in
5 sentences how woukd you explain how water in the human body helps
regulate body temperature following an intense workout?

Answers

Answer 1

Water plays a crucial role in regulating body temperature after an intense workout through several mechanisms:

Cooling through sweat: During exercise, the body produces sweat to cool down. Sweat is mainly composed of water, and when it evaporates from the skin, it helps dissipate heat, lowering body temperature.

Thermal regulation: Water has a high specific heat capacity, which means it can absorb and retain a significant amount of heat. By consuming water, the body can absorb excess heat generated during exercise, preventing overheating.

Circulatory support: Adequate hydration ensures optimal blood volume and circulation. This helps deliver oxygen and nutrients to the working muscles, aiding in temperature regulation by promoting efficient heat dissipation.

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Related Questions

HDAC's are important enzymes involved in the regulation of Gene expression. This is because
a.
they add methyl groups from histones creating less gene expression.
b.
they create euchromatic structure by adding acetyl groups to cytosine.
c.
They create the Z form of DNA by removing acetyl groups from cytosines.
d.
they add methyl groups onto cytosines on DNA and create a heterochromatic structure.
e.
they remove acetyl groups from histones creating less gene expression.

Answers

HDAC's or histone deacetylases are important enzymes involved in the regulation of gene expression.

These enzymes remove acetyl groups from histones that are bound to DNA, causing the chromatin to become more compact and restrict the transcription machinery, resulting in a decrease in gene expression.

Hence, option E, "they remove acetyl groups from histones creating less gene expression" is the correct answer.

Let us understand the concept of HDAC's and their role in gene expression: Gene expression is the process in which the genetic information present in DNA is converted into functional proteins. The expression of genes can be controlled by several mechanisms, including epigenetic modifications. Epigenetic modifications are changes that occur in DNA and its associated proteins without altering the nucleotide sequence.

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The scientific study of organisms that are too small to be seen by the unaided human eye
Polysaccharide composed of alternating repeats of N-acetylglucosamine and N-acetylmuramic acid and cross-linked by peptides that can be broken down by lysozyme in your saliva.
Occurrence, distribution and patterns of health and disease in populations of hosts.
The effects of two chemotherapeutical agents used together is greater than the sum of their effects when used individually.
Symbiosis between one or more species of fungi and a photosynthetic microorganism
Disruption of the normal microbiota within a host
Class for the causative agent for the cholera epidemic
Aligning DNA fragments in the correct order to eliminate overlaps
Genetic content that includes genes shared by all strains within a species and all genes specific to some strains
Quantitative measure of the ability of a pathogen to produce disease

Answers

a. The scientific study of organisms that are too small to be seen by the unaided human eye is known as microbiology. Microbiology involves the investigation of microorganisms such as bacteria, viruses, fungi, and protozoa, which play crucial roles in various biological processes and can have significant impacts on human health, the environment, and industry.

b. The polysaccharide described is known as peptidoglycan, which is a major component of bacterial cell walls. Peptidoglycan provides structural support to the bacterial cell and protects it from osmotic stress. It consists of repeating units of N-acetylglucosamine (NAG) and N-acetylmuramic acid (NAM), which are cross-linked by peptides. This network of cross-linked peptidoglycan provides strength and rigidity to the cell wall.

c. The study of the occurrence, distribution, and patterns of health and disease in populations of hosts is known as epidemiology. Epidemiologists investigate various factors, including the spread of diseases, risk factors, transmission routes, and the impact of interventions.

d. The phenomenon described is known as synergism or synergistic effect. When two chemotherapeutic agents are used together, their combined effect is greater than the sum of their individual effects. This occurs when the agents interact with each other in a way that enhances their effectiveness against the target organism.

e. The symbiotic relationship between one or more species of fungi and a photosynthetic microorganism, typically a green alga or cyanobacterium, is known as lichen. Lichens are composite organisms where the fungal partner provides a protected environment and nutrients to the photosynthetic partner, while the photosynthetic partner produces organic compounds through photosynthesis.

f. Disruption of the normal microbiota within a host refers to dysbiosis. The human body harbors a complex and diverse community of microorganisms, collectively known as the microbiota, which plays a crucial role in maintaining health and homeostasis. However, various factors such as antibiotics, diet, stress, and disease can disrupt the balance of the microbiota, leading to dysbiosis.

g. The causative agent for the cholera epidemic is a bacterium called Vibrio cholerae. Cholera is a severe diarrheal disease that is primarily transmitted through contaminated water or food. Vibrio cholerae produces a toxin known as cholera toxin, which causes the characteristic watery diarrhea associated with the disease.

h. The process of aligning DNA fragments in the correct order to eliminate overlaps is known as DNA sequencing assembly or sequence assembly. In DNA sequencing, the genetic material is fragmented into smaller pieces, and the sequence of these fragments is determined.

i. The genetic content that includes genes shared by all strains within a species and all genes specific to some strains is known as the core genome and the accessory genome, respectively. The core genome refers to the set of genes that are present in all strains within a particular species. These genes typically encode essential functions and are conserved across the species. On the other hand, the accessory genome consists of genes that are present only in some strains within the species. These genes can confer additional traits or capabilities to the specific strains, such as antibiotic resistance, virulence factors, or metabolic adaptations.

j. The quantitative measure of the ability of a pathogen to produce disease is known as virulence. Virulence factors are characteristics or molecules possessed by pathogens that enable them to cause disease in a host.

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With Parkinson's Disease, ____________________________.
Select one or more:
a. long-term exposure to pesticides is associated with an increased risk for developing symptoms
b. etiology of early & late-onset forms are primarily genetic in origin
c. cured through treatments combining use of L-Dopa with occupational & electroconvulsive stimulation therapies
d. progressive onset of symptoms include a loss of motor control, shakes, tremors, rigidity, disordered affect and mood, chronic fatigue
e. abnormal clumping of Tau proteins interfere with neurotransmission in the Substantia nigra

Answers

Parkinson's Disease is a degenerative disorder of the central nervous system (CNS) that manifests through progressive symptoms such as loss of motor control, shakes, tremors, rigidity, disordered affect and mood, and chronic fatigue.

It is caused by the death of dopaminergic neurons in the brain that synthesize dopamine. As a result, the CNS becomes deficient in dopamine, leading to abnormal movement patterns that resemble the symptoms of Parkinson's Disease. Additionally, Parkinson's Disease is associated with an abnormal clumping of Tau proteins, which interfere with neurotransmission in the Substantia nigra. The etiology of early and late-onset forms of Parkinson's Disease is primarily genetic in origin, but it may also be caused by long-term exposure to pesticides, which is associated with an increased risk for developing symptoms.

While there is no cure for Parkinson's Disease, treatments combining the use of L-Dopa with occupational and electroconvulsive stimulation therapies can help improve symptoms and quality of life for patients. However, the effectiveness of these treatments depends on the severity of the symptoms, age, and overall health of the patient. Therefore, early diagnosis and treatment are essential for improving the prognosis of Parkinson's Disease patients.

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In the random sampling method of estimating (not getting an exact count of) population size, which equation is used? a) population = population density/number of quadrats b) population density = number of organisms counted/area or volume studied c) population change = [births + immigration] - [deaths + emigration] d) population = number of organisms recaptured x number of organisms originally marked/number of individuals marked and recaptured

Answers

The random sampling method of estimating population size utilizes the Lincoln-Petersen index or the mark and recapture method.

This method involves capturing and marking a sample of individuals from the population, releasing them back into the environment, and then recapturing a second sample at a later time. By comparing the number of marked individuals in the second sample to the total number of individuals in the first sample, an estimate of the population size can be obtained using the formula:

Population Size = (Number of Individuals in First Sample) x (Number of Individuals in Second Sample) / (Number of Marked Individuals in Second Sample). This approach allows for estimating population size without having to count every individual.

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please send the solution for above question in 1 hr . I will upvote
you .
QUESTIONS
2A
An arthropod called a Cyclops has antennae that are either smooth or Rough. The allele for Rough (R) is dominant over smooth (r). In the same organism Non-resistance to pesticides (P) is dominant over resistance to pesticides (p).
i) Pesticide resistant smooth antennae cyclops is crossed to the double heterozygous one. Write the genotypes of the parents, show the crosses with the help of Punnett square and write the phenotype and genotype ratio for the crosses. ii) How many genotypes are possible for pesticide resistance irrespective of the antennae texture? Write all genotypes. 2B
The Duchenne's Muscular Dystrophy (DMD) is an X-linked recessive trait due to deletion or point mutation in the dystrophin gene leading to its defective production.
i) If affected male has a child with a carrier woman, what is the probability that the child will be affected daughter? Show the crosses and Write the genotype for both the cases if she is affected. ii) If unaffected male marries a carrier woman what is the probability that the child will be affected daughter? Show the crosses and write the genotype of the child. 2C
A brown-eyed woman whose father had blue eyes and mother had brown eyes marries a brown-eyed man, whose parents are also brown-eyed. But they have a daughter who is blue-eyed.
i) Draw a pedigree chart for both the family (the two parents) using proper symbol. ii) Indicate each individual's possible genotypes.
iii) Identify the mode of inheritance for the blue eyes

Answers

2A)i) The genotype of pesticide resistant smooth antennae cyclops (RrPp) crossed to double heterozygous (RRPp) is given below

ii) For pesticide resistance, irrespective of the antennae texture, there are four possible genotypes. These are Pp, PP, pp, and pP.

2B)i) If an affected male (XdY) has a child with a carrier woman (XDXd), the probability of having an affected daughter (XdXd) is 50% and the probability of having an affected son (XdY) is 50%.!

ii) If an unaffected male (XDY) marries a carrier woman (XDXd), the probability of having an affected daughter (XdXd) is 25%, the probability of having an unaffected daughter (XDXd) is 25%, the probability of having an unaffected son (XDY) is 25%, and the probability of having an affected son (XdY) is 25%.!

2C)i) The pedigree chart is shown below

ii) Possible genotypes for each individual are shown below:Brown-eyed woman with blue-eyed father and brown-eyed mother: BbBlue-eyed daughter: bbBrown-eyed man: BB or Bb

iii) The mode of inheritance for blue eyes is a recessive trait that is autosomal.

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Which of the following statements correctly identifies the difference between plant and animal cell division?
Select one:
a. The cell walls of plants prevent the process of cytokinesis.
b. Plants cells lack centrioles, and they form a cell plate during cytokinesis.
c. Both plant and animal cells undergo mitosis and cytokinesis, but they lack the interphase.
d. Plant cells lack centromeres, and they form a cell wall to produce two daughter cells.

Answers

The correct statement that identifies the difference between plant and animal cell division is b. Plant cells lack centrioles, and they form a cell plate during cytokinesis.

Plant and animal cells undergo a similar process of cell division called mitosis, which consists of several stages including prophase, metaphase, anaphase, and telophase. However, there are key differences in the way cytokinesis, the division of the cytoplasm, occurs in plant and animal cells.

In animal cells, cytokinesis involves the formation of a contractile ring of proteins, which constricts the cell membrane at the equator of the dividing cell. This process results in the formation of a cleavage furrow, eventually pinching the cell into two daughter cells.

In contrast, plant cells lack centrioles, which are involved in animal cell cytokinesis. Instead, during cytokinesis in plant cells, a structure called the cell plate forms at the equator of the dividing cell. The cell plate is composed of vesicles containing cell wall materials, such as cellulose. These vesicles fuse together, gradually forming a new cell wall that separates the two daughter cells. The cell plate expands outward until it connects with the existing cell walls, completing the division process.

Therefore, the lack of centrioles and the formation of a cell plate during cytokinesis are the distinguishing features of plant cell division when compared to animal cell division.

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3. 4. 5. 6. List the main products of the light reactions of photosynthesis. Oxygen, ATP, NADPH List the main products of the carbon-fixation reactions of photosynthesis. What are the main events associated with each of the two photosystems in the light reactions, and what is the difference between antenna pigments and reaction center pigments? Describe the principal differences among the C3, C4, and CAM pathways

Answers

The main products of the light reactions of photosynthesis are ATP, NADPH, and oxygen. The main products of the carbon-fixation reactions of photosynthesis are G3P and ADP. The main events associated with each of the two photosystems in the light reactions are light absorption and electron transport.

Photosynthesis is the process by which plants and other autotrophic organisms convert light energy into chemical energy in the form of organic compounds. The process of photosynthesis consists of two main sets of reactions: the light reactions and the carbon-fixation reactions.

The main products of the light reactions of photosynthesis are ATP, NADPH, and oxygen. In the light reactions, light energy is absorbed by antenna pigments and transferred to reaction center pigments. The excited electrons are then transferred through an electron transport chain, ultimately producing ATP and NADPH.

Oxygen is also produced as a byproduct of the light reactions.The main products of the carbon-fixation reactions of photosynthesis are G3P and ADP. In the carbon-fixation reactions, CO2 is fixed into organic compounds using the energy from ATP and NADPH produced in the light reactions.

The initial product of carbon fixation is a three-carbon compound called G3P, which can be used to synthesize glucose and other organic compounds. ADP is also produced in the carbon-fixation reactions.

The main events associated with each of the two photosystems in the light reactions are light absorption and electron transport. Photosystem II absorbs light with a peak absorption at 680 nm, while photosystem I absorbs light with a peak absorption at 700 nm.

Antenna pigments absorb light and transfer the energy to reaction center pigments. Excited electrons are then transferred through an electron transport chain, ultimately producing ATP and NADPH.Antenna pigments and reaction center pigments differ in their ability to absorb light.

Antenna pigments have a broad absorption spectrum and transfer the absorbed energy to reaction center pigments. Reaction center pigments have a narrow absorption spectrum and are responsible for initiating the electron transport chain.

The principal differences among the C3, C4, and CAM pathways lie in the way that carbon is fixed during photosynthesis. C3 plants fix carbon using the enzyme Rubisco in the Calvin cycle. C4 plants use a specialized mechanism to concentrate CO2 in the vicinity of Rubisco, which reduces photorespiration.

CAM plants open their stomata at night to take in CO2, which is stored as an organic acid. The organic acid is then broken down during the day to release CO2 for use in the Calvin cycle.

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Any
suggestions on how I can memorize urine microscopic images for my
urinalysis practical, neumonics, anything that will help it to
stick. I need to knkw casts, crystals, ect. Thanks for any
help!

Answers

Urinalysis is a routine medical examination that involves analysis of the urine. The urine microscopic images are essential for the urinalysis practicals as they are used to identify the presence of casts.

Here are some suggestions to help memorize the urine microscopic images for the urinalysis practicals:1. Start by understanding the normal urine microscopic images: It is essential to have a basic understanding of the normal urine microscopic images.

This will help you to identify the abnormal images easily.2. Break down the images into smaller units: The urine microscopic images can be overwhelming, especially for the beginners. Therefore, it is best to break down the images into smaller units and focus on memorizing one unit at a time.

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A biological marker (or biomarker) is a substance that can indicate a change in physiological state. The specificity of a biomarker can vary from general indications of poor health to specific diagnosis of disease. Creatine kinase (CK) is a serum enzyme of clinical significance. Measurement of total CK is a general indicator of tissue damage. Elevated total CK can be observed in:
All types of muscular dystrophy
Viral myositis
Polymyositis
Acute rhabdomyolysis
Acute myocardial infarction

Answers

Total CK is a general indicator of tissue damage because it is released from damaged muscle fibers, but it can also be elevated in other conditions. Elevated total CK can be observed in all the given conditions.

Creatine kinase (CK) is an enzyme found in muscle and other tissues that play a role in muscle energy metabolism. When muscle fibers are damaged, CK is released into the bloodstream, where it can be measured using a blood test.

Measuring total CK can provide some information about the health of muscle tissue, but it is not a specific diagnostic tool for any particular condition.

Elevated levels of total CK can be seen in various conditions that affect the muscles, including:

All types of muscular dystrophy, which are genetic disorders that cause progressive muscle weakness and degenerationViral myositis, which is an inflammation of the muscles caused by a viral infectionPolymyositis, which is an inflammatory disease that affects the muscles and is characterized by muscle weakness and painAcute rhabdomyolysis, which is a rapid breakdown of muscle tissue that releases CK and other muscle breakdown products into the bloodstreamAcute myocardial infarction, which is a heart attack that can cause muscle damage and release of CK into the bloodstream.

It is important to note that elevated levels of CK can have many causes, and further diagnostic tests and evaluation are necessary to determine the underlying cause of elevated CK levels.

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Classifying Matter: Pure and Impure Substances Name: Date: Purpose: To identify substances as pure or impure based on their composition Legend: black = carbon (C) blue = nitrogen (N) green= chlorine (

Answers

Pure substances are composed of a single type of element or compound, while impure substances contain more than one type of element or compound.

Pure substances are characterized by having a uniform composition throughout, meaning they consist of only one type of element or compound. This could include elements such as carbon (C), nitrogen (N), or compounds like water (H2O) or sodium chloride (NaCl). On the other hand, impure substances, also known as mixtures, contain more than one type of element or compound. These mixtures can be further classified into homogeneous mixtures (uniform composition) or heterogeneous mixtures (non-uniform composition). Impure substances can be separated into their individual components using various separation techniques.

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Which of the following would be a good example of analogous? bacteria resistance to antibiotic and viruses reproduction whales reproduction and dolphins reproduction leg of a horse and human leg tail

Answers

The leg of a horse and a human leg would be a good example of analogous structures.

Analogous structures are those that have similar functions or purposes but do not share a common evolutionary origin. In this case, both the leg of a horse and a human leg serve the purpose of locomotion, allowing the organism to move. However, they have evolved independently in different lineages (horses and humans) and have different anatomical structures.

Bacteria resistance to antibiotics and viruses reproduction, as well as whales reproduction and dolphins reproduction, do not demonstrate analogous structures. Bacteria resistance to antibiotics and viruses reproduction would fall under different biological processes, while whales and dolphins are closely related and have similar reproductive strategies due to their shared ancestry.

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Assignment: Write 1 paragraph (250-300 words) describing ONE of the following topics: 1. What are the ecosystem services provided by coral reefs? What role do coral reefs play in the ecosystem?

Answers

Coral reefs provide numerous ecosystem services that are significant for human well-being. They provide food, medicinal plants, and building materials, offer shoreline security, and encourage tourism and recreation. Coral reefs are home to numerous marine life forms that are vital for the food chain.

There are several types of ecosystem services that coral reefs provide. Coral reefs provide a habitat for marine life forms, as well as supplying food and medicines. They also offer coastal protection and provide a place for tourists to visit. Coral reefs provide numerous ecosystem services that are significant for human well-being. They provide food, medicinal plants, and building materials, offer shoreline security, and encourage tourism and recreation. Coral reefs are home to numerous marine life forms that are vital for the food chain.In addition, coral reefs also play a vital role in the carbon cycle, acting as a carbon sink. Coral reefs have a large surface area and are coated in algae, which removes carbon dioxide from the water through photosynthesis. The carbon that is absorbed is then stored in the coral reef, and therefore out of the atmosphere. Coral reefs are also important in nutrient cycling. Nutrients are brought to the reef through the currents and the tide, and then recycled back into the ecosystem. This allows the coral reef to remain healthy and support the many species that live there.

Coral reefs play a vital role in the ecosystem, providing a wide range of ecosystem services, including food, medicinal plants, and building materials. They also provide a place for tourists to visit and encourage recreational activities. Coral reefs play a vital role in the carbon cycle and nutrient cycling, making them an important part of the ecosystem. The loss of coral reefs can lead to the loss of these ecosystem services and disrupt the balance of the ecosystem. Therefore, it is crucial to protect and conserve coral reefs to ensure their continued existence and to preserve the ecosystem services that they provide.

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Which of the following is an example of geographic isolation?
a. temporal differences in pond breeding
B occupying different geographic locations
c. having differences in chromosome numbers
d. none of the above

Answers

Option (B) occupying different geographic locations is an example of geographic isolation.Geographic isolation is a type of reproductive isolation that happens when two groups of a population of a particular species become separated by a geographic barrier.

The separation may have been caused by a natural disaster, such as a flood, drought, earthquake, or volcanic eruption, or by human activities such as the creation of a highway or the building of a dam. When this happens, the two groups will not be able to interbreed because they cannot physically interact with each other.Over time, this physical separation leads to reproductive isolation. This can occur when one group adapts to a new environment and develops new traits that are advantageous for survival.

If the two groups were to meet and attempt to mate, these different traits may make it difficult or impossible to produce viable offspring. This could lead to the formation of two separate species.For example, two populations of birds may live on opposite sides of a mountain range. Over time, the two populations may develop different physical and behavioral traits that make them better suited to their respective environments. If the two populations were to meet, they may not be able to interbreed due to these differences, and two separate species may evolve. A geographic barrier has led to reproductive isolation between the two populations, resulting in speciation.

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Could you please explain your answer thanks
What is the complementary DNA strand to: 3¹ AGCTAGCTAGCTAAAGCT 5'

Answers

The complementary DNA strand to 3' AGCTAGCTAGCTAAAGCT 5' is 5' TCGATCGATCGATTTCGA 3'.

DNA strands consist of two complementary strands that pair together through specific base pairing rules. In DNA, adenine (A) pairs with thymine (T), and cytosine (C) pairs with guanine (G). To find the complementary DNA strand, we need to determine the bases that pair with the given sequence.

The given DNA sequence is 3' AGCTAGCTAGCTAAAGCT 5'. To find the complementary strand, we replace each base with its complementary base. Adenine (A) pairs with thymine (T), thymine (T) pairs with adenine (A), cytosine (C) pairs with guanine (G), and guanine (G) pairs with cytosine (C). Applying these base pairing rules, we get the complementary DNA strand as 5' TCGATCGATCGATTTCGA 3'. The new sequence is complementary to the given DNA strand and follows the base pairing rules, ensuring the proper pairing of bases in the DNA double helix.

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the longest living immunoglobulins are IgG1 and IgG4 of 21 days and other types of immunoglobulins have even shorter life span. Yet, people who have been vaccinated or recovered from natural infection of COVID-19 have been found to have neutralizing antibodies in circulation for up to 6 months. Can you provide an explanation for this phenomenon

Answers

The phenomenon provided in the question can be explained by multiple factors, including the generation of long-lived plasma cells, the presence of memory B cells, and ongoing antigen exposure or stimulation.

When the body is exposed to a pathogen, such as the SARS-CoV-2 virus, B cells produce antibodies to fight the infection. While most immunoglobulins have relatively short lifespans, the immune response to COVID-19 involves the generation of long-lived plasma cells. These plasma cells are capable of continuously producing specific antibodies for an extended period.

Additionally, memory B cells play a crucial role in maintaining immunity. These cells "remember" the pathogen and can quickly respond to reinfection. Memory B cells can undergo activation and differentiation into antibody-secreting plasma cells when they encounter the virus again. This process helps to sustain the production of neutralizing antibodies over time.

Furthermore, ongoing exposure to viral antigens or periodic booster vaccinations can contribute to the presence of detectable neutralizing antibodies in circulation for an extended period. Continuous antigen exposure can stimulate the immune system to produce new plasma cells, while booster vaccinations can reinforce the immune response and replenish antibody levels.

It's important to note that individual variations in immune responses can also influence the duration of antibody presence. Factors such as age, overall health, and the severity of the initial infection or vaccination can affect antibody production and longevity.

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You have been given the accession no NM_003183.6. a. List the name of protein domain(s) coded by this gene. b. Delete the exon which starts from 456 to 586 nucleotides. Find out and write down the protein domain(s) coded by this shorter sequence. Prove your findings with related images. c. When you delete exon positioned at 456 to 586, does this protein sequence remain in frame? Explain your answer. d. Which software(s) did you use for your answers? Write down the name(s) and aim(s) for each software Search for "3AXK' protein at PDB database; a. From which organism is this protein? b. How many beta strands and alpha helixes are found in this protein? c. How many subunits found in this protein? d. Paste a print screen of the 3D structure of this protein whit space fill style, coloured subunits at black background.

Answers

a. The protein 3AXK is obtained from the organism, "Homo sapiens." b. The protein has 6 beta strands and 9 alpha helices. c. The protein has four subunits in total. d. The 3D structure of the protein 3AXK.

a. The name of the protein domain coded by the given gene, NM_003183.6 is "integrin beta tail domain."

b. When the exon that starts from 456 to 586 nucleotides is deleted, the protein domain coded by this shorter sequence is the "Beta-tail domain." Here's the pictorial representation of the protein domains coded by the given gene:   

c. No, the protein sequence does not remain in the frame when the exon positioned at 456 to 586 is deleted. It results in a frameshift mutation as the codon is changed from GGT to TGC. So, it ultimately affects the downstream codons. 

d. The software that can be used for this answer is ExonPrimer. It is an effective tool for designing exon-specific PCR primers. 3AXK protein at the PDB database.

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What is fragile X-syndrome? What are the molecular events that
underlie it?

Answers

Fragile X syndrome is a genetic disorder that causes intellectual disability.

The underlying molecular events in fragile X syndrome is caused by a mutation in the FMR1 gene.

What is Fragile X syndrome?

Intellectual disability and other behavioral or developmental difficulties are common effects from fragile x syndrome's genetic disorder. It tends to affect both genders equally, although males may display more severe symptoms overall than females do.

Fragile x mental retαrdation 1 (FMR1) gene holds its primary responsibility for molecular conditions behind this syndrome.

The gene is found located on the X chromosome, carrying specific DNA sequences that experience repeat expansion where CGG trinucleotide enlargement frequently occurs across those with diagnosis of this condition.

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Excluding vaccination, describe a range of agents that can be used to treat viral infections, using named example viruses and target sites of drug action.

Answers

Viral infections cannot be treated effectively by antibiotics, which are reserved for bacterial infections.

There are, however, other treatments that can be used to alleviate the symptoms of a viral infection and in some cases, prevent or reduce the severity of the infection.

There are a number of agents that can be used to treat viral infections, including antiviral drugs,

interferons, and immunoglobulins.

Antiviral drugs are used to treat viral infections by targeting the virus itself or the host cell that is infected with the virus.

Some antiviral drugs work by inhibiting viral replication.

For example, acyclovir and valacyclovir are used to treat herpes simplex virus (HSV) and varicella-zoster virus (VZV) infections by inhibiting DNA replication.

Another example is oseltamivir, which is used to treat influenza by inhibiting viral neuraminidase,

an enzyme that is required for viral replication.

Other antiviral drugs work by blocking viral entry or release.

One example of this is enfuvirtide, which is used to treat HIV infection by blocking the fusion of viral and host cell membranes.

Interferon alfa is used to treat hepatitis B and C infections.

Immunoglobulins are antibodies that are produced by the body in response to viral infection.

They can be used as a treatment for viral infections because they can neutralize the virus and enhance the body’s immune response.

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Essay style question;
Compare and contrast the pharmacology of alcohol and cannabis
under the following headings: (a) pharmacological effects, (b)
mechanisms of action, (c) adverse effects, (d) depend

Answers

Individual responses to alcohol and cannabis can vary, and the overall effects and risks depend on factors such as dosage, frequency of use, route of administration, and individual susceptibility.

a. Pharmacological Effects: Alcohol and cannabis have distinct pharmacological effects. Alcohol is a central nervous system depressant that initially causes relaxation, lowered inhibitions, and euphoria. Its primary psychoactive component, delta-9-tetrahydrocannabinol (THC), produces various effects including euphoria, relaxation, altered perception of time, and increased appetite.

b. Mechanisms of Action: Alcohol primarily acts on the brain by enhancing the effects of gamma-aminobutyric acid (GABA), an inhibitory neurotransmitter, while inhibiting glutamate, an excitatory neurotransmitter. This leads to the overall depressant effects of alcohol. Cannabis interacts with the endocannabinoid system in the brain, primarily by binding to cannabinoid receptors (CB1 receptors).

c. Adverse Effects: Alcohol misuse can lead to numerous adverse effects, including liver damage, cardiovascular problems, addiction, cognitive impairment, and increased risk of accidents and injuries.

d. Dependency: Both alcohol and cannabis have the potential for dependency. Alcohol use disorder is a recognized condition characterized by a strong craving for alcohol, loss of control over its consumption, and negative consequences due to drinking.

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i
dont remember how to solve this step by step
1) Some studies indicate that Brontosaurus (a very large dinosaur) weighed about 15,400kg. Let's assume μ = 15,400 and o = 1200kg. a) Calculate Pr{Y> 17,000} b) Now assume you have a sample of n = 10

Answers

a) P_r{Y > 17,000} ≈ 0.0918

b) P_r{Y > 17,000} for n = 10 dinosaurs is lower than the probability in part (a).

c) The probability in part (b) is lower because larger sample size reduces variability and provides a more accurate estimate of the population mean.

a) P_r{Y > 17,000} = P_r{(Y - μ) / σ > (17,000 - 15,400) / 1200}

= P_r{Z > 1.33} ≈ 0.0918

b) For a sample of size n = 10, the distribution of the sample mean Y' follows a normal distribution with mean μ and standard deviation σ/√n. Therefore, Pr{Y > 17,000} can be calculated using the sample mean and sample standard deviation.

c) The probability Pr{Y > 17,000} for a single observation is lower than the probability Pr{Y > 17,000} for a sample of size n = 10. This is because when taking a larger sample, the variability decreases and the sample mean becomes a more precise estimate of the population mean. Consequently, the probability of observing extreme values (such as Y > 17,000) decreases as the sample size increases.

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Question 6 Some sharks have embryos enclosed in an egg sac inside the mother's body. The embryo receives nutrition from its mother. After full embryonic development, the mother shark gives birth to live young. What is this called? a) Ovoparous. b) Viviparous. c) Ovoviviparous.

Answers

The correct option from the given statement is (c) Ovoviviparous.  Ovoviviparous is a type of reproduction in which the mother shark holds fertilized eggs inside her body until they hatch.

After full embryonic development, the mother shark gives birth to live young. In Ovoviviparous, the embryo receives nutrition from its mother as it grows inside her. It's essential to remember that the eggs are never exposed to the outside environment. Sharks, snakes, reptiles, and other animals may all give birth in this manner.

Sharks, rays, and skates, in particular, are oviparous, ovoviviparous, or viviparous, depending on their species.Viviparous is a term used to describe sharks that produce living young rather than eggs. The baby sharks get their nourishment from the mother shark's body in this instance. The embryo grows within the mother's womb in this case, and there is no external egg covering.

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Circle "Increase" or "Decrease" to show the effect the following signaling molecule or type of signaling molecule would have on cellular CAMP concentration. a) Epinephrine b) Epinephrine Antagonist c) Phosphodiesterase (PDE) Increase / Decrease Increase 1 Decrease Increase 1 Decrease

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The answer to the given problem is as follows: CAMP concentration would increase with Epinephrine and decrease with Epinephrine Antagonist and Phosphodiesterase (PDE).

The signaling molecule epinephrine is known to stimulate the cellular CAMP concentration, whereas Epinephrine Antagonist and Phosphodiesterase (PDE) both work to decrease the cellular CAMP concentration. Therefore, Epinephrine increases the CAMP concentration in the cell, while Epinephrine Antagonist and Phosphodiesterase (PDE) decrease the CAMP concentration in the cell.  

Cyclic adenosine monophosphate (CAMP) is a key molecule that regulates cellular processes. It serves as a secondary messenger, transmitting signals from the exterior of a cell to the interior, initiating a series of events that cause the cell to change its behavior.

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forearm model - posterior view 0 F Ti
Identify the structure that is flagged by the blue arrow in the image above: + What is the most distal joint that the flagged muscle crosses? What action does th

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The structure identified by the blue arrow in the image is the extensor digitorum muscle.

The extensor digitorum muscle is a crucial muscle located on the posterior side of the forearm. It plays a vital role in the movement and control of the fingers.

This muscle originates from the lateral epicondyle of the humerus and the proximal ulna, and it runs down the forearm.

The most distal joint that the extensor digitorum muscle crosses is the metacarpophalangeal (MCP) joint.

The MCP joint is situated between the metacarpal bones (the long bones of the hand) and the proximal phalanges (the bones of the fingers).

When the extensor digitorum muscle contracts, it performs the action of extension at the MCP joint.

Extension at the MCP joint refers to straightening or stretching of the fingers. When the extensor digitorum muscle contracts, it pulls the fingers in an extended position, allowing them to open up or reach a fully straightened position.

Another muscle that produces the same movement at the MCP joint is the extensor indicis muscle. The extensor indicis muscle runs parallel to the extensor digitorum muscle and has a similar function.

It originates from the posterior ulna and inserts into the extensor expansion of the index finger. When the extensor indicis muscle contracts, it assists the extensor digitorum muscle in extending the index finger at the MCP joint.

In summary, the structure flagged by the blue arrow in the image is the extensor digitorum muscle. It crosses the metacarpophalangeal (MCP) joint, which is the most distal joint it affects.

The action performed by the extensor digitorum muscle at this joint is extension, allowing the fingers to straighten.

Another muscle that also produces this movement at the MCP joint is the extensor indicis muscle, specifically in the index finger. These muscles work together to provide control and movement for the extension of the fingers.

The question should be:

Forearm model - posterior view 0 F TiIdentify the structure that is flagged by the blue arrow in the image above: + What is the most distal joint that the flagged muscle crosses? What action does the flagged muscle have at this distal joint? Name one other muscle that also produces this movement at this distal joint.

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Fill in the blanks: In cats, there is a gene which produces ticked fur (bands of different colors on each hair) called Agouti (H). The recessive allele (h) for this gene produces hair which is a solid color from end to end. In addition, there is a coat color gene which has a recessive albino allele (a) which, in the homozygote, prevents the production of any coat color pigment, resulting in a white cat with pink eyes, the traditional albino. An albino female cat is mated to a solid brown male cat. All of their offspring are Agouti. The males and females among these offspring are allowed to freely intermate, producing a flock of F2 kittens. What is the gene interaction involve for this characteristic?__________________

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The gene interaction involve for this characteristic is Incomplete Dominance as agouti's dominant allele and the albino recessive allele expressed incomplete dominance, which resulted in all the offspring being Agouti.

The gene interaction involves for the characteristic, which is described in the problem that the allele for ticked fur is dominant, while the recessive allele is for solid-colored hair. The coat color gene has a recessive albino allele that, when homozygous, prevents the production of any coat color pigment, resulting in a white cat with pink eyes, the traditional albino.

An albino female cat is mated to a solid brown male cat. All of their offspring are Agouti. The males and females among these offspring are allowed to freely intermate, producing a flock of F2 kittens.Since the gene for ticked fur is dominant, all the offspring were Agouti. Since the F1 offspring was heterozygous for the dominant Agouti gene, the genotype was Aa, and the phenotype was Agouti.

Furthermore, since the albino gene is recessive, the genotype for the albino female cat was aa. The solid brown male cat was not albino, which means that he did not have an aa genotype but rather an AA or Aa genotype. The Aa and AA genotypes would produce the Agouti phenotype, while the aa genotype would produce the albino phenotype.In this case, the gene interaction involves incomplete dominance, which is a form of intermediate inheritance in which one allele for a specific trait is not fully dominant over the other allele, resulting in a combined phenotype or a new third phenotype that is a combination of the traits of the two alleles.

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Not all brains are the same. What makes us cognitively superior (smarter) than the other species?
a) Comparatively small areas of the brain dedicated to the association areas.
b) Comparatively large areas of the brain dedicated to the primary cortical areas V1, A1, S1, etc...
c) Comparatively small areas of the brain dedicated to the primary cortical areas in V1, A1, S1, etc...

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The answer to this question is b) Comparatively large areas of the brain dedicated to the primary cortical areas V1, A1, S1, etc...

When compared to other species, human beings can be seen to have a larger brain with greater number of neurons and more complex connections among them. A considerable portion of this large brain is dedicated to the primary cortical areas V1 (visual), A1 (auditory), S1 (somatosensory), including other sensory areas. These areas get information from the environment and process it. This constitutes the groundwork for high-level cognitive processes like perception, attention, memory, and reasoning. This enhanced capacity and complexity of the primary cortical areas allow humans to perceive, analyze, and respond to the environment in more refined ways than other species.

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If leutenizing hormone were inhibited from being release in a human male, which of the following events would not occur? the development of male secondary characteristics Osperm production and maturation release of GnRH from the hypothalamus release of FSH from the pituitary growth hormone production

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If leutenizing hormone (LH) were inhibited from being released in a human male, the event that would not occur is the release of GnRH (gonadotropin-releasing hormone) from the hypothalamus.

In the male reproductive system, the hypothalamus releases GnRH, which stimulates the anterior pituitary gland to secrete luteinizing hormone (LH) and follicle-stimulating hormone (FSH). LH plays a crucial role in male reproductive function by stimulating the production of testosterone in the testes, leading to the development of male secondary characteristics such as facial hair, deepening of the voice, and muscle development.

If LH release is inhibited, it would disrupt the hormonal cascade, preventing the release of testosterone and subsequent events dependent on testosterone. However, the inhibition of LH release does not directly affect the release of GnRH from the hypothalamus.

Therefore, the event that would not occur if LH release is inhibited is the release of GnRH from the hypothalamus. The development of male secondary characteristics, sperm production and maturation, release of FSH from the pituitary, and growth hormone production can still occur, but they may be affected indirectly due to the disruption in testosterone production resulting from the inhibited LH release.

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Learning objective: Use a drawing to demonstrate the interactions occurring immunohistochemistry Compare the IHC and ELISA, what is the similarity vs difference? The IHC performed in this research involved the following reagents: Substrate Rat anti-mouse CD45R Rat anti-mouse CD3 Human anti-rat IgG w/enzyme attached Add these reagents, and additional molecules needed, to the tissue/cells below to demonstrate what is actually occurring during the IHC analysis.

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In the immunohistochemistry (IHC) analysis, the reagents used include substrate, rat anti-mouse CD45R, rat anti-mouse CD3, and human anti-rat IgG with an attached enzyme.

Immunohistochemistry (IHC) and enzyme-linked immunosorbent assay (ELISA) are both immunological techniques used to detect specific antigens or proteins. They share similarities in their principle and the use of antibodies for detection but differ in their application and format.

Similarity:

Both IHC and ELISA involve the use of antibodies to specifically bind to target antigens or proteins. In both techniques, a primary antibody is used to capture the target, followed by the addition of a secondary antibody conjugated with an enzyme or a detection molecule.

Difference:

1. Application: IHC is primarily used for visualizing and localizing antigens or proteins in tissue sections or cells, providing spatial information. ELISA is commonly used for quantitative measurement of antigens or proteins in solution, providing information on concentration.

2. Format: IHC is performed on tissue sections or cells attached to a solid support, such as a glass slide, while ELISA is typically performed in microplate wells.

3. Detection: In IHC, the presence of the target antigen or protein is visualized using a chromogenic substrate that reacts with the enzyme-conjugated secondary antibody. In ELISA, the detection is typically based on a colorimetric or fluorescent signal generated by the enzyme-substrate reaction.

In the IHC analysis mentioned, the reagents mentioned, including substrate, rat anti-mouse CD45R, rat anti-mouse CD3, and human anti-rat IgG with an attached enzyme, are added to the tissue or cells. These reagents facilitate the binding and detection of specific antigens or proteins, allowing the visualization and localization of the target molecules within the tissue or cells.

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"Based on the information given, are there toxicologu studies that
may he avioded because of special circumstances. three situations
are listed below, please list the rationale of they exist and any
sp
8 Here are some descriptions of a few molecules. Based on the information given, are there toxicology studies that may be avoided because of special circumstances. If so, then the cost and perhaps time of development may be reduced. Discuss these special dispensations, if they exist and the rationale why those studies may not be necessary; that is, everybody-even the risk averse deem these studies not useful in risk assessment. Large peptide molecule for the treatment of pruritus (itch) The therapeutic is applied topically and studies have shown that no measurable drug reaches the systemic circulation. The intended patient population includes adult and elderly males and females. Large protein molecule (human-specific) administered intravenously for the treatment of Inflammatory Bowel Disease. The intended patient population includes males and females above the age of 18 years. A small peptide (arginine-histidine-alanine-tyrosine) for the topical treatment of decubitus ulcers. When applied to dermal ulcers, some drug does reach the systemic circulation. The intended patient population is typically the elderly, though a lot of off label use is included in younger patients (eg. Quadriplegic or persistent coma).

Answers

Non-clinical systemic toxicity testing is required as the systemic exposure is anticipated.

There are toxicology studies that may be avoided because of special circumstances for the three molecules. Here are the reasons:For a large peptide molecule for the treatment of pruritus (itch):It is applied topically and studies have shown that no measurable drug reaches the systemic circulation. The intended patient population includes adult and elderly males and females. Therefore, non-clinical systemic toxicity testing is not required.For a large protein molecule (human-specific) administered intravenously for the treatment of Inflammatory Bowel Disease:Since the intended patient population includes males and females above the age of 18 years and the molecule is human-specific, non-clinical systemic toxicity testing can be avoided.

Moreover, animal studies may not be necessary.For a small peptide (arginine-histidine-alanine-tyrosine) for the topical treatment of decubitus ulcers:When applied to dermal ulcers, some drug does reach the systemic circulation. The intended patient population is typically the elderly, though a lot of off label use is included in younger patients (eg. Quadriplegic or persistent coma). Therefore, non-clinical systemic toxicity testing is required as the systemic exposure is anticipated.

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Genetic information is stored in DNA. DNA consists of four types of [A] joined through a sugar-phosphate backbone. In the process of [B] the information in DNA is copied into mRNA. During [C] the mRNA is a template for the synthesis of protein. A sequence of three bases, called a codon, specifies an [D]. The codons are read by the anti-codons of [E] molecules in the process of translation. Fill in the blanks A. B. C. D. E.

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Genetic information is stored in DNA. DNA consists of four types of nucleotides joined through a sugar-phosphate backbone.

In the process of transcription, the information in DNA is copied into mRNA. During translation the mRNA is a template for the synthesis of protein. A sequence of three bases, called a codon, specifies an amino acid. The codons are read by the anti-codons of tRNA molecules in the process of translation.

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The following diagram represents the semi-permeable plasma membrane of a cell. 0W0W0WOWOWOWOWOWOWOW Molecule K Extracellular space Molecule J Structure X Structure Y Intracellular space MAKAGU_____ KAPORAN 10.04 JWOWOWOWOWOK 22a) a) PE PEN i. Name the process used by Structure Y to transport Molecule J from the intracellular space to the extracellular space. (1 mark) SC ii. Describe the process named in part i above, in relation to the transport of molecule J. (2 marks) ABIU x₂x² # E E ABC DC 123 i. ii. 22b) 0-0 0-0 0-0 0-0 0-0 0-0 0-0 0-0 0-0 0-0 0-0 0-0 0-0 0-0 0-0 0-0 0-0 0-0 b) Describe the fluid mosaic model of the plasma membrane. (2 marks) hohoh S ------------------------------- ---- ------- -------- ----------- ------------- -------

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The process used by structure Y to transport Molecule J from the intracellular space to the extracellular space is Exocytosis.

The process of exocytosis named in part a above, in relation to the transport of molecule J, is the movement of the materials from the cytoplasm of the cell to the exterior of the cell.

The movement of these materials is achieved through the fusion of secretory vesicles (transport vesicles) with the plasma membrane of the cell, causing the secretion of the contents of the vesicles into the extracellular space.

This process is important in cells that produce and export substances like hormones, enzymes, neurotransmitters, and other secretory products.

The fluid mosaic model describes the cell membrane as being fluid in nature, because the components of the membrane can move laterally within the bilayer.

The model also explains that the membrane is selectively permeable, meaning that it allows certain molecules to enter and leave the cell while preventing others from doing so.

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Other Questions
B. On the line provided, give meanings for the following abbreviations, then write each abbreviation next to its explanation below:Part One:1. AB _____________________________________________________________________________2. C-section _________________________________________________________________3. CIS _____________________________________________________________________________4. CVS ____________________________________________________________________________5. Cx ______________________________________________________________________________6. D & C ___________________________________________________________________________7. FSH ____________________________________________________________________________8. G ______________________________________________________________________________9. GYN ____________________________________________________________________________10. hCG or HCG _____________________________________________________________________a. _______ Pituitary gland secretion that stimulates the ovariesb. _______ Pregnancy hormonec. _______ Lower, neck-like portion of the uterusd. _______ Study of women and disorders of the female reproductive systeme. _______ Spontaneous or induced termination of pregnancyf. _______ Localized cancer growthg. _______ Sampling of placental tissue for prenatal diagnosish. _______ Surgical incision of the abdominal wall to deliver a fetusi. _______ A pregnant womanj. _______ Procedure to widen the cervix and scrap the lining of the uterusPart Two:1. HDN ____________________________________________________________________________2. HPV ____________________________________________________________________________3. HRT ____________________________________________________________________________4. HSG ____________________________________________________________________________5. IUD _____________________________________________________________________________6. IVF _____________________________________________________________________________7. LEEP ___________________________________________________________________________8. LH _____________________________________________________________________________9. multip __________________________________________________________________________10. 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Draw the circuit diagram 2. calculate the armature current 3. calculate the induced emf. Q2. A 230-V d.c. shunt motor has an armature resistance of 0.5 Q and field resistance of 115 2. At no load, the speed is 1200 r.p.m and the armature current 2.5 A. On application of rated load, the speed drops to 1120 r.p.m. Determine the line current and power input when the motor delivers rated load. Q2. A 230-V d.c. shunt motor has an armature resistance of 0.5 Q and field resistance of 115 2. At no load, the speed is 1200 r.p.m and the armature current 2.5 A. On application of rated load, the speed drops to 1120 r.p.m. Determine the line current and power input when the motor delivers rated load. 4. You have a credit card with an APR of 22%. The card requires a minimum monthly payment of 14% of the balance. You have a balance of $7,400. You stop charging and make only minimum monthly payments. (a)Find a formula for the balance, B, after t monthly payments. Simplify the decay factor to 4 decimal places. (b)According to the formula you just created, what percentage of the credit cards balance is paid off each month? Give your answer as a percentage rounded to two decimal places. Write your result in a complete sentence. Which of the following is an example of geographic isolation?a. temporal differences in pond breedingB occupying different geographic locationsc. having differences in chromosome numbersd. none of the above Which of the following is the most affected in people with sickle-cell anemia? O the partial pressure of oxygen in airO the vol % of CO2 in bloodO the partial pressure of CO2 in the tissuesO the partial pressure of CO2 in the lungs O the acidity of the blood plasmaO the acidity inside the red blood cells O the Bunsen solubility coefficient for oxygen O chloride shift idont remember how to solve this step by step1) Some studies indicate that Brontosaurus (a very large dinosaur) weighed about 15,400kg. Let's assume = 15,400 and o = 1200kg. a) Calculate Pr{Y> 17,000} b) Now assume you have a sample of n = 10 Short fetches of wind across the lakes typically produce longer and more intense lake effect bands of snow than do long fetches. True False Economic theory suggests that we can increase productivity by:Question 1 options:Specializing according to absolute advantage.Specializing according to comparative advantage.Specializing according to either absolute or comparative advantage.It is impossible to increase productivity. Detimine the minimum plate thickness requied for plane strain conditions to prevail at the crack tio for the following steels and calculate the plastic zone size, use v=0.3, (a) Stee A, Kc = 100MPavm and yield strength =660MPa. (b) Sizal 3,hic = 180MPav m and yield strength =350MPa.