a) Mary is 23-year-old and she weights 54kg. She is not a vegetarian. Calculate the recommendation for her protein intake. (Hint: you should calculate based on the protein intake for a healthy adult) (3 marks). b) Suppose she follows the recommendation calculated in a) in a 24-hours period. During that time she excreted 5g of nitrogen as urea. Calculate her state of nitrogen balance. Show your calculation (3 marks). c) Does Mary obtain a positive or negative nitrogen balance? Suggest THREE possible reasons for her nitrogen states (4 marks). d) Describe FOUR features in a Mediterranean diet. Provide THREE reasons to support if this diet is suitable for a six-year-old child (10 marks).

Answers

Answer 1

a) Based on the protein intake recommendation for a healthy adult, Mary's protein intake should be calculated.

b) By following the recommendation calculated in a), Mary excreted 5g of nitrogen as urea. Her state of nitrogen balance can be calculated based on this information.

c) Mary's nitrogen balance can be determined by evaluating whether she has a positive or negative nitrogen balance and considering possible reasons for her nitrogen states.

d) The Mediterranean diet is characterized by specific features, and it is important to assess its suitability for a six-year-old child based on three supporting reasons.

a) Mary's protein intake recommendation should be calculated based on the protein intake for a healthy adult.

b) Mary's state of nitrogen balance can be determined by evaluating the amount of nitrogen excreted.

c) Mary's nitrogen balance can be positive or negative, depending on various factors.

d) The Mediterranean diet has distinct features, and its suitability for a six-year-old child can be assessed based on supporting reasons.

a) Mary's protein intake recommendation should be calculated based on the protein intake for a healthy adult. This recommendation ensures that she consumes an adequate amount of protein for her age and weight. Protein is essential for numerous functions in the body, including tissue repair, enzyme production, and immune system function. By calculating her protein intake, Mary can maintain a balanced diet that supports her overall health and well-being.

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Related Questions

Spemann organizer is a dorsal structure . Planar cell polarity involves interaction with ECM. Hedgehog and Wingless are both secreted factors that establish the anterior/posterior identity of parasegments in Drosophila.

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The Spemann organizer is a dorsal structure involved in embryonic development. Planar cell polarity is a cellular process that includes interactions with the extracellular matrix (ECM). Hedgehog and Wingless are secreted factors that play roles in establishing the anterior/posterior identity of parasegments in Drosophila.

The Spemann organizer is a region in the developing embryo that plays a critical role in dorsal-ventral patterning. It was first identified by the German embryologist Hans Spemann and his student Hilde Mangold during their experiments with salamander embryos. The Spemann organizer is responsible for inducing the formation of the dorsal mesoderm and axial structures during embryonic development. It secretes various signaling molecules, such as Chordin and Noggin, which inhibit the activity of BMP (Bone Morphogenetic Protein) signaling and promote the development of dorsal structures.

Planar cell polarity refers to the coordinated orientation of cells within a tissue plane. It involves the alignment and polarization of cells along a specific axis, which is essential for the proper organization and function of tissues. Planar cell polarity is regulated by a complex network of signaling pathways and interactions with the extracellular matrix. The ECM provides cues and signals that guide the polarization of cells, influencing their orientation and behavior.

Hedgehog and Wingless are two secreted factors involved in establishing the anterior/posterior identity of parasegments in Drosophila, a commonly studied model organism in developmental biology. Hedgehog signaling pathway plays a key role in patterning the anterior/posterior axis by establishing concentration gradients of Hedgehog protein, which then activates target genes in a concentration-dependent manner. Wingless, also known as Wnt, is another signaling pathway that helps establish the anterior/posterior identity by regulating gene expression in specific regions of the embryo.

In summary, the Spemann organizer is a dorsal structure involved in embryonic development, planar cell polarity involves interactions with the extracellular matrix, and Hedgehog and Wingless are secreted factors that play important roles in establishing the anterior/posterior identity of parasegments in Drosophila.

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Which statement regarding facultative anaerobes is true?
a. They can survive in the presence or absence of oxygen.
b. They require oxygen to survive.
c. They require the absence of oxygen to survive.
d. They cannot metabolize glucose.
e. They require carbon dioxide to survive.

Answers

Facultative anaerobes can survive in the presence or absence of oxygen.

The correct answer is (a) They can survive in the presence or absence of oxygen. Facultative anaerobes are microorganisms that have the ability to switch between aerobic and anaerobic metabolism based on the availability of oxygen. In the presence of oxygen, they can perform aerobic respiration to generate energy.

However, in the absence of oxygen, they can switch to anaerobic metabolism, such as fermentation, to produce energy. This versatility allows facultative anaerobes to survive and thrive in environments with varying oxygen levels, making them adaptable to different conditions.

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Question 9 (3 points) Define "carrying capacity". Can the carrying capacity of a population change? Explain. A

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Carrying capacity is the maximum population that a particular ecosystem can sustain over a prolonged period under specific environmental conditions. It varies based on different factors like availability of resources, competition for resources, and other environmental factors. The carrying capacity of a population can change, depending on the changes in the environment.

For instance, if there is a decline in the availability of food or water, the carrying capacity would decrease, and if there is an increase in the availability of food and other resources, the carrying capacity would increase. The carrying capacity of a population can also change due to external factors like natural disasters, diseases, and human activities like deforestation, pollution, hunting, and climate change.

For example, if a forest that supports a particular population is destroyed, the carrying capacity of that ecosystem would decrease, and the population would decline. In conclusion, the carrying capacity of a population can change based on various internal and external factors that affect the ecosystem.

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If a population reaches the carrying capacity of the environment, O food and other resources will increase O the population will decline rapidly O unrestrained growth will occur O the population size

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If a population reaches the carrying capacity of the environment, the population size will fluctuate around this level (option d).

The carrying capacity of an environment is the maximum number of individuals of a particular species that an environment can support based on the resources available. If the population exceeds this carrying capacity, there may be a decline in resources, leading to a decrease in the population size. In contrast, if the population is below the carrying capacity, there may be room for growth until the carrying capacity is reached.

However, once the population reaches the carrying capacity, it is unlikely to continue to grow at the same rate. The availability of resources may fluctuate due to environmental factors such as weather patterns or natural disasters, causing the population to fluctuate in response. For example, if a drought occurs, there may be a decrease in the availability of water and food, leading to a decline in the population. Similarly, if there is an abundance of resources, the population may increase until it reaches the carrying capacity again.

Overall, once a population reaches the carrying capacity of the environment, the population size will fluctuate around this level due to the availability of resources and other environmental factors. It is important for populations to remain at or below the carrying capacity to ensure the continued health and survival of the species.

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The full question is given below:

If a population reaches the carrying capacity of the environment:

a. unrestrained growth will occur.

b. the population will decline rapidly.

c. food and other resources will increase.

d. the population size will fluctuate around this level.

Connective Tissues -- Select from the list of tissues below and match to their description. Mark only the numbers as the answer. 1- Blood 2- Adipose Tissue 4- Dense Regular Connective Tissue 5- Hyaline Cartilage 6- Osseous Tissue 3- Areolar Tissue 1 Tendons and ligaments. (Many long fibers)............ 2 Supporting rings in trachea............ 3 Found covering ends of long bones............ 4 Solid matrix of calcium salts.............. 5 White, glassy appearance.. 6 Serves as insulation material... 7 Most common tissue found in the skin. (Very unorganized).. 8 Most rigid supporting tissue......... 1449 5 5 _6_ 5 2 2

Answers

The matching of connective tissues with their description are:1. Tendons and ligaments. (Many long fibers) --- 4- Dense Regular Connective Tissue2. Supporting rings in trachea --- 5- Hyaline Cartilage3. Found covering ends of long bones --- 5- Hyaline Cartilage4. Solid matrix of calcium salts --- 6- Osseous Tissue5. White, glassy appearance --- 5- Hyaline Cartilage6. Serves as insulation material --- 2- Adipose Tissue7. Most common tissue found in the skin. (Very unorganized) --- 3- Areolar Tissue8. Most rigid supporting tissue --- 6- Osseous Tissue

Connective tissue is a group of tissues that support and connect various tissues and organs of the body. It contains three basic components: specialized cells, protein fibers, and ground substance. Connective tissues are of various types, some of them are mentioned below:Hyaline Cartilage: Hyaline cartilage is a flexible tissue that acts as a cushion between bones. It is found in the supporting rings of the trachea and larynx, covering the ends of long bones, and at the end of the ribs. It is characterized by a white, glassy appearance. The main function of hyaline cartilage is to provide a smooth surface for joint movement.

Dense Regular Connective Tissue: This tissue consists of many long fibers that are tightly packed together. It is found in tendons and ligaments and provides strong attachment points between bones and muscles. It also helps to transmit forces from one bone to another.

Areolar Tissue: Areolar tissue is a loose connective tissue that is found between other tissues and organs of the body. It is made up of collagen and elastin fibers, which provide support and elasticity to the surrounding structures.Osseous Tissue: Osseous tissue, also known as bone tissue, is the most rigid supporting tissue in the human body. It is made up of a solid matrix of calcium salts that provide structural support and protection to the body's internal organs.

Adipose Tissue: Adipose tissue is a type of connective tissue that serves as an insulating material in the body. It is made up of specialized cells called adipocytes that store energy in the form of fat. Adipose tissue is found throughout the body and helps to regulate body temperature and protect internal organs.

Therefore, the answer to the given question is, 1 Tendons and ligaments. (Many long fibers) --- 4- Dense Regular Connective Tissue2 Supporting rings in trachea --- 5- Hyaline Cartilage3 Found covering ends of long bones --- 5- Hyaline Cartilage4 Solid matrix of calcium salts --- 6- Osseous Tissue5 White, glassy appearance --- 5- Hyaline Cartilage6 Serves as insulation material --- 2- Adipose Tissue7 Most common tissue found in the skin. (Very unorganized) --- 3- Areolar Tissue8 Most rigid supporting tissue --- 6- Osseous Tissue

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Which statement is false about respiratory tract infections? a. Pneumonia immunisations must be repeated every year b. Influenza can lead to pneumonia c. Rhinosinusitis can be caused by both bacteria and viruses d. The common cold can be caused by parainfluenza viruses e. Immunisation does not provide complete protection against influenza

Answers

The false statement about respiratory tract infections is:

a. Pneumonia immunisations must be repeated every year.

Pneumonia immunizations do not need to be repeated every year. Once vaccinated against pneumonia, the immunity provided by the vaccine can last for several years or even a lifetime, depending on the specific vaccine and individual factors. It is not necessary to repeat pneumonia immunizations annually, unlike influenza vaccinations that require annual updates due to the evolving nature of the influenza virus.

The other statements are true:

b. Influenza can lead to pneumonia. Influenza infection can cause complications such as pneumonia, particularly in individuals with weakened immune systems or underlying health conditions.

c. Rhinosinusitis can be caused by both bacteria and viruses. Rhinosinusitis, inflammation of the nasal passages and sinuses, can be caused by both bacterial and viral infections. The majority of cases are viral in nature, but bacterial infections can also occur.

d. The common cold can be caused by parainfluenza viruses. Parainfluenza viruses are one of the many viruses that can cause the common cold, along with rhinoviruses and other respiratory viruses.

e. Immunization does not provide complete protection against influenza. While influenza immunization can significantly reduce the risk of contracting the flu and its complications, it does not offer 100% protection. The effectiveness of the vaccine can vary depending on factors such as the match between the vaccine strains and circulating strains, individual immune response, and other variables. However, immunization remains an important preventive measure to reduce the severity and spread of influenza.

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What is the mechanism of action of contraceptive pills? Describe
they interfere the uterine and ovarian cycles. Include: how do they
prevent ovulation? Pls don't copy paste from other chegg answers, I

Answers

Contraceptive pills contain synthetic estrogen and progesterone hormones that prevent ovulation and also alter the cervical mucus and lining of the uterus.

Contraceptive pills are used to prevent pregnancy. It contains synthetic estrogen and progesterone hormones which interfere with the ovarian and uterine cycles in females. It prevents ovulation by inhibiting the production of follicle-stimulating hormone (FSH) and luteinizing hormone (LH), which are responsible for the growth and maturation of follicles in the ovary. By doing so, the ovary does not release an egg, and therefore fertilization does not occur. Also, contraceptive pills thicken the cervical mucus, which makes it difficult for sperm to enter the uterus. If by chance the egg is released, the pills also alter the lining of the uterus, which makes it less receptive to the fertilized egg. Thus, the egg is not implanted, and pregnancy is avoided.Contraceptive pills contain synthetic estrogen and progesterone hormones that prevent ovulation and also alter the cervical mucus and lining of the uterus.

Contraceptive pills are highly effective in preventing pregnancy when taken correctly. It is essential to take them at the same time every day to ensure maximum protection. However, they do not protect against sexually transmitted infections (STIs).

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Infection with certain viruses inhibits SnRNA processing in eukaryotic cells. Explain why this favors the expression of viral genes in the host cell.

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Infection with certain viruses that inhibit SnRNA (small nuclear RNA) processing in eukaryotic cells can favor the expression of viral genes in the host cell due to the role of SnRNAs in pre-mRNA splicing.

SnRNAs are essential components of the spliceosome, a complex involved in removing introns from pre-mRNA molecules and joining exons together during mRNA maturation. By inhibiting SnRNA processing, the virus interferes with this splicing machinery, leading to aberrant splicing or reduced efficiency of splicing in the host cell.

As a result, the virus can benefit from this disruption in several ways:

1. Increased production of viral proteins: The inhibition of SnRNA processing can lead to the production of abnormal mRNA transcripts. These transcripts may contain alternative splicing patterns or retain introns, resulting in the synthesis of viral proteins with altered structures or functions. This can enhance the expression of viral genes and promote viral replication within the host cell.

2. Evasion of host immune response: The altered mRNA splicing caused by SnRNA processing inhibition can generate viral proteins that evade detection by the host immune system. By producing non-canonical protein isoforms, the virus can potentially escape immune surveillance, allowing it to persist and continue its replication cycle.

Overall, the inhibition of SnRNA processing by certain viruses disrupts normal cellular mRNA splicing, leading to altered protein expression patterns that favor the expression of viral genes. This provides the virus with a selective advantage for successful replication and evasion of the host immune response.

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Name 5 molecular mechanisms of biological problem .
and write me a few point about 1
Write me a topic of molecular machanisom of a biological problem .Also,some details about the topic .

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The five molecular mechanisms of biological problems are DNA replication, transcription, translation, signal transduction, and apoptosis. These mechanisms are fundamental processes that ensure genetic fidelity, regulate gene expression, enable protein synthesis, mediate cellular responses to signals, and maintain tissue homeostasis.

1. DNA Replication: DNA replication is a crucial molecular mechanism in biological systems that ensures the faithful duplication of genetic information during cell division. It involves the unwinding of the DNA double helix, synthesis of new complementary strands by DNA polymerases, and proofreading mechanisms to maintain accuracy. DNA replication is tightly regulated to prevent errors and maintain genomic stability.

2. Transcription: Transcription is the process by which genetic information encoded in DNA is transcribed into RNA molecules. It involves the binding of RNA polymerase to a specific DNA sequence called the promoter, followed by the synthesis of an RNA molecule that is complementary to the DNA template strand. Transcription is regulated by various factors, including transcription factors and epigenetic modifications, and plays a vital role in gene expression and cellular functions.

3. Translation: Translation is the process by which RNA molecules are decoded to synthesize proteins. It occurs in ribosomes, where transfer RNAs (tRNAs) bring specific amino acids to the ribosome, guided by the codons on the mRNA. The ribosome catalyzes the formation of peptide bonds between amino acids, leading to the synthesis of a polypeptide chain. Translation is regulated by various factors, including initiation factors, elongation factors, and termination factors, and is critical for protein synthesis and cellular function.

4. Signal Transduction: Signal transduction is a complex molecular mechanism that enables cells to respond to external stimuli. It involves the transmission of signals from the cell surface to the nucleus or other cellular compartments, leading to changes in gene expression, protein activity, or cell behavior. Signal transduction pathways often involve the binding of ligands to cell surface receptors, activation of intracellular signaling cascades, and modulation of transcription factors or enzymes.

5. Apoptosis: Apoptosis, also known as programmed cell death, is a molecular mechanism that regulates cell survival and tissue homeostasis. It involves a series of tightly controlled events, including the activation of caspases, DNA fragmentation, and membrane blebbing. Apoptosis can be triggered by various internal and external signals, such as DNA damage, oxidative stress, or developmental cues. Dysregulation of apoptosis can contribute to various diseases, including cancer and neurodegenerative disorders.

Understanding these molecular mechanisms is crucial for unraveling the complexities of biological systems and developing targeted interventions to address various biological problems. Each mechanism plays a vital role in cellular processes and contributes to the overall functioning and regulation of living organisms.

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Question A double-stranded DNA molecule with the sequence shown below produces, in vivo, a polypeptide that is five amino acids long. TAC ATG ATC ATT TCA CGG AAT TTC TAG CAT GTA ATG TAC TAG TAA AGT GC

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The double-stranded DNA sequence TAC ATG ATC ATT TCA CGG AAT TTC TAG CAT GTA ATG TAC TAG TAA AGT GC produces a polypeptide that is five amino acids long: Met-Tyr-Stop-Ile-Ser.

The DNA sequence is transcribed into messenger RNA (mRNA) through a process called transcription. The mRNA is then translated into a polypeptide during protein synthesis. Each three-nucleotide sequence, called a codon, codes for a specific amino acid. By analyzing the DNA sequence provided, the corresponding mRNA sequence would be AUG UAC UAG AUA AGU CGA UUA AAG AUC GUA CAU UAC AUC UAG, which would be translated into the polypeptide sequence Met-Tyr-Stop-Ile-Ser-Arg-Leu-Lys-Ile-Val-His-Tyr-Ile-Stop.

In summary, the given DNA sequence undergoes transcription and translation processes to produce a polypeptide that consists of five amino acids: Met-Tyr-Stop-Ile-Ser. The sequence of the DNA determines the sequence of the mRNA, which, in turn, determines the sequence of the polypeptide during protein synthesis.

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Journal Review for: Phylogeny of Gekko from the Northern Philippines, and Description of a New Species from Calayan Island DOI: 10.1670/08-207.1
In terms of the molecular data
1. What type of molecular data was used? Describe the characteristic of the gene region used and how did it contribute to the findings of the study.
2. What algorithms were used in the study and how were they presented? If more than 1 algorithm was used, compare and contrast the results of the algorithms.
In terms of the morphological data
3. Give a brief summary of the pertinent morphological characters that were used in the study. How where they presented?
4. Phylogenetic studies are usually supported by both morphological and molecular data. In the journal assigned, how was the collaboration of morphological and molecular data presented? Did it create conflict or was it able to provide sound inferences?
Separate vs. Combined Analysis
5. Identify the substitution model utilized in the paper.
6. In the phylogenetic tree provided identify the support value presented (PP or BS). Why does it have that particular support value?
7. Did the phylogenetic analysis utilize separate or combined data sets? Explain your answer.

Answers

1. The type of molecular data used in the paper “Phylogeny of Gekko from the Northern Philippines, and Description of a New Species from Calayan Island” is mitochondrial and nuclear genes. The molecular phylogenetic analysis was based on 3469 base pairs of two mitochondrial genes (12S and 16S rRNA) and one nuclear gene (c-mos).

Mitochondrial DNA is generally used in phylogenetic analysis because it is maternally inherited and has a high mutation rate. In contrast, nuclear DNA evolves at a slower rate and is biparentally inherited.
2. In this paper, the maximum parsimony (MP) and Bayesian inference (BI) algorithms were used. MP was presented as a strict consensus tree, and BI was presented as a majority rule consensus tree. MP is a tree-building algorithm that seeks to minimize the total number of evolutionary changes (such as substitutions, insertions, and deletions) required to explain the data. In contrast, BI is a statistical method that estimates the probability of each tree given the data. It is known to be a powerful tool for inferring phylogenies with complex evolutionary models. In this study, the two algorithms produced similar topologies, suggesting that the tree topology is robust.
3. The morphological data used in the study included the number of scales around the midbody, the presence of a preanal pore, the number of precloacal pores, and the length of the fourth toe. These morphological characters were presented as a table that shows the values for each species.
4. In this study, both molecular and morphological data were used to infer the phylogeny of the Gekko species. The phylogenetic tree was based on the combined data set of molecular and morphological data, which was presented as a majority rule consensus tree. The combined analysis provided sound inferences, and there was no conflict between the two datasets.
5. The substitution model utilized in the paper was GTR+I+G. This is a general time reversible model that incorporates the proportion of invariable sites and a gamma distribution of rates across sites.
6. In the phylogenetic tree provided, the support value presented is PP (posterior probability). This particular support value was used because Bayesian inference was used to construct the tree. PP values range from 0 to 1 and indicate the proportion of times that a particular clade is supported by the data.
7. The phylogenetic analysis utilized combined data sets. The authors explained that the combined analysis is a powerful tool that can increase the accuracy and resolution of phylogenetic trees, especially when the datasets are not in conflict with each other.

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2. Explain why ampicillin acts as an functions in bacteria. antibiotic, and the mechanism whereby the ampi gene [2]

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Ampicillin is an antibiotic that acts by inhibiting bacterial cell wall synthesis. It belongs to the class of antibiotics called penicillins and specifically targets the enzymes involved in the construction of the bacterial cell wall.

The mechanism of action of ampicillin involves interfering with the transpeptidation step of peptidoglycan synthesis. Peptidoglycan is a crucial component of the bacterial cell wall responsible for maintaining its structural integrity. It consists of alternating units of N-acetylglucosamine (NAG) and N-acetylmuramic acid (NAM), cross-linked by short peptide chains. Ampicillin works by binding to and inhibiting the transpeptidase enzymes known as penicillin-binding proteins (PBPs). These enzymes are responsible for catalyzing the cross-linking of the peptide chains in peptidoglycan. In summary, ampicillin acts as an antibiotic by inhibiting bacterial cell wall synthesis through the inhibition of transpeptidase enzymes.

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Name at least 3 specific facts that archaeologists have discovered about the Great Pyramid of Khufu? (2 points)

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Archaeologists have made several significant discoveries about the Great Pyramid of Khufu, also known as the Pyramid of Cheops.

Here are three specific facts:

1. Construction Techniques: Archaeologists have found evidence that the Great Pyramid was built using a technique called "quarry marks." These marks are inscriptions made by the pyramid builders to indicate the specific quarry location of the stones. This discovery provides insights into the construction methods and organization of the workforce involved in building the pyramid.

2. Internal Structure: Exploration of the pyramid's interior has revealed a complex network of passages and chambers. One of the most remarkable discoveries is the "King's Chamber," located near the pyramid's center. This chamber contains a granite sarcophagus but no evidence of a mummy. The purpose of the chamber remains a subject of debate among archaeologists and Egyptologists.

3. Boat Pits: In 1954, archaeologist Kamal el-Mallakh discovered five boat pits near the Great Pyramid. These pits contained disassembled boats believed to be funerary barges associated with Khufu's burial rituals. One of the boats, known as the Khufu Ship, has been meticulously reconstructed and is now on display near the pyramid complex.

These discoveries offer valuable insights into the construction techniques, internal structure, and burial rituals associated with the Great Pyramid of Khufu, contributing to our understanding of ancient Egyptian civilization and monumental architecture.

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Describe the property of lipids that makes them a better energy source than proteins or carbohydrates. Refer to bond energy in your description.

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Lipids are an excellent source of energy as they are the primary components of cellular membranes and carry out various functions in the human body. Lipids also have the highest energy density of all macronutrients and can generate more energy than carbohydrates or proteins per unit of weight.

Lipids are energy-dense due to the high number of carbon-hydrogen bonds that they contain. They also have lower levels of oxygen compared to carbohydrates and proteins, which means that they can generate more energy per molecule. The reason why lipids have more energy per molecule is that carbon-hydrogen bonds store more energy than oxygen-hydrogen bonds found in carbohydrates and proteins. As a result, when the body breaks down lipids, more energy is released than when carbohydrates and proteins are broken down.Lipids are also insoluble in water, and this property enables them to be stored in adipose tissues.

They can be broken down and released into the bloodstream to provide a long-lasting source of energy when there are no other energy sources available to the body. As a result, lipids can be stored for more extended periods and used by the body as an energy source when carbohydrates and proteins are not available.

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Exposure of zebrafish nuclei to cytosol isolated from eggs at metaphase of mitosis resulted in phosphorylation of NEP55 and L68 proteins by cyclin-dependent kinase 2. NEP55 is a protein of the inner nuclear membrane, and Les is a protain of the nuclear lamina. What is the most lkely role of phosphorylation of thase proteins in the process of mintois? a. They are incolved in chromosome condensation b. They are involved in migration of centrospmes to coposite sides of the nucleus. c. They are involved in the disassembly of the nuclear envelope
d. They eriafie the anachment of apindle mierecutoules to knetochares

Answers

The phosphorylation of NEP55 and L68 proteins by cyclin-dependent kinase 2 in zebrafish is most likely involved in the disassembly of the nuclear envelope during mitosis.

The process of mitosis involves several key events, including the condensation of chromosomes, the migration of centrosomes to opposite sides of the nucleus, the disassembly of the nuclear envelope, and the attachment of spindle microtubules to kinetochores. Among the given options, the most likely role of the phosphorylation of NEP55 and L68 proteins is in the disassembly of the nuclear envelope.

NEP55 is a protein of the inner nuclear membrane, while L68 is a protein of the nuclear lamina. Phosphorylation of these proteins by cyclin-dependent kinase 2 suggests that they are targeted for modification during mitosis. Phosphorylation events are known to play a crucial role in regulating the disassembly of the nuclear envelope, allowing for the separation of the nuclear contents from the cytoplasm and facilitating chromosome segregation. Therefore, the phosphorylation of NEP55 and L68 proteins is likely involved in the disassembly of the nuclear envelope, which is a critical step in mitotic progression.

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What is the risk that Optometry could pose to the public?
What could go wrong?
What dangerous substances/machines/tools/ techniques might be used?

Answers

Optometry, like any healthcare profession, carries certain risks that could potentially pose a threat to the public.

While the overall risk is relatively low, there are some potential concerns that should be addressed. One potential risk is misdiagnosis or incorrect prescriptions. Optometrists play a crucial role in assessing vision health and prescribing corrective measures such as glasses or contact lenses. If there are errors in the examination or prescription process, it could lead to suboptimal vision correction or even exacerbate existing eye conditions.

Another risk involves the improper use of medical instruments or equipment during eye examinations. For instance, incorrect handling or calibration of machines used for measuring intraocular pressure (tonometry) or examining the back of the eye (ophthalmoscopy) could result in inaccurate readings or potential harm to the patient.

Additionally, there is a risk of adverse reactions or complications related to certain substances used in optometric procedures. For instance, during eye examinations, eye drops containing dilating agents are sometimes used to facilitate examination of the retina. While adverse reactions to these eye drops are rare, there is a minimal risk of allergic reactions or other side effects.

It's important to note that optometrists undergo extensive training and follow strict protocols to mitigate these risks and ensure patient safety. Regular audits, quality control measures, and adherence to professional standards help minimize the chances of errors or dangerous situations arising.

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Which base normally pairs with this structure: O a. Thymine O b. Adenine O c. Cytosine O d. Guanine

Answers

The base that normally pairs with the structure given is adenine (b). In DNA bases, adenine (A) normally pairs with thymine (T), and guanine (G) pairs with cytosine (C).  Option b is correct answer.

These base pairs are formed through hydrogen bonding. Adenine and thymine form two hydrogen bonds, while guanine and cytosine form three hydrogen bonds.

In the given structure, the specific base that pairs with it is not provided. However, based on the options given, adenine (A) is the correct choice. Adenine is one of the four nitrogenous bases found in DNA bases, and it forms a complementary base pair with thymine (T). Thymine contains a structure that can hydrogen bond with adenine, forming two hydrogen bonds between them.

Therefore, when adenine is present in one DNA strand, its complementary base pair in the opposite strand will be thymine. This base pairing is essential for the accurate replication and transcription of DNA, ensuring the proper transmission of genetic information.

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Which of the following classes has the most species of Subphylum Vertebrata? Petromyzontida Actinopterygii Amphibia Reptilia O Mammalia 1.5 pts Question 75 The evolutionary novelty that evolved after the ancestors of Myxini and Petromyzontida and is present in Chondrichthyes, Actinopterygii, and Sarcopterygii is/are O gills O paired fins O keratinized skin the transverse line a heavily armored skin.

Answers

Actinopterygii has the most species of Subphylum Vertebrata. Actinopterygii (ray-finned fishes) has the most species of Subphylum Vertebrata with more than 30,000 known species.

In comparison, Mammalia has approximately 5,500 species, Reptilia has 10,000 species, Amphibia has 7,000 species, and Petromyzontida has only 34 species. Actinopterygii is characterized by bony, ray-like spines that support their fins and by a swim bladder for buoyancy. They are divided into two categories based on the location of their fins; those with the fins nearer to their tail are called teleosts, while those with the fins further back on their body are called chondrosteans.

Chondrosteans are considered primitive bony fishes, while teleosts are considered the most advanced bony fishes. Evolutionary novelties that developed after the ancestors of Myxini and Petromyzontida and are found in Chondrichthyes, Actinopterygii, and Sarcopterygii include gills and paired fins. Therefore, the correct answer is option B: paired fins.

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facilitated diffusion require? enzymescarrier proteinslipid carrierscarbohydrate carrierslipid or carbohydrate carriers

Answers

Facilitated diffusion is a process of passive transport that requires carrier proteins or channels to facilitate the movement of specific molecules across a cell membrane.

Facilitated diffusion is a type of passive transport that allows specific molecules to move across a cell membrane from an area of higher concentration to an area of lower concentration. Unlike simple diffusion, which relies on the concentration gradient and the physical properties of molecules, facilitated diffusion requires the assistance of carrier proteins or channels.

Enzymes are one type of carrier protein involved in facilitated diffusion. They can bind to specific molecules and undergo a conformational change to transport them across the membrane. Enzymes are often involved in the transport of small molecules, such as ions or sugars.

Carrier proteins are another important component of facilitated diffusion. These proteins have specific binding sites for particular molecules. When the molecule binds to the carrier protein, it undergoes a change in shape, allowing it to pass through the membrane and be released on the other side. Carrier proteins are involved in transporting larger molecules, such as amino acids or larger sugars.

In addition to carrier proteins, facilitated diffusion can also utilize lipid or carbohydrate carriers. Lipid carriers, such as lipoproteins, can transport lipid-soluble molecules across the membrane. Carbohydrate carriers, on the other hand, are specialized proteins that transport carbohydrates, such as glucose, across the membrane.

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Which of the following statements about venous blood pressure are true? Select ALL correct answers: A) Pressure in veins is much lower than pressure in arteries B) Arterial blood pressure fluctuates as the heart squeezes and relaxes, but venous blood pressure does not C) Valves in veins help prevent blood from flowing backward D) Contraction of skeletal muscles in the legs helps move venous blood back up towards the heart

Answers

All the following statements about venous blood pressure are true. Therefore, the correct options are: A, C, and D. Venous blood pressure is the pressure that is exerted by the circulating blood against the walls of veins, particularly the venae cavae and pulmonary veins.

There are various facts that surround this type of blood pressure and they include:

A. Pressure in veins is much lower than pressure in arteries: this statement is true. This is because the pressure in arteries is higher than that in veins. Arteries have a high-pressure rate that is due to the high volume of blood flowing through them in a short amount of time. Blood vessels act as conduits of blood circulation, and veins' thin walls help in the flow of low-pressure blood flow.

Therefore, statement A is correct.

B. Arterial blood pressure fluctuates as the heart squeezes and relaxes, but venous blood pressure does not: This statement is false. Venous blood pressure fluctuates when the heart squeezes and relaxes, and this is referred to as pulse pressure. However, venous pressure fluctuates more gently than arterial pressure. Therefore, statement B is incorrect.

C. Valves in veins help prevent blood from flowing backward: This statement is true. The venous system has valves that function to prevent the backward flow of blood and assist in the return of blood to the heart. The vein's wall contracts the blood and the valves in the veins block the blood from flowing backward. Therefore, statement C is correct.

D. Contraction of skeletal muscles in the legs helps move venous blood back up towards the heart: This statement is true. Venous blood returns to the heart through muscular contractions. The contraction and relaxation of the muscle's veins move the blood back to the heart, as the valves prevent blood from flowing backward. Therefore, statement D is correct.

The correct options that explain venous blood pressure are A, C, and D.

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Identify the true statement describing Celiac disease.
Select one:
a. gluten in wheat, barley and rye triggers an autoimmune reaction within the small intestine, leading to inflammation and malnutrition
b. Celiacs can eat gluten freely once they have been properly vaccinated
c. inflammation destroys the large intestinal wall, leading to severe and persistent chronic pain
d. severe forms of this condition are usually treated with surgery

Answers

Celiac disease, also known as celiac sprue or gluten-sensitive enteropathy, is a genetic autoimmune disease that affects around one percent of the population and occurs in response to consuming gluten, which is a protein found in wheat, barley, and rye.

Gluten triggers an immune response in the small intestine, causing inflammation, which damages the villi and causes malabsorption of nutrients.

Option a is the true statement that describes Celiac disease. The consumption of gluten, which is found in wheat, barley, and rye, triggers an autoimmune response within the small intestine, leading to inflammation and malnutrition. Celiac disease is a genetic autoimmune disorder that affects approximately one percent of the population. Gluten triggers an immune response in the small intestine, which causes inflammation, which damages the villi and leads to malabsorption of nutrients.

Celiac disease symptoms vary from person to person and can include diarrhea, abdominal pain, bloating, fatigue, weight loss, and anemia. The only treatment for celiac disease is to follow a gluten-free diet, which means avoiding all foods that contain gluten. Gluten-free oats, fruits, vegetables, and proteins can be consumed by individuals with celiac disease. Vaccines are not a cure for celiac disease, nor can they help to mitigate the symptoms. Surgery is not typically required for celiac disease treatment, but severe cases may require medical intervention.

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Why are the shape, orientation and location of the protein encoded by mc1r gene important in the fulfillment of its role?
Using the diagram below, describe the chain of events of protein synthesis of the MC1R protein. Starting from the mc1r gene (point A), indicate the molecules and details of the role of the process involved in each of the numbered steps 1-6.
Using the same diagram, describe the pathway which is triggered at point 7. Include in your answer the molecules and processes involved in each of the numbered steps 7-11.

Answers

The shape, orientation, and location of the protein encoded by the MC1R gene are important for its role because they determine the protein's functionality and interaction with other molecules. The specific shape of the protein allows it to bind to specific molecules, such as melanocyte-stimulating hormone (MSH), and activate signaling pathways involved in pigmentation regulation.

In protein synthesis (steps 1-6), the MC1R gene is transcribed into mRNA (step 1), which is then processed and transported out of the nucleus (step 2). The mRNA binds to ribosomes (step 3), and the ribosome reads the mRNA sequence to synthesize the corresponding amino acids (step 4). These amino acids are linked together to form a polypeptide chain (step 5), which folds into a specific 3D structure to become the MC1R protein (step 6).

In the pathway triggered at point 7, the MC1R protein interacts with MSH (step 7), leading to activation of the cAMP signaling pathway (step 8). This pathway activates enzymes, such as protein kinase A (PKA), which phosphorylate downstream proteins (step 9). Phosphorylated proteins initiate a series of cellular responses, such as the production of melanin, which determines skin and hair pigmentation (step 10). These responses ultimately lead to changes in pigmentation, such as tanning or red hair color (step 11).

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Which statement about protein sequencing by mass spectrometry is TRUE? A1. Cleavage by proteases like trypsin is used to ensure the protein fragments are charged. A2. The difference in mass between two fragments will often correspond to the mass of one amino acid. The first stage of tandem MS/MS occurs in aqueous solution, and the second stage occurs in gas phase. A4. The mass-charge ratio in mass spectrometry is roughly constant for all polypeptides. AS. None of the above are true. CQ4-19 (WSf Polypeptide backbone geometry) Which of these amino acids contain TWO chiral carbons each? AI. I and V. A2. L and V. A3. I and T. A4. L and T. AS. None of the above pairs of amino acids contain two chiral carbons each. A3.

Answers

The statement that is TRUE about protein sequencing by mass spectrometry is A2. The difference in mass between two fragments will often correspond to the mass of one amino acid.

Mass spectrometry is a powerful technique used for protein sequencing, where proteins are fragmented and their masses are analyzed. The mass difference between two adjacent fragments can often be attributed to the presence of a single amino acid residue, allowing for the determination of the protein sequence. This mass-based approach is widely used in proteomics research.

Regarding the question about amino acids containing two chiral carbons each, the correct answer is A3. I and T. Isoleucine (I) and threonine (T) are the amino acids that have two chiral carbons each in their structure. Chiral carbons are carbon atoms that are bonded to four different groups, resulting in the possibility of having two different spatial arrangements (R and S configurations). This property gives rise to optical isomerism in amino acids and plays a crucial role in their biological activity.

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GEFEL I 8 EE E C The structure shown in this image represents which part of a cell? Integral protein Integral protein Endoplasmic membrane Questions Filter (10) Y Pore Channel Polar head (hydrophilic)

Answers

The structure shown in this image below represents the Plasma membrane.

What is the plasma membrane?

Plasma membrane is the outer membrane of a cell. It is a phospholipid bilayer that separates the cell from its environment.

The plasma membrane is responsible for regulating the movement of substances into and out of the cell. It also plays a role in cell signaling and cell adhesion.

Integral proteins are proteins that are embedded in the membrane of a cell. They can be either transmembrane proteins, which extend all the way through the membrane, or peripheral proteins, which are attached to the surface of the membrane.

The above answer is based on the full question below;

The structure shown in this image represents which part of a cell? Pore Channel Integral protein Integral protein Polar head hydrophilic Fatty acid tal (hydrophobic)

A Nucleus

B) Lysosomes

C) Plasma membrane
D) Endoplasmic membrane

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Imagine that you are a scientist who develops a qPCR-based diagnostic kit for Covid-19. As such you should design the primers targeting the genome of SARS-Cov-2. In this project, you will design two primer pairs that can target the Spike (S) gene and Envelope (E) gene of the virus. The related gene sequences could be found from NCBI with the gene IDs 1489668 and 43740570, respectively. Once to find the genes please design your primers with Primer-Blast, analyze them considering the primer parameters, and analyze the self-dimer and hetero-dimer possibilities by using the IDT oligo-analyzer.

Answers

The two primer pairs for the Spike (S) gene and Envelope (E) gene of the virus, were designed using Primer-Blast and analysed for primer parameters and possible dimer formation.

Polymerase chain reaction (PCR) is a widely used technique for amplifying the specific DNA sequences by a factor of 10^6-10^9. qPCR is a modified version of PCR in which the amplified DNA is quantified in real-time with the help of fluorescent dyes or probes. In the given project, two primer pairs were designed that can target the Spike (S) gene and Envelope (E) gene of the virus using Primer-Blast. The primers were analysed for primer parameters, such as melting temperature (Tm), GC content, length, and specificity using the IDT oligo-analyzer.

Primer pairs were also checked for the possibility of dimer formation, such as self-dimer and hetero-dimer by the same method. The primers designed for qPCR amplification should have the ability to generate specific, reproducible, and reliable results that can help in the accurate detection of the virus. The designed primers could be used in qPCR-based diagnostic kits for Covid-19 and can play a significant role in the early detection and prevention of the virus's spread.

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Assume the diameter of the field of vision in your microscope is 3 mm under low power. If on Bacillus cell is 3um, how many bacillus cells could fit end to end across the field? How many 20 um yeast cells could fit across the field? Show your work.

Answers

Under low power with a field of vision diameter of 3 mm, approximately 1000 Bacillus cells could fit end to end across the field. This calculation is based on the assumption that the Bacillus cells are each 3 μm in size.

By dividing the diameter of the field (converted to micrometers) by the size of the Bacillus cell, we obtain the number of cells that can fit. In the case of yeast cells measuring 20 μm in size, the same calculation indicates that approximately 150 yeast cells could fit across the field.

It's important to note that these calculations assume a perfect arrangement of cells without any overlap or gaps, which may not be entirely accurate in real-world microscopy.

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Zoology experiment: The Predator-prey Interactions Between Zebrafish and Daphnia
1. Six 1-L beakers were filled with aged tap water.
2. To test the effect of light on the survival of Daphnia, the 6 beakers were divided equally into 2 treatments: light & dark. Beakers assigned to the dark treatment were covered w/ aluminum foil.
3. One zebrafish (about 2-3 cm) starved for 24 hours was placed in each beaker.
4. Fifty (50) Daphnia sp. individuals were added in each beaker containing the starved zebrafish. The top of the beakers assigned to the dark treatment were covered with aluminum foil.
5. One hour after, the zebra fish was scooped out & the no. of surviving Daphnia in each set-up were counted.
QUESTIONS:
1. What would be your hypothesis in this experiment?
2. What is your basis for formulating that hypothesis?
3. What do you think will happen to the survival rate of Daphnia when exposed to its predator under well-lit environment? In a completely dark set-up?

Answers

In this experiment, I hypothesize that the presence of a zebrafish predator (e.g. a starved zebrafish) will have a negative impact on the survival rate of Daphnia, which will be greater when exposed to light than in a completely dark set-up.

This is based on the fact that a well-lit environment will facilitate better visibility for the zebrafish, and thus higher predation efficiency. This is in contrast to a completely dark set-up, where the zebrafish will not be able to detect the Daphnia as easily, and so predation efficiency will be lower.

As the presence of zebrafish in the environment will effectively be a top-down control that determines the population size of Daphnia, it is likely that the observed change in the Daphnia’s survival rate will be greater when the zebrafish is experienced in a light environment, as opposed to a dark environment.

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A "symptom" is an objective finding which is discovered during
the physical examination.
A. True
B. False

Answers

The statement "A 'symptom' is an objective finding which is discovered during the physical examination" is false.

A symptom is defined as any subjective evidence or change in a patient's physical or mental condition, such as discomfort, pain, or fatigue, that is experienced by the patient and not observable by the physician. It is essential to note that the patient describes or reports a symptom rather than the physician discovering it during the physical examination.

Signs are objective measures discovered by a physician during a physical examination that can be seen, heard, measured, or felt. Signs can be obtained through laboratory tests, radiological imaging, or other diagnostic procedures.The differentiation between signs and symptoms is crucial because they have different diagnostic values. A patient's symptoms can direct the clinician toward a diagnosis, whereas signs assist in verifying or confirming the suspected diagnosis, which aids in the development of an appropriate management plan.

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QUESTION 28 A small population of Alrican Green monkeys is maintained for scientific medical research on the island of St. Kis Scienfaits discover that an alle be) in the population may be the cause of susceptibility to a herpes virus that infects T cels. Heterozygous monkeys (H1, H2) as well as homozygout (12, H2) monkeys are qually susceptible. This virus is known to be lethal in that it causes Tool lymphomas (cancer). A genetic screen of al 100 mionkeys held in captivity revealed that the H2 alele was present at a frequency of 0.7 The actual number of monkeys that are homozygous for this allelo (H2H2) is 25 Using the Hardy Weinberg equilibrium variables what is the expected number of homozygous monkeys (1212) in this population? QUESTION 29 A small population of African Green monkeys is maintained for scientfic medical research on the island of St Kits Scientists discover that an allelo (2) in the population may be the cause of susceptibility to a herpes virus that infects Tools Heterozygous monkeys (H1, H2) as well as homorygoun (2.2) monkeys are equally susceptible. This virus is known to be lethal in that it causes col lymphomas (cancer) A goale screen of all 100 monkeys held in captivity revealed the the H2 ailele was present at a frequency of 07. The actual rumber of monkeys that are homozygous for this all (H22) is 25 Using Hardy-Weinberg variables, how many monkeys in this population would be expected to be susceptible to the virus? 3) what is the frequency of the H1 allele 4) is the population in hardy weinberg equilibrium?

Answers

28) A small population of African Green monkeys is maintained for scientific medical research on the island of St. Kits. Scientists discover that an allele (H2) in the population may be the cause of susceptibility to a herpes virus that infects T cells.

Heterozygous monkeys (H1, H2) as well as homozygous (H2, H2) monkeys are equally susceptible. This virus is known to be lethal in that it causes Tool lymphomas (cancer). A genetic screen of all 100 monkeys held in captivity revealed that the H2 allele was present at a frequency of 0.7. The actual number of monkeys that are homozygous for this allele (H2H2) is 25.

The frequency of H2 in the population = p = 0.7. Therefore, the frequency of H1 in the population = q = 1 - 0.7 = 0.3We know that p2 + 2pq + q2 = 1 (Hardy-Weinberg equilibrium equation)The frequency of H2H2 monkeys can be given as q2 * total number of individuals in the population= 0.3 * 0.3 * 100= 9. Expected number of homozygous monkeys (H2H2) in this population = 9

29) A small population of African Green monkeys is maintained for scientific medical research on the island of St. Kits. Scientists discover that an allele (H2) in the population may be the cause of susceptibility to a herpes virus that infects T cells. Heterozygous monkeys (H1, H2) as well as homozygous (H2, H2) monkeys are equally susceptible. This virus is known to be lethal in that it causes col lymphomas (cancer). A genetic screen of all 100 monkeys held in captivity revealed the H2 allele was present at a frequency of 0.7. The actual number of monkeys that are homozygous for this allele (H2H2) is 25.

The frequency of H2 in the population = p = 0.7. Therefore, the frequency of H1 in the population = q = 1 - 0.7 = 0.3Heterozygous frequency = 2pq = 2 × 0.7 × 0.3 = 0.42Homozygous dominant frequency = p2 = 0.72 = 0.49Homozygous recessive frequency = q2 = 0.32 = 0.09Expected number of individuals susceptible to the virus = (0.42 + 0.09) * 100 = 51

Frequency of H1 = q = 1 - p = 1 - 0.7 = 0.3Is the population in Hardy-Weinberg equilibrium. No, the population is not in Hardy-Weinberg equilibrium.

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INTEGRINS AND THE ECM
Some have speculated that Abraham Lincoln had Marfan’s syndrome.
A) Describe the clinical features of Marfan’s syndrome
B) Which glycoprotein is mutated in Marfan’s syndrome? What is the function of that glycoprotein
normally in the body and where is it found?

Answers

A) Marfan syndrome is an inherited condition of the connective tissues that affects multiple organ systems in the body. The severity of the condition can range from mild to severe. B) The primary function of fibrillin-1 is to help regulate the formation of elastic fibers in the ECM.

INTEGRINS AND THE ECM:

Some have speculated that Abraham Lincoln had Marfan’s syndrome.

A) Clinical Features of Marfan Syndrome:

Marfan syndrome is an inherited condition of the connective tissues that affects multiple organ systems in the body. The severity of the condition can range from mild to severe.

Clinical features of Marfan syndrome are as follows:

Cardiovascular: Aortic root dilation, aortic dissection, mitral valve prolapse, tricuspid valve prolapse, pulmonary artery dilation

Ocular: Ectopia lentis, myopia, cataracts, retinal detachment

Skeletal: Joint hypermobility, scoliosis, pectus excavatum, pectus carinatum, arachnodactyly, dolichocephaly, tall stature, disproportionate limbs, reduced upper segment/lower segment ratio, acetabular protrusion, kyphosis, dural ectasia

Pulmonary: Spontaneous pneumothorax

B) Glycoprotein that is mutated in Marfan’s Syndrome:

Marfan syndrome is caused by a mutation in the FBN1 gene that produces fibrillin-1, a structural protein component of the extracellular matrix (ECM).

It affects the body’s connective tissue, causing it to weaken and stretch.

The primary function of fibrillin-1 is to help regulate the formation of elastic fibers in the ECM.

Fibrillin-1 is abundant in the aorta, heart valves, ciliary zonules, and other connective tissues.

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