If there is no family history of a particular disease but a child has the disease as well as the responsible genetic mutation in every cell, then it may have arisen due to a(n) ________ mutation early during development.

The correct answer is C) neurotransmitters. Neurotransmitters are neurochemicals that work between neurons' spaces. As the name describes, they transmit information between nerve cells.

Answer:

cellular respiration

C6H12O6 + 6O2 → 6CO2 + 6H2O.

reactants are C6H12O6 + 6O2

I. e sugar and oxygen.

Answer:

vestigial

Explanation:

Answer:

How old is she?

Explanation:

Answer:

bet

Explanation:

whats her name?

Answer:

Phagocytes would lose the capability to digest bacteria. There would be decreased levels of endocytosis occurring.

Explanation:

hope this helped

Phagocytes would lose the capability to digest bacteria. There would be decreased levels of endocytosis occurring.

Phagocytosis is the process by which a cell uses its plasma membrane to engulf a large particle, giving rise to an internal compartment called the phagosome. It is one type of endocytosis.

Phagocytosis is the rocess by which certain living cells called phagocytes ingest or engulf other cells or particles. The phagocyte may be a free-living one-celled organism, such as an amoeba, or one of the body cells, such as a white blood cell.

Phagocytosis can be divided into four main steps: (i) recognition of the target particle, (ii) signaling to activate the internalization machinery, (iii) phagosome formation, and (iv) phagolysosome maturation.

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Answer:

Memory cells are specalised types Nerve Cells .

The fourth histidine residue in carbonic anhydrase serves multiple roles, including providing a fourth ligand for the Zn(II), enhancing the binding of the three histidine residues to the metal ion, and neutralizing the negative charge on the hydroxide ion bound to Zn(II).

The fourth histidine residue in carbonic anhydrase plays a crucial role in the enzyme's function. One of its functions is to provide a fourth ligand for the Zn(II) ion when the hydroxide ion reacts with carbon dioxide (CO2). This coordination is important for the catalytic activity of the enzyme. Additionally, the fourth histidine residue helps in enhancing the binding of the three other histidine residues to the Zn(II) ion. This coordination strengthens the interaction between the enzyme and the metal ion, facilitating its catalytic function. Another role of the fourth histidine residue is to neutralize the negative charge on the hydroxide ion bound to the Zn(II) ion. By neutralizing this charge, the histidine residue stabilizes the hydroxide ion, which is a key component in the enzyme's catalytic mechanism. However, the fourth histidine residue in carbonic anhydrase does not directly participate in proton shuttling, provide a positive charge that enhances the binding of the carbon dioxide substrate, or make the transition state less unstable. These functions are primarily carried out by other active site residues and the overall protein structure of carbonic anhydrase.

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1) Sauropods were long-necked dinosaurs that grew to enormous sizes; 2) Stegosaurus was a spiked tail dinosaur with parasagittal plates on the dorsal surfaces of their necks, bodies, and tails ; 3) Allometric growth is the growth of certain body parts at different rates; 4) asteroid impact caused the extinction of the dinosaurs.

1)The following hypotheses try to explain the combination of characteristics seen in sauropods:

Selective pressures for increased height - because sauropods were herbivorous dinosaurs, they had to reach high for food, hence long necks. The selective pressures for a longer neck might have been due to the amount of food they could access which is proportional to the height they could reach.

Reduction in selective pressures on the forelimbs - as a result of the elongation of the necks, the forelimbs are no longer needed for feeding as the neck could do the job.

Mechanical requirements of large body size - when the size of the animal increases, there are certain mechanical requirements to support the body size which is why sauropods had huge legs and long tails.

2)The following hypotheses attempt to explain the function of the osteoderms on Stegosaurus:

Protection - the plates might have functioned as defensive structures for the animal. The spikes on their tails could be used as weapons to fend off predators.

Regulation of temperature - Stegosaurus may have been ectothermic, which means that they relied on external sources of heat to regulate their body temperatures. Osteoderms may have helped in regulating body temperature by either absorbing or releasing heat.

Signals for communication - the plates could have served as signals for communication between different Stegosauruses.

3) Allometric growth is the growth of certain body parts at different rates, leading to changes in the overall body shape of an organism. In marginocephalians, the frills and horns on their skulls grew at different rates, which led to the naming of several different species.

4) The impact hypothesis for the K-T mass extinction suggests that an asteroid impact caused the extinction of the dinosaurs. This hypothesis was originally proposed in the 1980s after the discovery of a layer of iridium, which is a rare element, in rocks that dated back to the time of the extinction.

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Answer:

D.) Nucleus

Explanation:

The Nucleus structure contains the highest conc of RNA..

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Answer:

Yes!

Explanation:

The process in which, water is evaporated from plant parts ( leaves and sometimes stem ) into the atmosphere, is called transpiration.

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Answer: Yes

Explanation:

The biochemical basis of cheek cell staining involves the use of specific dyes that bind to cellular components, allowing visualization under a microscope. Cheek cells are eukaryotes, as they possess a nucleus and membrane-bound organelles.

Step-by-step explanation:

In summary, the biochemical basis of cheek cell staining involves using dyes that bind to cellular structures, making them visible under a microscope. Cheek cells are eukaryotes, as they have a nucleus and other membrane-bound organelles.

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The biochemical basis of cheek cell staining process is to make the cells more visible under a microscope by using a dye or stain that binds to specific cellular structures, such as the nucleus or the cytoplasm.

The most commonly used stain for cheek cell samples is methylene blue, which binds to acidic structures in the cell, such as the nuclei and cytoplasmic granules.

Cheek cells are eukaryotic cells, meaning they have a well-defined nucleus and other membrane-bound organelles such as mitochondria and endoplasmic reticulum. They are not prokaryotes, which lack a nucleus and other membrane-bound organelles.

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The frequency of short-tailed individuals in this population is 0.2 or 20%. In a population in Hardy-Weinberg equilibrium, the frequencies of alleles and genotypes remain constant from generation to generation.

The frequency of an allele in a population is defined as the proportion of all alleles of that gene in the population that are of that particular type. The frequency of a genotype is defined as the proportion of individuals in the population that have that particular genotype.

Let's use the following notation to represent the frequency of alleles in the population:

p = frequency of the dominant allele (G)

q = frequency of the recessive allele (g)

According to the problem statement, the allele for long tails (G) is completely dominant to the allele for short tails (g). This means that individuals with the GG genotype and individuals with the Gg genotype will both have the long-tailed phenotype, while only individuals with the gg genotype will have the short-tailed phenotype.

We are given that 64% of the fish display the dominant phenotype, which means that the frequency of individuals with at least one G allele (i.e., the frequency of individuals with the GG or Gg genotype) is 0.64. We can use this information to set up the following equation

p^2 + 2pq = 0.64

where p^2 represents the frequency of individuals with the GG genotype, 2pq represents the frequency of individuals with the Gg genotype, and 0.64 represents the frequency of individuals with at least one G allele.

Since the frequency of the recessive allele (q) is simply 1 - p, we can substitute this into the equation above and simplify:

p^2 + 2p(1-p) = 0.64

p^2 + 2p - 2p^2 = 0.64

p^2 - 2p + 0.64 = 0

(p - 0.8)^2 = 0.16

p - 0.8 = ± 0.4

p = 0.8 ± 0.4

Since p represents the frequency of the dominant allele (G), and q represents the frequency of the recessive allele (g), we can calculate the frequency of short-tailed individuals (gg genotype) as follows:

q = 1 - p = 1 - 0.8 = 0.2

Therefore, the frequency of short-tailed individuals in this population is 0.2 or 20%.

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The relatively high frequency of the sickle-cell allele in areas where falciparum Malaria is endemic is an example of natural selection.

Malaria is a disease that spreads through the bites of infected female Anopheles mosquitoes. Symptoms include high fever, headache, chills, and flu-like symptoms. Malaria can be fatal in some instances. The sickle-cell alleleSickle-cell anemia is a disorder caused by a single nucleotide substitution in the beta-globin gene that results in the production of abnormal hemoglobin proteins.

Sickle-cell anemia is caused by the inheritance of two copies of the recessive sickle-cell allele.The relatively high frequency of the sickle-cell allele in areas where falciparum Malaria is endemic is an example of natural selection.

Natural selection is the process by which advantageous genetic variations become more prevalent in populations over time. In regions where falciparum malaria is endemic, individuals who are heterozygous for the sickle-cell allele have an advantage in fighting the disease, and as a result, the sickle-cell allele is more common.

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Answer:

Ribosomes

Explanation:

They are responsible for protein synthesis. Both eukaryotic and prokaryotic cells need proteins to live.

The human immune virus cannot reproduce without host cells.

The human immune virus is an RNA virus that only becomes active in human hosts.

When the virus gains access into a suitable host, it embeds its genome into that of the host and as such, the genetic machinery of the host is utilized to multiply the genome of the virus.

With time, the virus gradually takes over the host's genome system and breaks down their immune system, making them susceptible to diseases.

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Answer:

There is unanimous agreement among the coupled climatecarbon cycle models driven by emission scenarios run so far

that future climate change would reduce the effi ciency of the

Earth system (land and ocean) to absorb anthropogenic CO2.

As a result, an increasingly large fraction of anthropogenic CO2

would stay airborne in the atmosphere under a warmer climate.

Explanation:

Answer:

Plants in hot, arid climates would have fewer stomata because they would need to prevent undesirable water loss from the plant leaves.

Explanation:

Stomata are small, microscopic and essential pores located in the epidermis of stems, leaves, and other plant parts, that regulate gas exchange and photosynthesis. They have two specialized cells called guard cells.

Stomata number differs based on the temperature and the water.

Therefore, the stomata number differs and affects plant health.

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Answer:

Lion is a carnivore because it eats only meat

eg - lion, tiger, leopard etc

eg - deer, cow, buffalo

eg, cat, crow

Answer:

they need a high ‘surface to volume’ ratio, which is good for exchanging materials between the inside and outside of cells. But this is probably not really the size-limiting reason, since cells vary enormously in size and surface area to volume ratios.

Explanation:

In a eucoelomate, the mesoderm typically originates from the embryonic germ layer that lies between the ectoderm and endoderm.

This layer is known as the mesoderm layer, and it gives rise to various organs and tissues, including the musculoskeletal system, circulatory system, and reproductive system. Eucoelomates are organisms that have a true coelom, which is a fluid-filled body cavity that is lined by mesodermal tissue. The coelom is an important feature that allows for greater freedom of movement and specialization of organs. During development, the mesoderm layer splits into two layers, which eventually form the inner and outer linings of the coelom. This process is called schizocoely, and it is characteristic of eucoelomates.

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Answer:

B.

Explanation:

Speaking realistically, any of these could hypothetically happen, but for the sake of this question, I would say {B. Favorable genetic traits become more common over time.}

A. Many families have Alzheimer's disease. From this we conclude that the disease is X-linked recessive.

B. Meiosis is the process of generating gamete cells. There are two choices: mutual break join and another non-reciprocal break copy.

Many families are affected by Alzheimer's disease. It cannot be autosomal inherited because it affects more men than women. Affected males in the second generation have many affected children with normal wives.

If it was autosomal dominant, the sick father would have passed it on to all his second-generation children, but that is not the case. Therefore, it cannot be autosomal dominant.

If it had been autosomal recessive, he would not have affected his children, as his wife appears normal. increase. Therefore, it is not an autosomal abnormality.

Not X-linked dominant, in which case the affected father passed his only X chromosome to his daughter, making her daughter sick. But in this case, the sons are involved. Therefore, it is not X-linked dominant.

From this we conclude that the disease is X-linked recessive.

B. Meiosis is the process of generating gamete cells. Mutations can occur during meiosis in various ways such as double-strand breaks, inversions and point mutations. Important mutation types include missense, nonsense, and point mutations. Cells use multiple repair mechanisms to repair DNA damage or mutations during meiosis. One of the most efficient mechanisms is double-strand break repair by homologous recombination (HR).

Here, the DNA is unwound at the mutated region by recA loading. Recombination with homologous pairs continues through the RecBCD complex.

Here are his two alternatives: reciprocal break join and another non-reciprocal break copy.

Repair normally occurs during her S phase of the cell cycle, when the cell divides its DNA. However, mutations that occur during meiosis, such as chromatid pairing and recombination, are not easily repaired. Cells stop dividing only when the mutation gets worse.

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The kitten period, which lasts from 0 to 6 months of age, is the youngest stage of a cat's existence, and mitosis is responsible.

Mitosis is responsible for the growth of the organism makes somatic cells for the body and is necessary for the organism's life span.

Meiosis, often known as this procedure, is a component of sexual reproduction in cats (and in humans, dogs etc).

Therefore, mitosis is responsible, for a kitten becoming a cat, which is the growing phase in a cat's life span.

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The organelles present in an animal cell are-endoplasmic reticulum, centrioles, vacuoles, lysosomes, mitochondria, and cell membrane.

The endoplasmic reticulum is continuous with the outer nuclear membrane. It facilitates the production of chemical compounds as well as their transport into and out of the nucleus.

During meiosis and mitosis, centrioles are crucial to cell division.

Vacuoles play a key role in the storage of waste materials and aid in regulating the pH of the cell.

The waste molecules are broken down into simpler substances by lysosomes.

The mitochondria are the powerhouse of the cell because they contribute to the production of ATP, which the cell uses as energy for a variety of tasks.

The cell's protective barrier is the cell membrane. It controls how things get into and out of the cell.

The complete question is:

Which organelles are found in an animal cell? check all that apply.

(A). Endoplasmic reticulum

(B). Centrioles

(C). Cell wall

(D). Vacuoles

(E). Lysosomes

(F). Mitochondrial

(G). Chloroplasts

(H). Cell membrane

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Answer:

Explanation:

To test the research question of whether birds preferentially prey upon moths with respect to their resting backgrounds, we could alter Kettlewell's experimental design in the following ways:

Increase the sample size: One issue with Kettlewell's original study was the small sample size. To increase the power of the study, we could increase the number of moths and birds tested. This could involve setting up multiple experimental arenas, each with a different background, and testing multiple birds and moths within each arena.

Randomize the backgrounds: In Kettlewell's original study, the experimental arenas were set up with specific backgrounds (e.g., light and dark tree trunks). To avoid potential confounding variables, we could randomize the backgrounds for each arena. For example, we could use a computer program to generate random backgrounds for each arena, ensuring that there is an equal distribution of light and dark backgrounds.

Blind the observers: Another potential issue with Kettlewell's study is observer bias, where the experimenters may unconsciously favor data that supports their hypothesis. To avoid this, we could blind the observers to the background and moth type. For example, we could set up the experiment in such a way that the observers cannot see the background or the type of moth being tested.

Control for other variables: We could also control for other variables that may impact bird predation, such as temperature, humidity, and wind speed. By controlling for these variables, we can ensure that any differences in moth predation are due to the resting background and not other extraneous factors.

Overall, by altering Kettlewell's experimental design in these ways, we can conduct a more robust study that addresses potential confounding variables and has higher statistical power.

The Gram stain provides critical information about the structure of the cell wall.

The stain works by first adding crystal violet, which stains all bacterial cells, and then iodine, which forms a complex with the crystal violet, making it more difficult to remove. In the next step, a decolorizing agent is applied to remove the crystal violet-iodine complex from the layer of the Gram-negative bacteria, leaving them colorless. The thicker layer of the Gram-positive bacteria retains the crystal violet-iodine complex and appears purple.

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