Which of these viruses is the least related to
Dengue Virus?
a.
West Nile Virus
b.
Yellow Fever
c.
SARS-CoV-2
d.
Zika Virus
e.
Tick-borne Encephalitis
Which protein of Dengue virus appears to be most

Answers

Answer 1

The virus which is least related to Dengue Virus is SARS-CoV-2. Dengue virus belongs to the genus Flavivirus, and the family Flaviviridae that also contains other viruses such as West Nile Virus.

Yellow fever, Japanese encephalitis virus, Tick-borne encephalitis, and Zika Virus.
SARS-CoV-2 is a member of the genus Betacoronavirus, and the family Coronaviridae.

It is the virus responsible for COVID-19 disease. COVID-19 disease has some similarities with dengue fever in terms of symptoms like fever, headache, muscle and joint pain, fatigue, and rash.
The protein of Dengue virus that appears to be the most antigenic is the E glycoprotein.

E glycoprotein is the major envelope protein of dengue virus. It plays an essential role in the virus' lifecycle, such as receptor binding, fusion, and virus maturation.
To know more about belongs visit:

https://brainly.com/question/28239924

#SPJ11


Related Questions

What is a functional characteristic of B cells that make them
different from innate immune cells?

Answers

B cells possess the unique ability to produce specific antibodies that recognize and neutralize antigens. This distinct characteristic sets them apart from innate immune cells.

A functional characteristic of B cells that distinguishes them from innate immune cells is their ability to produce specific antibodies. B cells are a type of adaptive immune cell responsible for the production of antibodies, which are specialized proteins that recognize and bind to specific antigens, such as pathogens or foreign substances.

When a B cell encounters an antigen that matches its specific receptor, it undergoes activation and differentiation, leading to the production of antibody molecules that can specifically recognize and neutralize the antigen. This process, known as humoral immunity, provides a highly specific defense mechanism against pathogens.

Unlike innate immune cells, such as macrophages or natural killer cells, which have broad recognition capabilities, B cells generate a diverse repertoire of antibodies that can target a wide range of pathogens.

To know more about antibodies, refer to the link:

https://brainly.com/question/445604#

#SPJ11

From the Olds and Milner experimnet paper . Describe a negative
control that was used in their design.

Answers

In the Olds and Milner experiment paper, a negative control that was used in their design is the use of rats that were not given any treatment. Negative controls are the group(s) in a research study that receive no treatment or receive treatment that should not have an effect on the outcome of the experiment.

The purpose of the negative control is to ensure that any observed effects are actually due to the treatment being tested, and not due to other factors such as chance, natural variation, or errors in the experimental procedures.In the case of the Olds and Milner experiment, the negative control was a group of rats that were not given any treatment, such as electrical stimulation or drugs.

This group was used to compare the behavior of the experimental group, which received electrical stimulation of the pleasure centre of the brain, and the group that received drugs, with the behavior of rats that received no treatment. By comparing the behavior of these groups, the researchers were able to determine whether any observed effects were due to the treatment being tested or due to other factors.

Learn more about variation here ;

https://brainly.com/question/17287798

#SPJ11

Bradford Hill viewpoints or "criteria" for a causal relationship for this specific exposure and disease combination. (2 points each) Click Save and Submit to save and submit. Click Save All Answers to save all answers.

Answers

The Bradford Hill viewpoints or "criteria" for a causal relationship are as follows:Strength of associationConsistencySpecificityTemporalityBiological gradientPlausibilityCoherenceExperimental evidenceAnalogy1.

Strength of association - the more likely it is that there is a causal relationship between the exposure and the disease.2. Consistency - The explanation for this criterion is that the association has been observed consistently across multiple studies.3.

Specificity - This criterion is met when a specific exposure is associated with a specific disease.4. Temporality - The main answer is that the exposure must occur before the disease.5. Biological gradient - This criterion is met when there is a dose-response relationship between the exposure and the disease.6. Plausibility - The explanation for this criterion is that there must be a plausible biological mechanism to explain the relationship between the exposure and the disease.7. Coherence - The main answer is that the relationship should be coherent with what is already known about the disease.8. Experimental evidence - This criterion is met if experimental studies support the relationship between the exposure and the disease.9. Analogy - This criterion is met if the relationship between the exposure and the disease is similar to that of other established relationships.

TO know more about that viewpoints visit:

https://brainly.com/question/1043206

SPJ11

1. its structure (tertiary/quaternary). Does it require a co-factor/co-enzyme? 2. if an enzyme: a) what class of enzyme is it?(Oxidoreductase Transferase Hydrolase Lyase Isomerase Ligase). b) what does this class of enzyme do? c) describe the specific reaction it catalyzes and where. d) the significance of this reaction e) is it allosterically controlled or regulated? 3. If a protein that is not an enzyme consider: a) Does it belong to a class of proteins (i.e. transcription factors, cell adhesion, receptor etc.) Explain this class. b) What does your protein do specifically? c) Where does it perform its task? d) Is it regulated or controlled? 4. are their mutant forms that contribute to disease or disorder? Explain.

Answers

The structure of proteins can be classified into four categories, namely Primary, Secondary, Tertiary, and Quaternary structure.

Enzymes require cofactors or coenzymes for their activity. 2. a) Oxidoreductase. b) It catalyzes oxidation-reduction reactions, which involve the transfer of electrons. c) NAD+ to NADH+ H+ (nicotinamide adenine dinucleotide) is the specific reaction catalyzed by the oxidoreductase enzyme, and it takes place in the mitochondria. d) The oxidation of NAD+ to NADH+ H+ is significant as it is a vital step in cellular respiration, and it allows the mitochondria to produce ATP. e) Yes, it is allosterically controlled or regulated.

3.

a) There are a number of classes of proteins, including but not limited to structural proteins, enzymes, transcription factors, and receptor proteins.

b) Enzymes catalyze chemical reactions in the body, while structural proteins provide support and structure to cells. Transcription factors bind to DNA and regulate gene expression, while receptor proteins recognize and respond to specific ligands.

c) Proteins carry out their functions in different cellular compartments, such as the cytoplasm, nucleus, or mitochondria, depending on their function.

d) Yes, it is regulated or controlled.

4. Mutant forms of proteins may contribute to disease or disorder. Protein misfolding is one of the major reasons for numerous neurodegenerative diseases, and prion diseases are caused by the accumulation of abnormal proteins.

To know more about proteins visit:

https://brainly.com/question/30986280

#SPJ11

1- Prior to its charging with an amino acid, how is the 3' end of a transfer RNA modified from its original structure as an RNA Pol III transcript? 2.Why is this modification so important in the function of the tRNA?
3. When it is not bound by the ribosome, a mature tRNA is usually bound in the cytoplasm by one of two proteins. What are these proteins and what is different about the tRNAs bound by each?

Answers

1. The 3' end of a tRNA is modified by adding a CCA sequence.

2. This modification allows tRNA to bind specific amino acids, enabling proper function in protein synthesis.  3. AARS and EF-Tu are the proteins that bind mature tRNA in the cytoplasm, facilitating amino acid attachment and ribosome interaction, respectively.

1. The 3' end of a transfer RNA (tRNA) is modified by the addition of a CCA sequence, which is not encoded in the original RNA Pol III transcript.

2. This modification is important for tRNA function because the CCA sequence serves as a binding site for amino acids during protein synthesis. It allows the tRNA to properly carry and transfer specific amino acids to the ribosome during translation.

3. The two proteins that can bind mature tRNA in the cytoplasm are aminoacyl-tRNA synthetases (AARS) and EF-Tu. AARS binds to tRNA before amino acid attachment and ensures the correct amino acid is attached to the tRNA. EF-Tu binds to aminoacyl-tRNA and delivers it to the ribosome during protein synthesis. The difference between tRNAs bound by each protein lies in their interaction: AARS recognizes the tRNA anticodon and ensures correct amino acid attachment, while EF-Tu recognizes the aminoacyl-tRNA complex and facilitates its proper positioning on the ribosome for protein synthesis.

learn more about tRNA here:

https://brainly.com/question/29544584

#SPJ11

What is transcription? What is translation?
What is a gene? What are codons? What steps happen to reduce the
length of RNA before it leaves the nucleus?
What do we call RNA after these steps have been

Answers

Transcription is the process in which genetic information encoded in DNA is converted into a complementary RNA sequence. Translation, on the other hand, is the process where the RNA sequence is used to synthesize proteins. A gene is a segment of DNA that contains the instructions for building a specific protein.

Codons are three-letter sequences of nucleotides in mRNA that specify particular amino acids or signaling functions. Before leaving the nucleus, RNA undergoes processing steps including capping, polyadenylation, and splicing. After these steps, the processed RNA is called mature mRNA.

1. Transcription:

Transcription is the first step in gene expression, where the DNA sequence is used as a template to produce a complementary RNA molecule. During transcription, an enzyme called RNA polymerase binds to the DNA at the promoter region and synthesizes a single-stranded RNA molecule, known as the primary transcript or pre-mRNA. The RNA molecule is synthesized in the 5' to 3' direction and is complementary to the DNA template strand.

2. Translation:

Translation is the process by which the information in mRNA is used to synthesize proteins. It occurs in the cytoplasm, specifically on ribosomes. Ribosomes read the mRNA sequence in sets of three nucleotides called codons. Each codon corresponds to a specific amino acid or a stop signal. Transfer RNA (tRNA) molecules carry the corresponding amino acids to the ribosome, where they are linked together to form a protein chain according to the mRNA sequence.

3. Gene:

A gene is a segment of DNA that contains the instructions for building a specific protein or performing a specific function. Genes are located on chromosomes and are made up of coding regions called exons and non-coding regions called introns. Genes play a crucial role in determining an organism's traits and functions.

4. Codons:

Codons are three-letter sequences of nucleotides in mRNA that encode specific amino acids or act as signaling sequences. There are 64 possible codons, including 61 codons that code for amino acids and 3 codons that serve as stop signals to terminate protein synthesis. The genetic code, known as the genetic code, specifies the relationship between codons and amino acids.

5. Steps to Reduce RNA Length:

Before leaving the nucleus, the primary transcript undergoes processing steps to produce mature mRNA. These steps include:

- Capping: The addition of a modified guanine nucleotide (5' cap) to the 5' end of the mRNA molecule. This cap helps protect the mRNA from degradation and is involved in mRNA export from the nucleus.

- Polyadenylation: The addition of a string of adenine nucleotides (poly-A tail) to the 3' end of the mRNA molecule. This tail aids in mRNA stability and export from the nucleus.

- Splicing: The removal of introns, non-coding regions, from the primary transcript. The exons, coding regions, are joined together to form a continuous mRNA sequence.

6. Mature mRNA:

After the processing steps, the mRNA molecule is referred to as mature mRNA. It is shorter in length than the primary transcript and contains only the exons that code for proteins. Mature mRNA is transported out of the nucleus and serves as a template for protein synthesis during translation in the cytoplasm.

To know more about Transcription refer here:

https://brainly.com/question/32921071#

#SPJ11

What is the meaning of "adding a methyl group to H3K9"? please
explain breifly

Answers

Adding a methyl group to H3K9 refers to the process of adding a methyl chemical group to the ninth lysine residue on the histone H3 protein. This process is known as histone methylation and it is a crucial epigenetic modification that controls gene expression by altering the way DNA is packaged in chromatin.

Histone methylation, including the methylation of H3K9, can either activate or repress gene expression depending on the location and number of methyl groups added. In general, the addition of methyl groups to H3K9 is associated with gene repression, whereas the removal of these methyl groups is associated with gene activation.

Histone methylation is a dynamic process that is regulated by various enzymes, including histone methyltransferases and demethylases. The addition or removal of a methyl group can alter the chromatin structure and accessibility, thereby regulating the expression of genes in different tissues and developmental stages.

Learn more about H3K9

https://brainly.com/question/31994076

#SPJ11

Recombination mapping has been fundamental in studying the arrangement of loci along chromosomes. Which of the following statements about recombination mapping is NOT correct?
A. Genome-wide association mapping can be combined with recombination mapping for better understanding of genetic bases of phenotypes
B. It cannot be used for breeding of animals
C. Generation time is an important factor for its feasibility
D. It cannot be used for asexual organisms
E. Measuring phenotypes is an important component

Answers

Recombination mapping has been fundamental in studying the arrangement of loci along chromosomes. The statement about recombination mapping that is not correct is "b)It cannot be used for breeding of animals."Reciprocal recombination between homologous chromosomes leads to the creation of recombinants.

Recombinants carry alleles for which recombination has occurred in the region between the genes. It is crucial to note that genetic recombination plays a vital role in mapping genes, genetic variation, and genetic evolution. Moreover, it allows the production of genetic maps, which can be used to construct physical maps.Generally, the benefits of recombination mapping are as follows:To detect DNA polymorphisms and map traits of interestTo discover genetic variation and the positions of genes that influence traitsTo determine the order and distances between genetic markersTo detect regions of the genome that are under evolutionary pressureTo determine the positions of genes on chromosomesGenome-wide association mapping can be combined with recombination mapping for better understanding of genetic bases of phenotypes. Measuring phenotypes is an important component in determining the genetic basis of phenotypes. Also, generation time is an important factor in determining the feasibility of recombination mapping.However, it cannot be used for asexual organisms as it needs sexual reproduction to bring about the generation of recombinants. Therefore, the statement about recombination mapping that is not correct is "It cannot be used for breeding of animals."

To know more about Recombination mapping visit:

https://brainly.com/question/10298507

#SPJ11

ogether, H and L chain variable regions form the antigen binding site of an antibody
molecule. Therefore, replacing the light chain (receptor editing) in an autoreactive clone with a new one will _____.
A) Maintain the same antigen specificity
B) Change the antigen specificity away from autoreactivity
C) Create an autoreactive antigen-binding site
D) Improve the binding affinity to the same antigen

Answers

The correct answer is B) Change the antigen specificity away from autoreactivity.

Replacing the light chain in an autoreactive clone with a new one through receptor editing allows for the generation of a different antigen-binding site. The variable region of the light chain, along with the variable region of the heavy chain, forms the antigen binding site of an antibody molecule. By introducing a new light chain, the antigen specificity of the antibody is altered, moving it away from autoreactivity. This mechanism helps to eliminate or reduce the binding of autoreactive antibodies to self-antigens and promotes the generation of antibodies with different antigen specificities, reducing the risk of autoimmune reactions.

learn more about:- antigen here

https://brainly.com/question/24384193

#SPJ11

Exam Section 1: Item 43 of 50 Mark 44. Fluid reabsorption by the proximal renal tubule can be increased by which of the following mechanisms in copilaries? A) Decreasing the hydrostatic pressure in the glomerular capilaries B) Decreasing the oncotic pressure in the peritubular capillaries C) increasing the hydrostatic pressure in the peritubular capillaries OD) Increasing the oncotic pressure in the peritubular capillaries

Answers

The correct answer is option D) Increasing the oncotic pressure in the peritubular capillaries.

Fluid reabsorption by the proximal renal tubule can be increased by the mechanism of increasing the oncotic pressure in the peritubular capillaries.

This occurs as a result of increased reabsorption of water and solutes from the tubule into the peritubular capillaries.

Fluid reabsorption in the proximal renal tubule: Fluid reabsorption occurs through the proximal tubules of the nephron.

It is the primary process that occurs in the proximal tubules, where up to 70% of the glomerular filtrate is reabsorbed.

Fluid reabsorption in the proximal renal tubule can be regulated by a variety of factors.

This includes increasing the oncotic pressure in the peritubular capillaries.

The hydrostatic pressure in the glomerular capillaries is increased in glomerular filtration. Peritubular capillaries are supplied by efferent arterioles that increase their hydrostatic pressure.

This leads to reabsorption of filtrate and maintenance of homeostasis.

To know more about oncotic pressure visit:

https://brainly.com/question/28588069

#SPJ11

Identify the route by which a virus enters and leaves the host
cell. Explain the process in your own words please. It does not
have to be long

Answers

When it comes to identifying the route by which a virus enters and leaves the host cell, it is important to first understand that viruses are not living organisms. They are infectious agents that can only reproduce within the host cell of a living organism.

As such, viruses have evolved to have specific mechanisms for entering and leaving host cells.

In terms of entry, viruses can enter host cells through a variety of means, depending on the type of virus and the host cell. Some viruses enter through the cell membrane by fusing with the membrane and then releasing their genetic material into the host cell.

Other viruses enter by being engulfed by the host cell in a process called endocytosis.

Once inside the host cell, viruses begin to hijack the cell's machinery to replicate their own genetic material.

This process can cause damage to the host cell and lead to the production of new viruses, which can then be released from the host cell through a process called budding.

During budding, the virus takes a piece of the host cell membrane as it leaves, which can help it evade the host's immune system.

The exact process of viral entry and exit can vary depending on the specific virus and host cell involved.

However, understanding these mechanisms is crucial for developing treatments and vaccines to prevent and treat viral infections.

To know more about virus visit;

brainly.com/question/13051239

#SPJ11

Animals underwent considerable diversification between 535-525 million years ago, during the Cambrian period of the Paleozoic Era. After this period diversification continued along with mass extinctions. Choose the statement that is accurate of this time in animal evolution. a. The first animals to move to land were the tetrapods, like tik-tah lik b. The arthropods were the first animal group thought to inhabit land c. Two surviving groups of vertebrates that inhabited land are the insects and amphibians d. Animals would not be able to inhabit land without the help of fungi

Answers

Fungi and plants were not yet established on land, so arthropods were entirely reliant on organic material that washed up on the shorelines. The arthropods' presence and consumption of this organic material possibly helped to change the land's environment, making it more hospitable for plant and fungi colonization.

During the Cambrian period of the Paleozoic Era, animals underwent significant diversification. After this time, diversification persisted alongside mass extinctions. The statement that accurately describes this time in animal evolution is "The arthropods were the first animal group thought to inhabit land."Option B is the correct option.Explanation:The arthropods were the first animal group thought to inhabit land. The first animals to move to land were tetrapods such as tiktaalik that lived much later, about 375 million years ago. The first animals to migrate to land were arthropods, which include spiders, scorpions, and insects, who were the first animals to invade land. Fungi and plants were not yet established on land, so arthropods were entirely reliant on organic material that washed up on the shorelines. The arthropods' presence and consumption of this organic material possibly helped to change the land's environment, making it more hospitable for plant and fungi colonization.

learn more about established here:

https://brainly.com/question/30555576

#SPJ11

If human teeth were made of bone in terms of cellular composition, development, and structure: how would this affect teeth function, and which strange and new dental pathologies would humans suffer?
(150 words minimum; no sources required)

Answers

If human teeth were made of bone in terms of cellular composition, development, and structure, it would affect teeth function and lead to strange and new dental pathologies that humans would suffer. Teeth made of bone would be harder, less flexible, and more brittle than our teeth.

This would cause the teeth to be more prone to fracturing, especially during biting and chewing. The structure of teeth would also change, causing the teeth to become less efficient at grinding and cutting food. One of the most notable pathologies that humans would suffer would be the loss of teeth, which would lead to the impairment of speech and difficulties eating. With bone teeth, the dental pulp inside the tooth would also change, leading to greater sensitivity to changes in temperature and more susceptibility to infection. The repair and maintenance of bone teeth would also be more challenging, as the development of tooth enamel would require a greater supply of calcium and phosphorus to meet the demands of an increasingly brittle and less efficient teeth structure.
In conclusion, the presence of bone in teeth would have a significant impact on the function, development, and structure of teeth, resulting in new dental pathologies and other complications. This, in turn, would make the maintenance of dental health more challenging for humans.

To know more about human teeth

https://brainly.com/question/29100502

#SPJ11

In type 1 diabetes the glucagon/insulin ratio is at a higher than normal level. Explain the changes that occur in the regulation of metabolic pathways as a consequence of this abnormal ratio and describe how this can account for the observed hyperglycaemia, hyperlipidaemia and ketoacidosis.

Answers

Type 1 diabetes mellitus (T1DM) is caused by the destruction of the pancreatic islet cells that produce insulin, resulting in an absence or inadequate production of insulin.

This leads to an increase in the glucagon/insulin ratio, which results in changes in metabolic pathways regulation. The glucagon/insulin ratio is at a higher than normal level in T1DM. The changes that occur in the regulation of metabolic pathways as a consequence of this abnormal ratio are given below:1. Hyperglycemia: Hyperglycemia occurs due to the lack of insulin, which causes an increased amount of glucose to accumulate in the bloodstream. Glucose is the main energy source for the body, and insulin helps cells absorb glucose.

In T1DM, the body produces too many ketones, which leads to an increase in acidity in the blood, known as ketoacidosis. Ketones are acidic, and the excessive production of ketones leads to the blood becoming too acidic, which can be life-threatening if not treated.T1DM patients can have several complications as a result of this abnormal ratio. It is essential that patients manage their glucose levels regularly, keep their diet healthy, and take insulin injections as prescribed to minimize the risk of these complications.

To know more about destruction visit:

https://brainly.com/question/1165953

#SPJ11

What is the opposite end of a DNA strand that begins with a 5
prime phosphate?
Group of answer choices:
3 prime hydroxyl
5 prime phosphate
5 prime hydroxyl
3 prime phosphate

Answers

The opposite end of a DNA strand that begins with a 5 prime phosphate is the 3 prime hydroxyl end. DNA is a double-stranded molecule in which two nucleotide chains spiral around one another.

The nucleotides are linked together by a phosphodiester bond between the phosphate group of one nucleotide and the 3’-OH group of the next. The directionality of a DNA strand refers to the orientation of the nucleotides within it. The 5’ end of a nucleotide contains a phosphate group attached to the 5’ carbon of the sugar molecule. The 3’ end, on the other hand, has a hydroxyl (-OH) group attached to the 3’ carbon of the sugar molecule.The process of transcription takes place in the 5’ to 3’ direction, so the 3’ end is the end where new nucleotides are added.

On the other hand, the 5’ end is the end where the phosphate group is located. The two strands in a DNA molecule run in opposite directions, with one running from 5’ to 3’ and the other running from 3’ to 5’. As a result, the opposite end of a DNA strand that begins with a 5’ phosphate is the 3’ hydroxyl end.

To know more about phosphate visit:

https://brainly.com/question/16094787

#SPJ11

& After diluting your culture 1:2500, you plate and get 154 colonies. what was the initial concentration? olm) olm

Answers

When we dilute a sample, we are reducing the number of organisms present in it. The amount of dilution can be calculated by dividing the original volume of the sample by the volume of the diluent added.

For example, a 1:10 dilution means that one unit of sample was diluted with nine units of diluent (usually water), resulting in a tenfold decrease in the number of organisms present.The initial concentration of the culture can be calculated as follows:The number of colonies that grew on the plate can be used to calculate the number of organisms present in the original culture.

Let's use C = N/V to find the initial concentration, where C is the concentration, N is the number of organisms, and V is the volume of the sample.Culture concentration × Volume of the culture = Number of organismsN1 × V1 = N2 × V2Where N1 is the initial concentration.

To know more about dilute  visit:

https://brainly.com/question/31521767

#SPJ11

Describe the organization of white and grey matter in
the spinal cord including the specific regional names of columns
and horns

Answers

The spinal cord consists of both white and grey matter. White matter surrounds the central grey matter and is organized into columns, while the grey matter is divided into horns.

The spinal cord is a cylindrical bundle of nerve fibers that extends from the base of the brain to the lower back. It is composed of white matter, which forms the outer region, and grey matter, which forms the inner region. White matter contains myelinated axons that transmit signals up and down the spinal cord. The white matter is organized into three main columns: the dorsal column, ventral column, and lateral column. These columns serve as conduits for sensory and motor information.

Grey matter, located centrally within the spinal cord, contains cell bodies, unmyelinated axons, and interneurons. It is shaped like a butterfly or an H, with anterior (ventral) and posterior (dorsal) horns on each side. The anterior horns contain motor neurons that send signals to the muscles, while the posterior horns receive sensory input from peripheral nerves. Additionally, there are lateral horns found in the thoracic and upper lumbar regions, which are associated with the autonomic nervous system.

Overall, the organization of the spinal cord includes white matter columns that facilitate communication between different levels of the central nervous system, and grey matter horns that play a vital role in motor control, sensory processing, and autonomic functions.

Learn more about spinal cord here:

https://brainly.com/question/29588686

#SPJ11

What are some important characteristics of the water molecule that make it useful in biological systems?
O Water is a bent molecule
O Water is an ionic compound
O Water can form hydrogen bonds
O Water is polar

Answers

The water molecule is a polar molecule that forms hydrogen bonds. It is an ionic compound. hence, all the options are correct.

The water molecule is a polar molecule, which means that it has a partial negative charge on one end and a partial positive charge on the other. This polarity is due to the unequal sharing of electrons between the hydrogen and oxygen atoms in the molecule. The partial negative charge on one end of the molecule is attracted to the partial positive charge on the other end, which allows water molecules to form hydrogen bonds with each other.

Hydrogen bonds are relatively weak attractive forces between a hydrogen atom in one water molecule and a bonding site on another water molecule. These bonds allow water molecules to pack closely together, which gives water its high surface tension and its ability to form droplets and sheets. The hydrogen bonds also allow water to dissolve a wide range of substances, which is important for many biological processes.

The fact that water is a polar molecule and can form hydrogen bonds makes it useful in biological systems because it can dissolve a wide range of substances and it can act as a solvent, transporting ions and other molecules throughout the body. The ability of water to form hydrogen bonds also allows it to maintain a relatively constant temperature and to store and release heat quickly. These properties make water essential for many biological processes, including cellular respiration, digestion, and transport.

Learn more about water

https://brainly.com/question/18681949

#SPJ11

Many nucleic acid biochemists believe that life on Earth began with cells having an RNA genome, but DNA then replaced RNA because the deoxyribose 2'-H makes DNA much more chemically stable. DNA also possesses T instead of U. Why might T be better than U to minimize errors in replicating the genetic material?

Answers

The replacement of U with T in DNA avoids this problem because T cannot undergo the same type of spontaneous deamination at the C4 position. This substitution thus increases the stability and fidelity of DNA as a genetic material.

The ribose sugar in RNA contains a 2' hydroxyl group (-OH) that can undergo spontaneous hydrolysis leading to RNA degradation. The deoxyribose sugar in DNA, on the other hand, is missing this hydroxyl group, making it more chemically stable. The replacement of RNA by DNA led to more stable genetic material and increased genetic fidelity, making DNA more favorable for storing and replicating genetic information.

The substitution of T for U in DNA further increased genetic stability. The base U in RNA can readily undergo spontaneous deamination at the C4 position to form base analogs such as uracil-5-oxyacetic acid (Uox) and uracil-5-carboxylic acid (Ucx). These base analogs can result in errors during DNA replication because they can pair with A instead of with G as is the case with U. This can lead to mutations that can be harmful or beneficial depending on the context in which they occur. The 5-methyl group in T also provides additional stability by helping to prevent unwanted chemical modifications of the base.

To know more about deoxyribose visit:

brainly.com/question/16522655

#SPJ11

Like all other rapidly growing cells, cancer cells must replicate their DNA and divide rapidly. However, also like all other rapidly growing cells, this can cause problems- what are these problems and how do cancer cells mitigate these problems?

Answers

Rapid DNA replication and division in cancer cells can result in a number of issues. The potential for errors during DNA replication, which can lead to genetic mutations, is one of the major obstacles.

These alterations may speed up the development of cancer and increase its heterogeneity.The strategies that cancer cells have developed to address these issues include:1. DNA repair pathways: To correct mistakes and maintain genomic integrity, cancer cells frequently upregulate DNA repair pathways. These repair processes, though, aren't always effective, which causes mutations to build up.2. Telomere upkeep: Telomeres, guardrails at the ends of chromosomes, guard against DNA deterioration and preserve chromosome integrity. To stop telomere shrinking and maintain telomere length, cancer cells activate telomerase or use alternative lengthening of telomeres (ALT) mechanisms.

learn more about replication here :

https://brainly.com/question/31845454

#SPJ11

3STA
Crystal structure of ClpP in tetradecameric form from
Staphylococcus aureus
indicate:
a- The number of subunits it consists of
b- The ligands it contains

Answers

The ClpP structure is made up of 14 subunits and contains several ligands that can be used to develop ClpP inhibitors.

The crystal structure of ClpP in tetradecameric form from Staphylococcus aureus indicates that it consists of 14 subunits and has two canonical heptameric rings. It is a serine protease whose active sites are situated inside a barrel-shaped particle. This particle is made up of two rings of seven identical subunits stacked on top of each other. The ligands it contains are Mg2+, AMP-PNP, and 20S proteasome inhibitor peptide. This data has been found useful for developing ClpP inhibitors that could be used as antibiotics to treat infections caused by S. aureus and other bacteria.

: The crystal structure of ClpP in tetradecameric form from Staphylococcus aureus reveals that it is composed of 14 subunits that form two canonical heptameric rings. It is a serine protease, with active sites situated inside a barrel-shaped particle. This particle is made up of two rings of seven identical subunits stacked on top of each other. The ligands present in the ClpP structure include Mg2+, AMP-PNP, and 20S proteasome inhibitor peptide. The data provided by this crystal structure is useful for the development of ClpP inhibitors that could be used as antibiotics to treat infections caused by S. aureus and other bacteria.

In conclusion, the ClpP structure is made up of 14 subunits and contains several ligands that can be used to develop ClpP inhibitors.

To know more about ClpP structure visit:

brainly.com/question/31097159

#SPJ11

Question 3 Which of the following statements is true of the male reproductive system? A The interstitial (Leydig) assist in sperm formation B The testes are temperature sensitive for optimal sperm pro

Answers

The testes are temperature sensitive for optimal sperm production.The testes are a pair of male reproductive organs, located within the scrotum. The testes are responsible for producing sperm and testosterone. Sperm production requires the testes to be held at a temperature slightly lower than body temperature, around 2-3°C lower.

This temperature is essential for optimal sperm production and quality. The testes are temperature sensitive organs that are very vulnerable to damage from high temperatures.Leydig cells or interstitial cells of the testes are located in the connective tissue surrounding the seminiferous tubules. These cells are responsible for producing and secreting testosterone. While testosterone is necessary for sperm production, the Leydig cells are not involved in the process of sperm formation. They only assist in the maturation of sperm, which takes place in the epididymis.

To know more about testosterone visit:

https://brainly.com/question/13061408

#SPJ11

What is the main structure used by integral membrane proteins to
go all the way across a membrane? What feature(s) of this structure
allows it to be used for this purpose

Answers

The main structure used by integral membrane proteins to traverse across a membrane is called a transmembrane domain. This domain possesses hydrophobic regions that enable it to embed within the lipid bilayer.

Integral membrane proteins are proteins that are embedded within the lipid bilayer of a cell membrane. These proteins perform various important functions, such as transporting molecules across the membrane and transmitting signals. To span the entire width of the membrane, integral membrane proteins typically contain a transmembrane domain.

The transmembrane domain is a structural feature of integral membrane proteins that consists of one or more stretches of hydrophobic amino acids. These hydrophobic regions are composed of nonpolar amino acids, which are repelled by the aqueous environment both inside and outside the cell. This property allows the transmembrane domain to insert itself into the hydrophobic core of the lipid bilayer, anchoring the protein within the membrane.

The hydrophobic nature of the transmembrane domain is crucial for its function. By interacting with the hydrophobic lipid tails of the membrane, it provides stability and ensures proper positioning of the protein within the bilayer. Additionally, the hydrophobic regions prevent water-soluble molecules from crossing the lipid bilayer, allowing the integral membrane protein to selectively transport specific substances across the membrane.

In summary, the transmembrane domain, with its hydrophobic regions, is the primary structure used by integral membrane proteins to traverse across a membrane. Its hydrophobic nature enables it to embed within the lipid bilayer, facilitating the protein's vital functions in cellular processes.

Learn more about transmembrane domain:

https://brainly.com/question/14927867

#SPJ11

(10 pts) Please answer the following questions based on your
knowledge of host-pathogen coevolution, the evolution of virulence
in pathogens, and the information provided about vertical and
horizontal

Answers

Parasites/pathogens are expected to evolve to be more virulent when they are transmitted horizontally (individual to individual) rather than vertically (parent to offspring through reproduction). This conclusion is based on the potential trade-offs between replication within hosts and transmission between hosts.

The evolution of virulence in parasites/pathogens is influenced by the trade-offs between their ability to replicate within hosts and their ability to transmit to new hosts. When transmission is predominantly vertical, occurring from parent to offspring through reproduction, there is a higher likelihood of coadaptation between the host and the pathogen.

In this scenario, the pathogen's fitness depends on the survival and reproductive success of its host, leading to a lower incentive for high virulence. High virulence could harm the host's reproductive success and, consequently, the transmission of the pathogen.

On the other hand, when transmission is mainly horizontal, occurring from individual to individual, the pathogen faces different selection pressures. The primary challenge for the pathogen in this case is to successfully infect and transmit to new hosts before the current host succumbs to the infection.

Horizontal transmission provides opportunities for the pathogen to encounter a broader range of hosts and exploit different ecological niches. Consequently, there is a higher likelihood of selection for higher virulence, as the pathogen benefits from maximizing its replication within each host and spreading to new hosts more effectively.

Overall, the trade-off between replication and transmission favors the evolution of higher virulence in pathogens that are transmitted horizontally. Horizontal transmission provides a larger pool of potential hosts, and pathogens that can exploit these opportunities by rapidly reproducing within hosts are more likely to succeed in spreading and establishing new infections.

Learn more about pathogens here:

https://brainly.com/question/31994092

#SPJ11

The complete question is:

Please answer the following questions based on your knowledge of host-pathogen coevolution, the evolution of virulence in pathogens, and the information provided about vertical and horizontal transmission. Considering potential trade-offs between replication within hosts and transmission between hosts, do you expect parasites/pathogens to evolve to be more virulent if they are transmitted vertically (parent to offspring through reproduction) or horizontally (individual to individual)? Explain how you came to this conclusion.

The Vostok ice core data... O All of the answers (A-C) B. Shows a clear NEGATIVE correlation between CO2 concentration and temperature Band C O C. Gives the natural range of variation in CO2 concentrations in the past 650,000 years O A. Tells us the age of Antarctica

Answers

The Vostok ice core data gives the natural range of variation in CO₂ concentrations in the past 650,000 years. The correct option is C.



The Vostok ice core data is used to study the changes in Earth's atmosphere and climate over the past 650,000 years. The ice cores are taken from deep in the ice sheet in Antarctica. The air bubbles trapped in the ice can tell us a lot about the composition of the atmosphere in the past.

Therefore, the main answer is "C. Gives the natural range of variation in CO₂ concentrations in the past 650,000 years."The ice cores from Vostok show us how the CO₂ concentrations have changed over the past 650,000 years. They have varied naturally between around 180 and 300 parts per million (ppm). This variation is largely due to natural factors such as volcanic eruptions and changes in the Earth's orbit and tilt. Therefore, it can be concluded that the Vostok ice core data gives the natural range of variation in CO₂ concentrations in the past 650,000 years.

The Vostok ice core data does not show a clear negative correlation between CO₂ concentration and temperature. It does tell us the age of Antarctica, but this is not one of the options given.

Therefore, the answer is C. Gives the natural range of variation in CO₂ concentrations in the past 650,000 years.

To know more about Vostok ice core, visit:

https://brainly.com/question/31850504

#SPJ11

Question 54 In what part of the kidney can additional water removed from the filtrate? The descending loop of Henle The proximal tubule The ascending loop of Henle The collecting duct

Answers

Additional water can be removed from the filtrate in the collecting duct of the kidney.

The collecting duct plays a crucial role in the final adjustment of urine concentration. It is responsible for reabsorbing water from the filtrate back into the bloodstream, thereby concentrating the urine. The permeability of the collecting duct to water is regulated by the hormone antidiuretic hormone (ADH), which determines the amount of water reabsorbed. When the body needs to conserve water, ADH is released, making the collecting duct more permeable to water and allowing for its reabsorption. Thus, the collecting duct is the site where the final adjustments to urine concentration occur by removing additional water from the filtrate.

learn more about:- collecting duct  here

https://brainly.com/question/25747764

#SPJ11

draw and label angiosperm mature female gametophyte (embryo sac). Label the following structures: funiculus, integuments, micropyle, egg cell, synergids, polar nuclei, antipodals, chalazal end.

Answers

The gametophyte generation is the dominant phase of the life cycle in bryophytes, pteridophytes, and gymnosperms, whereas in angiosperms, the sporophyte phase is dominant.

The gametophytes in angiosperms are smaller and more reduced than those in other groups. Angiosperms have two gametophytes, the male gametophyte (pollen grain) and the female gametophyte (embryo sac).The following are the structures that are labelled in angiosperm mature female gametophyte (embryo sac)Funicle: This is a stalk that connects the ovule to the placenta. The funicle is also known as the ovule's umbilical cord.Integuments: These are two layers of protective cells that envelop the nucellus of the ovule.Micropyle: A small opening in the integument near the embryo sac is known as the micropyle. This opening allows for the entry of the pollen tube during fertilization.Egg cell: The egg cell is a haploid female gamete that is found in the embryo sac's synergid cells.Synergids: These are two cells that are positioned near the egg cell in the embryo sac.Polar nuclei: These are two nuclei in the centre of the embryo sac that fuse to create a triploid nucleus in angiosperms.Antipodals: These are three cells that are located at the opposite end of the embryo sac from the egg cell.Chalazal end: This is the embryo sac's basal region. This area is located near the funicle and is opposite the micropyle.

Learn more about generation here:

https://brainly.com/question/30928503

#SPJ11

(a) Outline the principles that determine the assignment of a Biosafety level or number to a GMO product. (4 marks) (b) Give four examples of a real or theoretical GMO for each biosafety level or number from each of the following categories: animals, plants, and microbes. Explain why your example belongs at the biosafety level you have assigned to it. (You can provide two separate examples from any one category).

Answers

(a) Principles that determine the assignment of a Biosafety level to a GMO product are as follows:Level 1: It is safe,Level 2: Microbes that are possibly pathogenic to healthy adults,Level 3: Microbes pose a severe risk of life-threatening disease.

Level 1: It is safe, and the microbes used are not known to cause diseases in healthy adults. There are no specific requirements for laboratory design. Gloves and a lab coat are the only personal protective equipment required.

Level 2: Microbes that are possibly pathogenic to healthy adults but can be treated by available therapies are used. Laboratory design must restrict the entry of unauthorized individuals and require written policies and procedures. Personal protective equipment such as lab coats, gloves, and face shields are required.

Level 3: Microbes that are either indigenous or exotic and pose a risk of life-threatening diseases via inhalation are used. The laboratory must be restricted to authorized persons, must have controlled entry, and must be separated from access points. Negative air pressure in the laboratory, double-entry autoclaves for waste sterilization, and other specific engineering features are required. Respiratory protection is a must.

Level 4: The most dangerous organisms that pose a severe risk of life-threatening disease by inhalation are used. It's almost entirely constructed of stainless steel or other solid surfaces, with zero pores or cracks. A separate building with no outside windows and filtered, double-door entry is required. All employees must don a positive-pressure air-supplied space suit. There should be a separate waste disposal system, and the air in the laboratory should be filtered twice before being released into the environment.

(b) Four examples of a real or theoretical GMO for each biosafety level or number from each of the following categories: Animals, Plants, and Microbes are as follows:

Level 1:Microbes: Bifidobacterium animalis Plant: Nicotiana tabacum Animal: Zebrafish (Danio rerio)

Level 2:Microbes: Lactococcus lactis Plant: Arabidopsis thaliana Animal: Mouse (Mus musculus)

Level 3:Microbes: Mycobacterium tuberculosis Plant: Oryza sativa Animal: Monkey (Macaca mulatta)

Level 4:Microbes: Ebola virus Plant: None Animal: None

The above-listed GMOs belong to specific Biosafety levels because the level is determined by the risk of the organism to the environment or individual. The higher the Biosafety level, the more severe the disease is, which is why Biosafety level 4 requires extremely strict procedures. The assigned Biosafety level is determined by assessing the organism's pathogenicity and virulence, as well as the possibility of infection through ingestion, inhalation, or other methods.

Learn more about Biosafety:

brainly.com/question/30564176

#SPJ11

Number the structures of the urinary system of vertebrates in order from the production of urine (1) to the elimination of urine (5).
_________ urethra
________ kidney
________ ureter
_______ urogenital opening
_______urinary bladder

Answers

The structures of the urinary system of vertebrates in order from the production of urine (1) to the elimination of urine (5) are as follows: Kidney  ,Ureter ,Urinary bladder ,Urethra ,Urogenital opening .

The urinary system is responsible for filtering waste products from the blood and removing them from the body in the form of urine.Filtering waste from the blood and excreting it from the body as urine is the responsibility of the urinary system.  Urine is produced in the kidneys, which filter blood and remove waste products. From the kidneys, urine travels through the ureters and into the urinary bladder, where it is stored until it is eliminated from the body through the urethra and urogenital opening.

To know more about blood  , visit;

https://brainly.com/question/920424

#SPJ11

The most common genetic cause of severe human obesity is heterozygous coding mutations in the melanocortin 4 receptor. Based on what you know about this POMC system, which region of the hypothalamus that integrates peripheral signals for homeostatic control could be disrupted by this mutation? a) Arcuate b) Lateral hypothalamus Oc) Ventromedial hypothalamus d) Dorsomedial hypothalamus e) All of the above

Answers

Therefore, the answer to the question is (a) Arcuate.

The POMC system includes a number of endogenous peptides and receptor genes that have a direct role in energy homeostasis. The hypothalamus has different nuclei that play a role in appetite, satiety, and energy homeostasis.

The most common genetic cause of severe human obesity is heterozygous coding mutations in the melanocortin 4 receptor.

In this context, the region of the hypothalamus that integrates peripheral signals for homeostatic control which could be disrupted by this mutation is the Arcuate (ARC).

Explanation:When it comes to energy balance, the hypothalamus plays a vital role. It is a brain area that includes a range of nuclei with various functions. The hypothalamus is known to control eating behavior and energy balance.

It receives signals from the peripheral organs and regulates food intake, body weight, and energy expenditure.

The hypothalamus has several distinct nuclei that play a crucial role in regulating feeding behavior, including the Arcuate (ARC), the lateral hypothalamus (LH), the dorsomedial hypothalamus (DMH), and the ventromedial hypothalamus (VMH).

The most common genetic cause of severe human obesity is heterozygous coding mutations in the melanocortin 4 receptor.

This receptor is found primarily in the hypothalamus and is involved in the control of appetite and energy homeostasis. Melanocortin 4 receptor signaling in the hypothalamus helps to control food intake and energy expenditure.

According to the given information, the POMC system is associated with the ARC nucleus, which is responsible for integrating peripheral signals that regulate food intake and energy expenditure.

Therefore, the answer to the question is (a) Arcuate.

To know more about genetic visit;

brainly.com/question/30459739

#SPJ11

Other Questions
Answer the following questions about the careers of medical billing and coding, occupational therapy, pharmacy, and physical therapy to help you pinpoint the fields that might be best suited to your skills and interests.What distinctions do you see among each of these fields?Which fields appeal to you? Why do they appeal to you?Which fields don't interest you? Why do you dislike about the field?Which fields would require the least patient interaction, and which would require the most?Next, think about you impressions of these fields before you started this course. Has your opinion changed now that you've learned about each field in greater detail in Lesson Seven? 13. Give the definition of entropy. Why did we create this quantity? 14. What is the relationship between entropy, heat, and reversibility? Listen Which of the following excerpts from a sales message is an example of social proof?1. Be the envy of your friends and coworkers by purchasing this first-generation product today. 2.Did you know that our product has almost twice the computing power of other handheld devices available today? 3.This device gives you the functionality of a laptop and the portability of a smartphone. 4.Act now-our special limited-time offer expires soon! 5.Fortune 500 companies have purchased over 200,000 of these devices for their top executives. What are the mechanisms for the formation of each microstructural feature for titanium alloys when they undergo SLM manufacturing 1. The parathyroid gland releases ______ when plasma calcium islow. This hormone then triggers ______ of bone tissue.a. PTH depositionb. Calcitonin destructionc. Calcitonin deposition The magnitudes of vectors u and v and the anglebetween the vectors are given. Find the sum ofu+v.|u|=24,|v|=24,=129 A child disturbs a wasp nest, is stung repeatedly, and goes into shock within minutes, manifesting respiratory failure and vascular collapse. This is MOST likely to be due to: 1. systemic anaphylaxis 2. serum sickness 3. an Arthus reaction 4. cytotoxic hypersensitivity 1.In the formula,D=(1r)D, what doesDrepresent? A.The level of linkage disequilibrium in the current generation B.The level of linkage disequilibrium in the next generation C.the recombination rate D.none of the above1. In the formula,D=(1r)D, why is the range ofr00.5?A.Recombination either doesn't happen or if it does, the maximum possibility of recombination at any given locus is no better than randomB. It depends on the sex ratioC. It depends on the population size D.none of the above 2.When alleles at one locus impacts the evolution of alleles at other loci we have a _ pattern of...A. linkage equilibrium B.linkage disequilibriumC. a coadapted gene complexD. outbreeding depressionE. none of the above3. this one is not "a coadapted gene complex" becauseigot it wrong. please help me get the right now In the formula,D=(1r)D, what does D represent? A.The level of linkage disequilibrium in the current generation B.The level of linkage disequilibrium in the next generationC. the recombination rate D.none of the above 4. this is not "the level of linkage disequilibrium in thr next generation" because i got it wrong so please help find the right one i will rate please A frictionless piston-cylinder device as shown in Figure Q4 contains 7.5 liters of saturated liquid water at 275kPa. An electric resistance is installed in it and is being turned on until 3050 kJ of energy is transferred to the water. Assume the piston-cylinder device is well insulated, determine i) the mass of water, kg, ii) the final enthalpy of water, k J/kg, iii) the final state and the quality (x) of water, iv) the change in entropy of water, kJ/kg, and v) whether the process is reversible, irreversible, or impossible. Sketch the process on Pv diagram with respect to the saturation lines. Class, lets discuss the categories that organisms can be grouped in based on their nutritional requirements. Find one microorganism, either a prokaryote or eukaryote, and describe the environment in which it lives. (Does it live underwater? On skin? In soil? Give as many details as possible!) To complete your initial post, you will then use the vocabulary we discussed to classify it based on its nutritional needs and environmental requirements. (Is it a halophile? A chemoheterotroph? Use as many terms as you can!) Which of the following statements is NOT true? Positive discipline has reduced absenteeism and grievances at some firms. Employees prefer positive discipline because it promotes respect from supervisors. Positive discipline is more time-consuming than progressive discipline. Supervisors who dislike disciplinary tasks prefer progressive discipline. A separately-excited DC motor is operating with the following parameters and conditions. Motor rated output: 40 kW Motor input voltage: 340 V Armature resistance: 0.5 ohm Field resistance: 150 ohm Motor speed: 1800 rpm Field current: 4A Motor current: 8A Calculate the motor torque in N-m) Differential EquationFind the general solution using the Integrating Factors Found by Inspection1. (x2y2+ I)dx + x4y2 dy = 02. y(x3 y5)dx x(x3 + y5)dy =0.Find the particular solution using the Integrating Factors Found by Inspection1. y(x3y3 + 2x2 y) dx + x3(xy3 2)dy =0; when x = 1, y=1.Can you solve all problem that I give pls. Let X+iY be a complex signal and its magnitude is given by Z=X + Y, and phase 0 = tan- (Y/X) if X0 and phase = tan- (Y/X) + if x < 0X-N(0,1) and Y-N(0,1).Use the MATLAB or on functions to create a Gaussian distributed random value of X. Repeat this procedure and form a new random value of Y. Finally, form a random value of Z and 0, respectively. Repeat this procedure many times to create a large number of realizations of Z and 0. Using these samples, estimate and plot the probability density functions of Z and 0, respectively. Find analytical distributions among what we learned in the lectures that seem to fit your estimated PDFs. To clarify, you need to submit your code, plots of sample distributions and analytical distributions (as well as names and parameters of the analytical distributions). Note: X-N(0,1) denotes random variable X follows a Gaussian distribution with mean 0 and variance 1. Be able to determine blood type genotypes and phenotypes inoffspring using parental information for the H/h locus and the IA/IB locus (impacts of epistasis). Consider an insulated duct (i.e. adiabatic wall). Now we let Helium gas steadily enters the duct inlet at 50C at a rate of 0.16 kg/s and heated by a 3-kW electric resistance heater. The exit temperature of helium will be: Calculate the enthalpy change for the reaction from thefollowing:A ---->B H = -188 kJ/mol2C + 6B ----> 2D +3E H = -95kJ/mol E You want to move in a system that connects points A, B, and C.1. Choose the type and diameter of pipe that you consider suitable for your fluid and design the system with at least 3 accessories and a control valve. They are not randomly placed, think about where to put them and why they would be useful or necessary at that point.2. Draw your ISO diagram specifying length of pipes and if there is change in height between points in the system.3. Determine the maximum flow that your system can take to the conditions that you established, do not forget to define the pressure or the DP (includes approach and calculations made).Briefly explain what was done to obtain the maximum possible flow rate in the system and write the magnitude obtained.4. Do you consider that the Q you estimated is adequate? why? Justify your answers.5. What value of K should we produce with the valve to lower the flow to 50%?6. The minor losses, were they negligible? Justify your answer.7. Determine the power required to move the fluid between two of the points in the system. Cytochrome bb/f is a multi-protein complex that has multiple functions. Which of the following is NOT a function of the cytochrome b/f complex? the two PQH2 traverse different paths within the complex Cytochrome b participates in cyclinc e- flow while cytochrome f participates in non-cyclic e- flow O receives e- from PQH2 and Fd O All of these answers are functions of the cytochrome bb/f complex O exists in the thylakoid membrane this question is genetics1-A non-disjunction is caused by a failure of chromosomes to separate properly during meiosis. Which non-disjunction listed below will cause (in 100% of cases) death of the zygote in the womb?Select one:a. Two copies of the Y chromosomeb. Two copies of the X chromosomec. Three copies of chromosome 1d. Three copies of chromosome 212- Which of the following processes, that take place in homological chromosomes, may cause a quantitative chromosomal aberrations in humans?Select one:1. Meiotic nondisjunction;2. Conjugation during mitosis;3. Conjugation during meiosis;4. Crossing over.