapy is still a relatively new treatment for sickle cell anemia, it shows promising results in improving quality of life for patients and reducing the need for blood transfusions and hospitalizations.

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in a gene pool, the Z allele frequency is 89% and the z allele frequency is 11%. Determine the frequency of heterozygous (Zz) individuals in the population

Answers

Answer:

Explanation:

The frequency of heterozygous individuals can be calculated using the Hardy-Weinberg equation:

p^2 + 2pq + q^2 = 1

where p is the frequency of one allele (Z in this case) and q is the frequency of the other allele (z).

Given that the Z allele frequency is 89%, we can calculate q as:

q = 1 - p

q = 1 - 0.89

q = 0.11

Using this value for q and the given value for p, we can calculate the frequency of heterozygous individuals as:

2pq = 2(0.89)(0.11) = 0.196

So, the frequency of heterozygous (Zz) individuals in the population is 0.196 or 19.6%.

A person typically obtains all chromosomes from their father.falsetrue

Answers

Answer: False

Explanation: Children randomly get one of each pair of chromosomes from their mother and one of each pair from their father.

1. How is lymph moved through the body?

The muscles pump it.
The spleen pumps it
The heart pumps it
The lungs pump it

Answers

Explanation:

Lymph moves through the body via the contraction of muscles surrounding the lymphatic vessels. The movement of the skeletal muscles during physical activity helps to squeeze the lymphatic vessels, propelling the lymph forward. Additionally, the valves within the lymphatic vessels prevent backflow and assist with the movement of lymph towards the heart. The spleen, heart, and lungs do not play a significant role in moving lymph through the body.

What might happen if you forget to remove the cover of the petri dish before exposing it to UV light for Lab #25- Effects of UV light?

Answers

If you forget to remove the cover of the petri dish before exposing it to UV light for Lab #25- Effects of UV light, the bacteria.

This is because the cover of the petri dish will block the UV light from reaching the bacteria, which is the main purpose of this lab experiment. In addition, if you forget to remove the cover of the petri dish, you may not be able to see the effects of the UV light on the bacteria. This is because the cover will prevent you from being able to observe any changes or growth patterns in the bacteria as a result of exposure to the UV light.
Moreover, forgetting to remove the cover of the petri dish may also lead to inaccurate results and conclusions. This is because you may assume that the bacteria is resistant to UV light, when in reality, it is simply because the cover blocked the UV light from reaching the bacteria.
Overall, it is important to carefully follow the instructions for Lab #25- Effects of UV light and to make sure that the cover of the petri dish is removed before exposing it to UV light. This will ensure that you obtain accurate and reliable results from your experiment.

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determine if the statement is true for glycolysis only, fermentation only, or both. Glycolysis Fermentation Statement Produces ATP = Uses NADH to produce NAD Occurs in cytosol Uses NAD to produce NADH = Produces organic acids or alcohol Drag and drop the correct answers into the boxes You can also click the correct answer, then the box where it should go.

Answers

The true statement for Glycolysis is only uses NAD to produce NADH, while using NADH to produce NAD+, and produces organic acids or alcohol only occur in fermentation. Both processes are occurs in cytosol and produce ATP. 

Glycolysis vs Fermentation


In the process of glycolysis, glucose was split into two pyruvate molecules, two NAD+ were reduced to two NADH + H+, and two ATP was produced. As the oxygen is not present, pyruvate will undergo a process called fermentation, which is NADH + H+ from glycolysis will be recycled back to NAD+ and produce either lactate or alcohol

The statement "Produces ATP" is true for both glycolysis and fermentation. Glycolysis is a process that breaks down glucose and generates ATP, while fermentation also generates a small amount of ATP.The statement "Uses NADH to produce NAD+" is true for fermentation only. In fermentation, NADH is oxidized to regenerate NAD+ for the continuation of glycolysis.The statement "Occurs in cytosol" is true for both glycolysis and fermentation, as both processes occur in the cytosol of the cell.The statement "Uses NAD to produce NADH" is true for glycolysis only. In glycolysis, NAD+ is reduced to NADH by gaining electrons from the breakdown of glucose.The statement "Produces organic acids or alcohol" is true for fermentation only. Fermentation produces organic acids (such as lactic acid) or alcohol (such as ethanol) as byproducts.


Therefore, Glycolysis is produces ATP, occurs in cytosol, and uses NAD to produce NADH, while fermentation is produce ATP, occurs in cytosol, uses NADH to produce NAD+, and produces organic acids or alcohol.

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__________ were middle paleozoic fish that had paired fins and scales on their bodies.

Answers

Placoderms were middle Paleozoic fish that had paired fins and scales on their bodies.

In general , Placoderms had a diverse range of body shapes and sizes, ranging from small, bottom-dwelling species to larger, predatory types. They are important in the evolutionary history of fishes because they represent an early stage in the development of jawed vertebrates.

Placoderms had a variety of body shapes and sizes, ranging from small, bottom-dwelling species to larger, predatory types.The evolution of placoderms and their jawed anatomy was a key step in the evolution of all jawed vertebrates, which includes all modern fish, amphibians, reptiles, birds, and mammals.

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I’m not sure on the answers

Answers

Answer:

1.thylakoids or atp(I'm not sure)

2.it is an output of stomach enzymes

For the codon: 5'- GAU-3', what are the last two nucleotides in the anticodon (5' to 3)? a. 5-CU-3 b. 5-GA-3 c. 5-AU-3 d. 5-AG-3' e. 5-UA-3 f. 5-UC-3

Answers

Option f is correct. For the codon: 5'- GAU-3', the last two nucleotides in the anticodon are 5-UC-3. First nucleotide: G (in the codon, it pairs with C). The codon's second nucleotide, U, pairs with A.

Aspartic acid (Asp) is encoded by the codon 5'-GAU-3'. We must apply the DNA and RNA base pairing principles in order to identify the final two nucleotides in the anticodon.

The three-nucleotide sequence in transfer RNA (tRNA) that is complementary to the codon in messenger RNA (mRNA) during protein synthesis is known as the anticodon.

Together with the first two nucleotides (5'-3') in the codon, the final two nucleotides in the anticodon create Watson-Crick base pairs. Consequently, the final two nucleotides of the anticodon (5'-3') for the codon 5'-GAU-3' would be 5-UC-3.

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Complete question

For the codon: 5'- GAU-3', what are the last two nucleotides in the anticodon (5' to 3)?

a. 5-CU-3

b. 5-GA-3

c. 5-AU-3

d. 5-AG-3'

e. 5-UA-3

f. 5-UC-3

Beyond Mendel 151 Exercise D: One gene, multiple alleles Codominance Genotype Blood type In human ABO blood type there are three blood type phenotypes possible: Pr, PP. PP, P P, and i (Table 3). Understanding these blood types is important in blood transfusions, and it is due to the acceptance of the glycoproteins on the blood cellular membranes coded by the / gene. Similar to Mendelian genetics, i is recessive to both and P. But what happens when both and Palleles are in a gene? As it turns out. both alleles are activated. So blood cells of the genotype produce different glycoproteins with both sugars. This double expression is known as codominance. 21. What is the expected ratio of blood types in children from a type AB mother and a type O father? Blood type is autosomal (not sex-linked).

Answers

The expected ratio of blood types in the children is 1:1 (Blood type A : Blood type B)

What is the expected ratio of blood types in children from a type AB mother and a type O father, considering blood type is autosomal?



To determine the expected ratio of blood types in children from a type AB mother and a type O father, we can perform a Punnett square analysis:

Step 1: Identify the genotypes of the parents. An AB mother has the genotype IAIB, while a type O father has the genotype ii.

Step 2: Create a Punnett square with the genotypes of the parents.

      IA   IB
 i  IAi  IBi
 i  IAi  IBi

Step 3: Analyze the Punnett square results. From the Punnett square, we see the following genotypes for the offspring:

- 2 IAi (Blood type A)
- 2 IBi (Blood type B)

Step 4: Determine the ratio of blood types. The expected ratio of blood types in the children is 1:1 (Blood type A : Blood type B). There will be no type AB or type O children in this case.

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A couple is expecting a baby. The father has white skin and the mother has dark brown skin.

1. What color of skin will their child most likely have?

2. What are other possibilities for this child's skin color?

Answers

Answer:

Explanation:

The skin colour of a child from a mixed-race couple with one parent having white skin and the other having dark brown skin is determined by complex genetic factors. The child's skin color will depend on the specific combination of genes inherited from each parent.

The child is likely to have a skin color that is a shade between the mother's dark brown skin and the father's white skin. However, the exact skin colour cannot be predicted with certainty, as there is a wide range of possible skin tones that can result from the mixing of different skin colors.

Other possibilities for the child's skin color include having skin that is closer in color to either the mother or the father, or having a skin tone that is a combination of both parents' skin colors. Additionally, the child's skin color may be influenced by other genetic factors, such as the presence of genes for freckles or other pigmentation traits. It is also possible for the child's skin color to change over time due to factors such as sun exposure and hormonal changes during puberty. Ultimately, the child's skin color is a complex trait that cannot be predicted with certainty based on the skin colors of the parents.

The specific skin color of the child cannot be determined with certainty as it depends on the genetic inheritance of multiple genes involved in determining skin color. The child could have a range of possible skin colors.

Skin color is a polygenic trait influenced by a combination of genes from both parents, and additional factors such as genetic variations from ancestral backgrounds may contribute to the variation in skin color.

It is not possible to predict the exact skin color of the child based solely on the skin colors of the parents.

The child could have a range of possible skin colors.

The child could inherit a combination of genetic variations from both parents, resulting in a skin color that may be lighter than the mother's, darker than the father's, or fall somewhere in between.

The child's skin color could be a blend of traits inherited from their parents, and additional factors such as genetic variations from ancestral backgrounds may also contribute to the variation in skin color.

The child's specific skin color will be a unique combination influenced by genetic inheritance and other factors.

Thus, the child's unique skin tone cannot be predicted with absolute precision because it depends on the genetic inheritance of several genes that affect skin tone. A variety of skin tones are possible for the child.

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Which molecules inhibit RNA polymerase to prevent transcription?
Select one:
a. activators
b. repressors
c. promoters
d. introns

Answers

While both repressors and introns can impact gene expression, they do so in different ways. Option B and D are correct answers.

There are molecules known as repressors that inhibit RNA polymerase and prevent transcription. These molecules bind to specific DNA sequences, called operators, near the promoter region of a gene.

Repressors can block the binding of RNA polymerase to the promoter region, or they can interfere with the ability of RNA polymerase to initiate transcription.

This results in the inhibition of gene expression. Introns, on the other hand, are non-coding regions within genes that are transcribed along with the coding regions, but are removed from the mRNA during processing.

Introns do not directly inhibit RNA polymerase or prevent transcription. However, they can play a regulatory role in gene expression by affecting alternative splicing or mRNA stability.

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Please help hurry I’ll mark brainly


The number of severe weather occurrences has risen over the past 30 years. Choose one severe weather event from the past 30 years, either hurricane,
tsunami, tomado, or monsoon and do some research. Form a well-constructed paragraph with the following information:
• Location, time frame, and name of event (if applicable).
• What was the cause of the severe storm?
• What impact, immediate and long term, did the severe weather have on the environment, vegetation, and humans in the area.
• If you feel humans are the main cause of global warming, do you think you contribute to global warming, and if so, how?
• Where do you stand on whether the U.S. government should be addressing global warming and climate change?

Answers

Answer:

One severe weather event that occurred in the past 30 years was Hurricane Katrina, which hit the Gulf Coast of the United States in August of 2005. It was caused by warm sea surface temperatures, low wind shear, and atmospheric disturbances. The hurricane had a devastating impact as it caused widespread flooding and extensive damage to infrastructure and homes. The immediate impacts included lack of access to basic necessities such as food, water, and medical care.  Many people experienced power outages which led to spoiled food and stoves that did not work. The long-term impacts included loss of wetlands and damage to the local economy, with some estimates putting the total cost of the storm at over $100 billion. I do believe humans are the main cause of global warming, and I contribute to it by using gas to fuel my car and using oil to heat my home. I believe that the U.S. government has a responsibility to address global warming and climate change as it has the potential to cause significant environmental, economic, and social impacts. It is imperative that policies that reduce greenhouse gas emissions and promote the use of renewable energy sources are enforced to help mitigate the effects of global warming.

A. Which of the following doesn't interact directly with DNA?
Group of answer choices
TATA-box binding protein
RNA polymerase II
transcription factors
the mediator complex
B. Changing an RNA sequence following transcription by removing segments is done by ______________.
deacetylation capping polyadenylation the spliceosome DNA binding RNA editing methylation acetylation translation

Answers

A. The mediator complex doesn't interact directly with DNA.

The other options, such as TATA-box binding protein, RNA polymerase II, and transcription factors, all interact directly with DNA during the process of transcription.
B. Changing an RNA sequence following transcription by removing segments is done by RNA Splicing
The spliceosome is responsible for changing an RNA sequence following transcription by removing segments. This process, known as splicing, removes introns from the pre-mRNA, leaving only the exons to form the mature mRNA molecule. The final mRNA thus consists of the remaining sequences, called exons, which are connected to one another through the splicing process.

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what is required when cells are putting together polymers by assembling monomers?

Answers

When cells assemble polymers by linking monomers, they require energy in the form of ATP (adenosine triphosphate), enzymes, and appropriate building blocks.

Cells use anabolic reactions to join monomers and create polymers, which require energy input. These anabolic reactions are usually endergonic, meaning that they require energy input to proceed. Cells also require specific enzymes that catalyze the formation of covalent bonds between monomers. These enzymes lower the activation energy required for the reaction, making it energetically favorable. Additionally, cells require the appropriate building blocks, which are usually obtained from food. For example, amino acids are used to build proteins, and nucleotides are used to build nucleic acids. Overall, assembling polymers from monomers is a complex process that requires energy, enzymes, and appropriate building blocks.

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Place the steps of carbohydrate breakdown in order. Note: you must place all steps in order to earn credit for this question. O Glycolysis occurs, breaking down sugars into 3-carbon molecules called pyruvate. Each pyruvaO te molecule is converted to a 2-carbon molecule called Acetyl CoA and some carbon dioxide is released as waste. O The citric acid cycle occurs which contributes to a proton gradient and carbon dioxide as waste. O Oxidative phosphorylation occurs, which creates a significant amount of ATP. O Carbohydrates are broken down into individual sugar monomers.

Answers

Carbohydrate breakdown is a complex process that begins with the breaking down of carbohydrates into individual sugar monomers.

Next, glycolysis occurs, breaking down the sugars into 3-carbon molecules called pyruvate. Each pyruvate molecule is then converted to a 2-carbon molecule called Acetyl CoA, with carbon dioxide as a waste product.

Subsequently, the citric acid cycle occurs, which helps to form a proton gradient and produces more carbon dioxide. Finally, oxidative phosphorylation takes place, resulting in the production of a significant amount of ATP. Altogether, this process produces energy from carbohydrates and helps to sustain life.

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Classify each definition or example as a somatic mutation, gametic (germline) mutation, of both. The mutation occurs m any cell except a germ cell, and thus the mutation does not affect the progeny of the individual. A particular tobacco leaf becomes discolored due to mutation halfway through the life of the plant Mutations can be caused by an alteration in the DNA sequence

Answers

In both the first and second examples are somatic mutations, while the third statement applies to both somatic and gametic (germline) mutations.

The classify of each definition or example as a somatic mutation, gametic (germline) mutation, of both are :

1. "The mutation occurs in any cell except a germ cell, and thus the mutation does not affect the progeny of the individual." - This statement refers to a somatic mutation. Somatic mutations occur in non-germ cells and only affect the individual in which they occur, not their offspring.


2. "A particular tobacco leaf becomes discolored due to mutation halfway through the life of the plant." - This example also describes a somatic mutation. The mutation occurred in a specific somatic cell within the tobacco plant, causing the discoloration of the leaf.


3. "Mutations can be caused by an alteration in the DNA sequence." - This statement applies to both somatic and gametic (germline) mutations. An alteration in the DNA sequence can lead to either type of mutation, depending on the cell in which the alteration occurs.

If the alteration is in a somatic cell, it would result in a somatic mutation, whereas if the alteration is in a germ cell, it would result in a gametic (germline) mutation, which can be passed on to the individual's offspring.

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The first statement "Mutation occurs in any cell except a germ cell, and thus the mutation does not affect the progeny of the individual" and the second statement "A particular tobacco leaf becomes discolored due to mutation halfway through the life of the plant" refers to somatic mutation and the third one "Mutations can be caused by an alteration in the DNA sequence" refers to both somatic mutation and germline mutation.

The first definition/example refers to a somatic mutation, as it occurs in any cell except a germ cell and does not affect the progeny. The second example also refers to a somatic mutation, as it affects only a particular tobacco leaf and not the germ cells. The last statement is a general statement about mutations and can refer to both somatic and gametic (germline) mutations, as they are both caused by alterations in the DNA sequence. The key difference between them is that somatic mutations occur in non-germ cells and do not affect the offspring, while gametic (germline) mutations occur in germ cells and can be passed on to the next generation.

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4. What material is moved through the body in the lymphatic system?

plasma
water
lymph
blood

Answers

Answer:

The material that is moved through the body in the lymphatic system is lymph.

I’m pretty sure the answer is lymph

should 3m hedge against adverse movements on foreign exchange rates? how should it do this?

Answers

Yes, 3M should hedge against adverse movements in foreign exchange rates.

They can do this by using derivatives such as forward contracts, futures contracts, or options.

3M should hedge against adverse movements in foreign exchange rates to minimize the risk of financial loss due to fluctuating currency values. To do this, they can employ a variety of financial instruments:

1. Forward contracts: 3M can enter into a private agreement with a counterparty to buy or sell a specific amount of foreign currency at a predetermined exchange rate on a future date. This helps lock in the exchange rate and avoid fluctuations.

2. Futures contracts: Similar to forward contracts, futures contracts allow 3M to buy or sell a specific amount of foreign currency at a predetermined exchange rate on a future date, but these contracts are standardized and traded on exchanges.

3. Options: 3M can purchase options to buy or sell foreign currency at a specific exchange rate on or before a specified date. Options provide flexibility, as 3M is not obligated to exercise the option if the exchange rate is unfavorable.

By utilizing these financial instruments, 3M can effectively manage their foreign exchange risk and protect their financial interests.

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