1 2 3 4 5 6 7 8 D 10 A. Albumin B. Electrolytes C. Fibrinogen D. Oxygen E. carbon dioxide F. immunoglobulins G. Water H. hormones & enzymes 1. urea & creatinine J. glucose, amino acids, & fats G_Makes up about 92% of plasma T Circulating regulatory substances Plasma cations and anions Constitutes more than half of total plasma protein A clotting protein made by the liver Proteins that aid in recognition and neutralization of pathogens Wastes produced by metabolic processes that are carried in the blood and then disposed of by kidneys or sweat glands Nutrients absorbed from the digestive system and then carried in the blood to be delivered to body cells Although it's always the least abundant, the lack of this protein could result in hemophilia Starvation usually affects the amount of this plasma protein, resulting in low plasma osmolarity

Answers

Answer 1

Given the following terms, we need to match them with their respective descriptions. Albumin B. Electrolytes C. Fibrinogen D.

Oxygen E. carbon dioxide F. immunoglobulins G. Water H. hormones & enzymes 1. urea & creatinine J. glucose, amino acids, & fats.G - Makes up about 92% of plasmaT - Circulating regulatory substancesPlasma cations and anions - ElectrolytesConstitutes more than half of total plasma protein - Albumin A clotting protein made by the liver .

Fibrinogen Proteins that aid in recognition and neutralization of pathogens - Immunoglobulins Wastes produced by metabolic processes that are carried in the blood and then disposed of by kidneys or sweat glands - 1. Urea & creatinineNutrients absorbed from the digestive system and then carried in the blood to be delivered to body cells - J. Glucose, amino acids, & fatsAlthough it's always the least abundant, the lack of this protein could result in hemophilia - Factor VIIStarvation usually affects the amount of this plasma protein, resulting in low plasma osmolarity - Albumin.

To know more about Albumin visit :

https://brainly.com/question/29544650

#SPJ11


Related Questions

2. Most of the calcium sensors fall into main families
characterized by having either ____ or ______ Ca 2+ binding
domains.

Answers

The presence of these domains allows proteins to regulate a wide range of cellular processes in response to changes in intracellular Ca2+ levels.

Most of the calcium sensors fall into main families characterized by having either EF-hand or C2 Ca2+ binding domains. EF-hand domains are the most abundant and widespread Ca2+ binding motif found in proteins.

These motifs consist of two helices separated by a short turn that contains four acidic residues arranged in a characteristic loop structure that coordinates the Ca2+ ion. The C2 domain is a structurally diverse Ca2+ binding domain found in numerous proteins with different functions, including signal transduction and membrane trafficking. In conclusion, EF-hand and C2 Ca2+ binding domains are the two main families of Ca2+ sensors.

The most abundant and widespread motif is the EF-hand domain, while the C2 domain is structurally diverse and found in many different proteins.

The presence of these domains allows proteins to regulate a wide range of cellular processes in response to changes in intracellular Ca2+ levels.

To know more about Ca2+ ion visit:

brainly.com/question/1511779

#SPJ11

If your procedure calls for "sterile" conditions and you will be aliquoting a bacterial culture or sample into several microcentrifuge tubes, what must be done to the pipette tips before you can use them in your procedure?

Answers

If your procedure calls for "sterile" conditions and you will be aliquoting a bacterial culture or sample into several microcentrifuge tubes, the pipette tips must be sterilized before you can use them in your procedure. Steps to sterilize pipette tips: To sterilize the pipette tips, autoclave them or use presterilized, disposable tips that have been purchased.

If your procedure calls for "sterile" conditions and you will be aliquoting a bacterial culture or sample into several microcentrifuge tubes, the pipette tips must be sterilized before you can use them in your procedure. Steps to sterilize pipette tips: To sterilize the pipette tips, autoclave them or use presterilized, disposable tips that have been purchased. Autoclaving is the most reliable method, but it requires specialized equipment and a thorough understanding of the process. Autoclaving is a technique used to sterilize equipment and solutions, which involves heating them to a high temperature and pressure to kill any microorganisms present.

The autoclave works by using steam to raise the temperature inside the chamber, and it can take up to 30 minutes for a cycle to complete. Afterward, the samples and pipette tips must be allowed to cool down before they can be used.It is also important to keep the pipette tips sterile after they have been sterilized. Before use, always hold the tips above the sample and make sure they do not touch anything else. If the tip touches anything, such as your hand or the rim of the tube, it is no longer sterile. Always change the tips between samples to avoid contamination from previous samples.

To know more about pipette visit:

https://brainly.com/question/31387225

#SPJ11

Which of the following is true concerning the scapula?
O the end of the spine projects as the expanded process called the coracoid
the coracold articulates with the clavicle
O the glenoid cavity is where the scapula and humerus articulate
O the lateral border of the scapula is near the vertebral column
the scapular notch is a prominent indentation along the inferior border

Answers

The true statement about scapula is "The glenoid cavity is where the scapula and humerus articulate".

The glenoid cavity is a shallow, concave socket located on the lateral side of the scapula. It is the site where the scapula articulates with the head of the humerus, forming the glenohumeral joint, commonly known as the shoulder joint. This joint allows for a wide range of movement of the arm.

The other options provided are not true concerning the scapula:

The end of the spine of the scapula projects as the expanded process called the acromion, not the coracoid.The coracoid process is a separate bony projection on the anterior side of the scapula and does not articulate with the clavicle.The lateral border of the scapula is farther away from the vertebral column, while the medial border is closer to it.The scapular notch refers to a small indentation on the superior border of the scapula, not the inferior border.

To learn more about scapula, here

https://brainly.com/question/30516789

#SPJ4

Plant rhabdoviruses infect a range of host plants and are transmitted by arthropod vectors. In regard to these viruses, answer the following questions:
a. Plant rhabdoviruses are thought to have evolved from insect viruses. Briefly describe the basis for this hypothesis? c. Recently, reverse genetics systems have been developed for a number of plant rhabdoviruses to generate infectious clones. What are the main components and attributes of such a system? (3 marks

Answers

a. The hypothesis that plant rhabdoviruses evolved from insect viruses is based on several pieces of evidence. Firstly, the genetic and structural similarities between plant rhabdoviruses and insect rhabdoviruses suggest a common ancestry.

Both groups of viruses possess a similar genome organization and share conserved protein motifs. Additionally, phylogenetic analyses have shown a close relationship between plant rhabdoviruses and insect rhabdoviruses, indicating a possible evolutionary link.

Furthermore, the ability of plant rhabdoviruses to be transmitted by arthropod vectors, such as insects, supports the hypothesis of their origin from insect viruses. It is believed that plant rhabdoviruses have adapted to infect plants while retaining their ability to interact with and utilize insect vectors for transmission. This adaptation may have occurred through genetic changes and selection pressures over time.

c. Reverse genetics systems for plant rhabdoviruses allow scientists to generate infectious clones of the virus in the laboratory. These systems typically consist of several key components:

Full-length cDNA clone: This is a DNA copy of the complete viral genome, including all necessary viral genetic elements for replication and gene expression. The cDNA clone serves as the template for generating infectious RNA.

Promoter and terminator sequences: These regulatory sequences are included in the cDNA clone to ensure proper transcription and termination of viral RNA synthesis.

RNA polymerase: A viral RNA polymerase, either encoded by the virus itself or provided in trans, is required for the synthesis of viral RNA from the cDNA template.

Transcription factors: Certain plant rhabdoviruses require specific host transcription factors for efficient replication. These factors may be included in the reverse genetics system to support viral replication.

In vitro transcription: The cDNA clone is used as a template for in vitro transcription to produce infectious viral RNA. This RNA can then be introduced into susceptible host plants to initiate infection.

The main attributes of a reverse genetics system for plant rhabdoviruses include the ability to manipulate viral genomes, generate infectious viral particles, and study the effects of specific genetic modifications on viral replication, gene expression, and pathogenicity. These systems have greatly facilitated the understanding of plant rhabdoviruses and their interactions with host plants and insect vectors.

To know more about RNA synthesis

https://brainly.com/question/33306079

#SPJ11

Which type of secretion occurs destroying the entire cell as it releases its product? a. endocrine secretion b. merocrine secretion c. apocrine secretion d. holocrine secretion

Answers

The correct answer is d. holocrine secretion, where the entire cell is destroyed during the release of its product.

Holocrine secretion is a type of secretion in which the entire cell is destroyed during the process of releasing its product. This occurs when the secretory cells accumulate and store their product within their cytoplasm until it reaches a certain level of maturity. Once the product reaches the desired level, the entire cell disintegrates, releasing the accumulated secretion along with the cell debris.

Examples of holocrine secretion can be found in certain glands of the body, such as the sebaceous glands in the skin. Sebaceous glands produce sebum, an oily substance that helps lubricate and protect the skin and hair. In the case of sebaceous glands, the secretory cells accumulate sebum within their cytoplasm until they burst, releasing the sebum and cell fragments onto the skin's surface.

In contrast, other types of secretion, such as endocrine secretion, merocrine secretion, and apocrine secretion, do not involve the destruction of the entire cell. Endocrine secretion refers to the release of hormones directly into the bloodstream, while merocrine secretion involves the release of secretory products through exocytosis without any cell damage. Apocrine secretion is characterized by the release of secretory products along with a portion of the cell membrane.

Learn more about sebaceous glands here:

https://brainly.com/question/30421221

#SPJ11

Which of the following statements is TRUE about transcription
initiation
complexes required by eukaryotic RNA Polymerase Il?
O a. TFIlD recognizes and binds multiple promoter elements
O b. Mediator ha

Answers

Eukaryotic RNA Polymerase II requires a transcription initiation complex to begin transcription. The transcription initiation complex is composed of transcription factors, RNA polymerase, and other proteins.

The complex is formed at the promoter region of the DNA strand, which is recognized by transcription factors. Transcription initiation complexes are essential for the proper functioning of RNA Polymerase II.The correct statement regarding transcription initiation complexes required by eukaryotic RNA Polymerase Il is a. TFIlD recognizes and binds multiple promoter elements. TFIlD, a general transcription factor, is responsible for recognizing and binding to the TATA box, an essential element of the promoter region. In addition to recognizing the TATA box, TFIlD also binds to other promoter elements, such as the initiator element and downstream promoter elements. This binding helps to stabilize the transcription initiation complex, allowing RNA polymerase to begin transcription. The mediator is another general transcription factor, but it does not bind directly to the promoter region.

Instead, it interacts with transcription factors and RNA Polymerase II to help regulate transcription and ensure that it proceeds correctly.In summary, the transcription initiation complex is essential for the initiation of transcription by RNA Polymerase II. TFIlD recognizes and binds to multiple promoter elements, while the mediator interacts with other transcription factors and RNA Polymerase II to help regulate the process.

Learn more about Eukaryotic here:

https://brainly.com/question/29119623

#SPJ11

Question 35 1 points Saved Assume you want to examine the reponse of a number strains to a 2.3.5 triphenyltetrazolium (TTC) agar overlay. Place the available options in the correct order (start to finish that would allow you to perform the test most effectively. 3. Place YPD agar medium with strains at 30°C 6. Assess any colour formation in the TTC overlay after an appropriate period of time 2 Wait to for TTC to set 1. ~ Inoculate strains on the surface of YPD agar medium in small patches 4. V Overlay molten TTC agarose 5. V Incubate the strains for 48-72 hours

Answers

The given procedure is aimed to examine the response of a number of strains to a 2.3.5 triphenyltetrazolium (TTC) agar overlay.

The correct order of steps to perform the test most effectively are as follows:

1. Inoculate strains on the surface of YPD agar medium in small patches.

2. Wait for TTC to set.

3. Place YPD agar medium with strains at 30°C.

4. Overlay molten TTC agarose.

5. Incubate the strains for 48-72 hours.

6. Assess any colour formation in the TTC overlay after an appropriate period of time.

Explanation:

When working with agar medium, the basic procedure is to create and sterilize an agar solution, then pour it into sterile Petri dishes and allow it to cool.

Once the agar medium has hardened, inoculate with the microorganisms and allow them to grow under specific conditions to test for characteristics or reactions.

In this question, the given procedure has 6 steps, and the correct order to perform the test most effectively is provided as follows:

Step 1: Inoculate strains on the surface of YPD agar medium in small patches.The first step is to inoculate strains on the surface of YPD agar medium in small patches. This will be used to examine the response of a number of strains to a 2.3.5 triphenyltetrazolium (TTC) agar overlay.

Step 2: Wait for TTC to set.Wait for the TTC to set after inoculating the strains on the surface of YPD agar medium. This step is critical for the success of the procedure.

Step 3: Place YPD agar medium with strains at 30°C.Place YPD agar medium with strains at 30°C. This step is important to provide the appropriate temperature for the strains to grow.

Step 4: Overlay molten TTC agarose.

Overlay molten TTC agarose over the inoculated strains. This step will help to examine the response of the number of strains to a 2.3.5 triphenyltetrazolium (TTC) agar overlay.

Step 5: Incubate the strains for 48-72 hours.After overlaying molten TTC agarose over the inoculated strains, incubate the strains for 48-72 hours. This will provide the time necessary for the strains to grow and produce results.

Step 6: Assess any colour formation in the TTC overlay after an appropriate period of time. After incubating the strains for 48-72 hours, assess any color formation in the TTC overlay after an appropriate period of time.

This step is important for evaluating the results of the experiment.

To know more about procedure visit:

https://brainly.com/question/27176982

#SPJ11

True or False?
The transfer of heat from one body to another takes place only when there is a temperature difference between the bodies

Answers

Answer: True

Explanation: heat, energy that is transferred from one body to another as the result of a difference in temperature. If two bodies at different temperatures are brought together, energy is transferred—i.e., heat flows—from the hotter body to the colder.

3. What did the boiling do to the enzyme? 4. Why did tube 4 have a negative reaction for starch and a negative reaction for sugar? What was this a negative control to show which part of the experiment

Answers

The boiling done to the enzyme denatured, or destroyed, it. When enzymes are exposed to heat, they begin to unravel and form new shapes that no longer enable it to carry out its intended biological function, in this case, the breakdown of starch and sugar.

This is why tube 4, the negative control, had a negative reaction for both starch and sugar--the boiling destroyed the enzyme, so the reaction was inhibited.

This negative control was necessary to show if the other tubes were reacting due to the enzyme or if they were doing so for some other reason. Without this negative control, it would have been difficult to determine if other tubes were reacting due to the presence of the enzyme.

know more about enzymes here

https://brainly.com/question/31385011#

#SPJ11

Question 12 2 pts Why should stains be used when preparing wet mounts of cheek cells and onion skin epidermis? Edit View Insert Format Tools Table 12pt Paragraph | BIU A' εν των : I **** P 0 word

Answers

Stains are used when preparing wet mounts of cheek cells and onion skin epidermis for several reasons:

Contrast enhancement: Staining the cells helps to improve the visibility of cellular structures and details that may be otherwise difficult to observe.

Unstained cells may appear translucent and lack sufficient contrast, making it challenging to differentiate different cellular components.

Cell identification: Stains can help distinguish different types of cells and cellular structures within the sample. For example, in cheek cells, staining can help identify epithelial cells and differentiate them from other contaminants or debris present in the sample.

Highlighting specific structures: Different stains selectively bind to specific cellular components or structures, allowing researchers to target and visualize specific features of interest.

Learn more about Stains here:

https://brainly.com/question/10245084

#JSP11

Humans can have type A blood, type B blood, type AB blood, or type o. Which of the following is a possible genotype for an individual with type B blood Answers A-D А ТА Br DAT

Answers

Among the given options, the possible genotype for an individual with type B blood is option B: B. This individual would have the genotype "BB" for the ABO blood group.

The ABO blood group system is determined by the presence or absence of specific antigens on the surface of red blood cells. In the case of type B blood, individuals have the B antigen present on their red blood cells.

The genotype for type B blood can be either homozygous (BB) or heterozygous (BO), as the B allele is responsible for producing the B antigen.

In this case, the genotype "BB" indicates that both alleles inherited by the individual are B alleles, resulting in the production of the B antigen on their red blood cells. This genotype is associated with type B blood.

To summarize, the possible genotype for an individual with type B blood is "BB."

To learn more about genotype click here: brainly.com/question/30784786

#SPJ11

A restriction endonuclease breaks Phosphodiester bonds O Base pairs H-bonds O Peptide bonds

Answers

A restriction endonuclease breaks phosphodiester bonds in DNA.

Restriction endonucleases, also known as restriction enzymes, are enzymes that recognize specific DNA sequences and cleave the DNA at those sites. These enzymes play a crucial role in molecular biology techniques, such as DNA cloning and genetic engineering.

The primary function of a restriction endonuclease is to cleave the phosphodiester bonds between nucleotides in the DNA backbone. These phosphodiester bonds connect the sugar-phosphate backbone of the DNA molecule and form the structural framework of the DNA strand. By cleaving these bonds, restriction endonucleases create breaks in the DNA strand, resulting in fragments with exposed ends.

The recognition and cleavage sites of restriction endonucleases are typically specific palindromic DNA sequences. For example, the commonly used restriction enzyme EcoRI recognizes the DNA sequence GAATTC and cleaves between the G and the A, generating overhanging ends.

It is important to note that restriction endonucleases do not break base pairs or hydrogen bonds. Base pairs are formed through hydrogen bonding between complementary nucleotide bases (adenine with thymine or uracil, and guanine with cytosine) and remain intact during the action of restriction endonucleases.

While peptide bonds are involved in linking amino acids in proteins, restriction endonucleases do not cleave peptide bonds as their target is DNA, not protein.

In summary, restriction endonucleases break the phosphodiester bonds that connect nucleotides in the DNA backbone, allowing for the manipulation and analysis of DNA molecules in various molecular biology applications.

To learn more about phosphodiester refer here:

https://brainly.com/question/23660733

#SPJ11

Write the sequence of the complementary strand of each segment of a DNA molecule. A. 5'TGGGTA-3' 3'-_____ -5' b. 5'-ACGCGGTC-3' 3'_____ -5' c. 5'-TCATTCAAG-3' 3'-_____-5' d. 5'-AAAGAGTGGAAAAAX-3'
3'-______-5'

Answers

The sequences of the complementary strands for each segment of the DNA molecule are as follows:

a. 5'TGGGTA-3' - 3'ACCCAT-5' (Option A)

b. 5'-ACGCGGTC-3' - 3'-TGCGCCAG-5' (Option B)

c. 5'-TCATTCAAG-3' - 3'-AGTAAGTTC-5' (Option C)

d. 5'-AAAGAGTGGAAAAAX-3' - 3'-TTTCTCACCTTTTTX-5' (Option D)

To find the complementary strand, you need to identify the base pairing rules in DNA: adenine (A) pairs with thymine (T), and cytosine (C) pairs with guanine (G). By applying these rules, you can determine the complementary sequence by swapping the bases accordingly. For example, in Option A, the original sequence 5'TGGGTA-3' pairs with 3'ACCCAT-5' as the complementary sequence. Similarly, the other options can be determined by applying the base pairing rules.

You can learn more about complementary strands at

https://brainly.com/question/1534778

#SPJ11

which of the following contain unusual eukaryotes which are
without microtubules and mitochondria
microsporidia
archaezoa
rhizopoda
apicomplexan

Answers

Archaezoa and Microsporidia are eukaryotes that are without microtubules and mitochondria.

Archaezoa and Microsporidia are two groups of eukaryotic organisms that lack microtubules and mitochondria.

1. Archaezoa: Archaezoa are a group of unicellular eukaryotes that were once classified as a kingdom within the domain Eukarya.

They are known for their unique characteristics, including the absence of typical eukaryotic organelles such as mitochondria and microtubules.

Instead of mitochondria, Archaezoa possess hydrogenosomes, which are specialized organelles involved in energy metabolism. These organisms exhibit diverse modes of nutrition, including both parasitic and free-living forms.

2. Microsporidia: Microsporidia are a group of intracellular parasitic eukaryotes. They are characterized by their small size and the absence of typical eukaryotic organelles like mitochondria and microtubules.

Instead, they possess unique structures called polar tubes, which are used to infect host cells.

Microsporidia rely on host cells for energy production and other essential cellular functions, as they lack the ability to generate ATP through oxidative phosphorylation in mitochondria.

Rhizopoda and Apicomplexa, on the other hand, do contain microtubules and mitochondria and are not classified as unusual eukaryotes in terms of these organelles.

Rhizopoda, also known as amoebas, are characterized by their ability to form temporary extensions of the cell membrane called pseudopodia, which aid in movement and feeding.

Apicomplexa are a diverse group of parasitic protozoa, including well-known parasites such as Plasmodium, the causative agent of malaria.

They possess a unique apical complex involved in host cell invasion and are known to have both microtubules and mitochondria.

To know more about "Eukaryotes" refer here:

https://brainly.com/question/8364511#

#SPJ11

This is a 5 part question.
In humans, not having albinism (A) is dominant to having albinism (a). Consider a
cross between two carriers: ax Aa. What is the probability that the first child will
not have albinism (A_)?

Answers

In humans, the presence of albinism (a) is a recessive trait while the absence of albinism (A) is dominant. Therefore, we can write Aa for individuals who are carriers of the albinism trait. Let us consider a cross between two carriers; ax Aa.

A Punnett square can be used to determine the probability of offspring phenotypes.

Ax  A  aAa  aa Phenotypic Ratio:3:1

The above Punnett square represents the cross between two carriers. The possible gametes that can be produced by the mother and father are represented along the top and left of the table, respectively.

The phenotypes are listed along the left and top of the table as well. The inside of the table contains the possible genotype combinations of the offspring.

The probability of the first child not having albinism (A_) can be determined by adding the probability of the child having the genotype Aa or AA. Since the absence of albinism (A) is dominant, an individual with the genotype AA will not have albinism.

The probability of a child having an Aa genotype is 2/4, which can be calculated by adding the probabilities of the first two squares in the Punnett square. The probability of a child having an AA genotype is 1/4, which can be calculated by looking at the bottom left square of the Punnett square.

Therefore, the probability of the first child not having albinism is (2/4 + 1/4) = 3/4.

To know more about humans visit;

brainly.com/question/11655619

#SPJ11

In compact bone, the bone cells receive nourishment through minute channels called Select one O a lacunae b. lymphatics costeons O d. lamellae De canaliculi During the thyroidectomy procedure, the sup

Answers

In compact bone, the bone cells receive nourishment through minute channels called canaliculi.

Compact bone is one of the types of bone tissue found in the human body. It is dense and forms the outer layer of most bones. Within the compact bone, there are small spaces called lacunae, which house the bone cells known as osteocytes. These osteocytes are responsible for maintaining the health and integrity of the bone tissue.

To receive nourishment, the osteocytes in compact bone rely on a network of tiny channels called canaliculi. These canaliculi connect the lacunae and allow for the exchange of nutrients, oxygen, and waste products between neighboring osteocytes and the blood vessels within the bone. The canaliculi form a complex network that permeates the compact bone, ensuring that all bone cells have access to vital resources for their metabolic processes.

Overall, the canaliculi play a crucial role in providing nourishment to the bone cells in compact bone, facilitating the exchange of substances necessary for cell function and bone maintenance. This network ensures the vitality and health of the bone tissue, supporting its structural integrity and overall function in the skeletal system.

Learn more about canaliculi:

https://brainly.com/question/30911234

#SPJ11

While shadowing doctors in the ER, a patient with a gun shot wound receives a blood transfusion. Surgeons take care of his wounds, but the blood transfusion was of the incorrect ABO type. Which of the following would not happen?
O a Type II hypersensitivity reaction
O significant production of complement anaphylotixins
O IgG mediated deposition of complement on the transfused RBCs
O the formation of MACS on the transfused RBCs
O Massive release of histamine
O The patient becomes very jaundice as transfused RBCs are lysed

Answers

In the case of an incorrect ABO blood transfusion, the most unlikely event is that the patient becomes very jaundiced as transfused RBCs are Lisdawati is blood? Blood is a specialized body fluid that delivers necessary substances.

The cells in the body steady a supply of oxygen for energy and the expulsion of carbon dioxide is essential. Blood provides a means for the transportation of these necessary substances, as well as cellular waste.

BO blood Groups: BO blood groups are the most important blood groups, which is determined by the presence of antigen A, B, or absence of antigen A and B on red blood cells, and antibodies in plasma (anti-A and anti-B).

To know more about incorrect visit:

https://brainly.com/question/31061687

#SPJ11

which of the following is/are likely to be fertile
a. allodiploids
b. allotetraploids
c. triplioids
d. all
e. none

Answers

Allotetraploids are likely to be fertile. Allotetraploids are organisms that have two complete sets of chromosomes derived from different species.

These organisms usually result from hybridization events between two different species followed by genome doubling. Due to having complete sets of chromosomes, allotetraploids often have balanced chromosomal composition, allowing for normal meiosis and fertility. On the other hand, allodiploids (a) and triploids (c) are less likely to be fertile. Allodiploids have two complete sets of chromosomes derived from different species, but they lack a complete set of chromosomes from either parent species. Triploids, on the other hand, have three complete sets of chromosomes, which can lead to problems during meiosis and reduced fertility.

learn more about:- chromosomes  here

https://brainly.com/question/30077641

#SPJ11

Which of the following about Km is true? a. Km can equal 0. b. Km is the substrate needed to achieve 25% Vmax. c. Km can inform binding affinity. d. Km can inform maximal velocity.

Answers

The answer that is true regarding Km is that Km can inform binding affinity. Km is also known as the Michaelis-Menten constant. The constant describes the relationship between the enzyme and the substrate.

It is used to determine the binding affinity of the enzyme for its substrate. In the case of enzymes, the binding affinity of a substrate and an enzyme is the strength of the interaction between the substrate and the active site of the enzyme. The lower the value of Km, the higher the binding affinity of the enzyme. A low Km indicates that the substrate and the enzyme can interact and form the enzyme-substrate complex quickly.

A high Km indicates that the substrate and enzyme are less efficient at forming the enzyme-substrate complex. Therefore, the correct answer to the question is option C, Km can inform binding affinity.

To know more about enzyme visit:-

https://brainly.com/question/31385011

#SPJ11

Provide the staphylococci species that is coagulase (+).
aureus is a spore-forming bacteria and can survive in high salt environment and tolerate a wide range of temperatures. T/F
Provide two specific drug resistant S. aureus strain that are highly problematic in clinical settings.
Provide the staphylococci species that is capable producing a superantigen.
Provide the names of five enzymes that are important for the pathogenesis of staphylococci.
Describe the mechanism of toxicity of enterotoxins from S. aureus.
What is the function of Fibrinolysin?
What are the major clinical diseases caused by S. aureus?
What is the mechanism of resistance due to PBP 2a expression?
What is the mechanism of resistance in VRSA?
Describe the hemolytic pattern of (a) alpha-, beta- and gamma-hemolysin.
Which specific streptolysin is immunogenic?
Which Streptococci species has hyaluronic acid containing capsule?
Which Streptococci species has sialic acid containing capsule?
Provide the names of three different bacteria that cause pneumonia.
Provide three different ways pneumolysin increases the virulence of S. pneumoniae.
Provide the names of four spore forming bacterial pathogens.
Provide the names of two different bacterial pathogens that produce lactic acid.
What type of virulence factor is diphtheria toxin and what is the mechanism of this exotoxin?
What are the two cell wall components that are specific to mycobacterium and not found in other Gram-positive pathogens?

Answers

Staphylococci species that is coagulase (+): Staphylococcus aureus is the staphylococci species that is coagulase (+). It is a gram-positive bacteria that is present in the human skin and nares. aureus can also survive on surfaces and equipment that have not been disinfected and people carrying this bacteria can act as carriers and spread it to others.

Specific drug-resistant S. aureus strains: MRSA and VISA (Vancomycin-Intermediate Staphylococcus Aureus) are two specific drug-resistant S. aureus strains that are highly problematic in clinical settings. S. aureus species capable of producing a super antigen: S. aureus is the species capable of producing a super antigen.

Enzymes that are important for the pathogenesis of staphylococci: The enzymes that are important for the pathogenesis of staphylococci are catalase, coagulase, hyaluronidase, lipase, and nuclease. Mechanism of toxicity of enterotoxins from S. aureus: Enterotoxins from S. aureus cause food poisoning, with symptoms such as vomiting, diarrhea, and abdominal cramps.

The enterotoxins have super antigenic properties which allow them to activate large numbers of T-cells. The activation of the T-cells leads to the release of cytokines that cause the symptoms of food poisoning.

Fibrinolysin: Fibrinolysin is an enzyme produced by S. aureus that breaks down fibrin clots. It can aid in the spread of the bacteria in the body by allowing them to move through clots and reach new areas.

Major clinical diseases caused by S. aureus: Some of the major clinical diseases caused by S. aureus are skin infections (such as boils and impetigo), pneumonia, bloodstream infections, and endocarditis. Mechanism of resistance due to PBP 2a expression: PBP 2a is a penicillin-binding protein that is not affected by beta-lactam antibiotics. The expression of PBP 2a leads to resistance to beta-lactam antibiotics such as penicillin and cephalosporins.

Mechanism of resistance in VRSA: Vancomycin-resistant S. aureus (VRSA) is resistant to vancomycin, which is usually the drug of last resort for treating S. aureus infections. The resistance is due to the acquisition of a plasmid that carries genes for resistance to both vancomycin and methicillin.

Hemolytic pattern of alpha-, beta-, and gamma-hemolysin: Alpha-hemolysin causes complete lysis of red blood cells, producing a clear zone around the colony. Beta-hemolysin causes partial lysis of red blood cells, producing a green zone around the colony. Gamma-hemolysin does not cause any lysis of red blood cells, producing no zone around the colony.

Specific streptolysin that is immunogenic: Streptolysin O is the specific streptolysin that is immunogenic. Streptococci species with hyaluronic acid-containing capsule: Streptococcus pyogenes is the species with hyaluronic acid-containing capsule.

Streptococci species with sialic acid-containing capsule: Streptococcus pneumoniae is the species with sialic acid-containing capsule.

Bacteria that cause pneumonia: Streptococcus pneumoniae, Haemophilus influenzae, and Legionella pneumophila are three different bacteria that cause pneumonia. Ways pneumolysin increases the virulence of S. pneumoniae: Pneumolysin increases the virulence of S. pneumoniae by promoting the lysis of host cells, activating complement, inducing inflammation, and inhibiting the immune response. Spore-forming bacterial pathogens: Bacillus anthracis, Clostridium botulinum, and Clostridium tetani are four spore-forming bacterial pathogens.

Bacterial pathogens that produce lactic acid: Lactobacillus and Streptococcus are two different bacterial pathogens that produce lactic acid. Virulence factor of diphtheria toxin and mechanism: Diphtheria toxin is an exotoxin that inhibits protein synthesis in eukaryotic cells. It is an A-B toxin, where the A subunit inhibits protein synthesis and the B subunit binds to the cell surface receptors.

Cell wall components specific to mycobacterium: Mycolic acid and arabinogalactan are the two cell wall components that are specific to Mycobacterium and not found in other Gram-positive pathogens.

To know more about Staphylococci visit:

https://brainly.com/question/31580481

#SPJ11

Which of the following is true of a mature mRNA in eukaryotes?
it contains a poly A tail it is translated in the nucleus all of the answer choices are correct it is comprised of introns spliced together

Answers

A mature mRNA in eukaryotes contains a poly A tail. The poly A tail is a sequence of adenine nucleotides that are added to the 3' end of the mRNA molecule, after transcription has been completed.

The poly A tail is important for the stability and export of the mRNA molecule from the nucleus to the cytoplasm, where it will be translated into protein.The other answer choices are incorrect:It is not translated in the nucleus. Translation, which is the process of protein synthesis, occurs in the cytoplasm of the cell after the mRNA molecule has been transported out of the nucleus.

It is not necessarily comprised of introns spliced together. Introns are non-coding regions of the DNA sequence that are removed from the pre-mRNA molecule during RNA splicing. The mature mRNA molecule that is transported to the cytoplasm does not contain introns.

option d is incorrect.All of the answer choices are not correct as option b and d are incorrect. option a is correct.

To know more about mRNA visit:

https://brainly.com/question/29316969

#SPJ11

The enzymes and cofactors necessary to carry out the PCR are added
A. Together with the liquids in the primer mixture for the reaction
B. With the shot or small balls of EdvoBead ™ PLUS
C. After the first few cycles inside the thermocycler
D. At the time the electrophoresis is done

Answers

The enzymes and cofactors necessary to carry out the Polymerase Chain Reaction (PCR) are added with the liquids in the primer mixture for the reaction.

PCR is a widely used molecular biology technique that allows for the amplification of specific DNA sequences. The key components required for PCR include a DNA template, primers, DNA polymerase, nucleotides, and cofactors. The enzymes and cofactors necessary for PCR are typically included in the PCR reaction mix. These components are added together with the liquids in the primer mixture for the reaction. The primer mixture contains the forward and reverse primers that are specific to the target DNA sequence to be amplified.

The enzymes involved in PCR include a heat-stable DNA polymerase, such as Taq polymerase, which can withstand the high temperatures required for denaturation during the PCR cycles. Cofactors, such as magnesium ions (Mg2+), are also included in the reaction mix as they are essential for the activity of the DNA polymerase. The PCR reaction mix is prepared before the reaction is initiated. It contains all the necessary components, including enzymes and cofactors, to enable DNA amplification. Once the reaction mix is prepared, it is added to the PCR tubes or wells, along with the DNA template and primers.

The PCR reaction then proceeds through cycles of denaturation, annealing, and extension within the thermocycler machine. The addition of enzymes and cofactors at this stage ensures their presence throughout the PCR process and enables efficient DNA amplification.

Learn more about enzymes here: https://brainly.com/question/30600790

#SPJ11

Briefly describe how the 3 different types of neurotransmitters are synthesized and stored. Question 2 Briefly describe how neurotransmitters are released in response to an action potential.

Answers

Neurotransmitters are chemical messengers that transmit signals across synapses from one neuron to another, as well as from neurons to muscles or glands.

They are classified into three categories, each of which is synthesized and stored differently. These categories are:Acetylcholine, monoamines, and amino acidsAcetylcholine is synthesized by combining choline and acetyl CoA in nerve terminals using the enzyme choline acetyltransferase (ChAT). Once synthesized, acetylcholine is stored in vesicles in nerve terminals.Monoamines are synthesized from dietary amino acids, such as phenylalanine, tyrosine, and tryptophan. Monoamines are synthesized using enzymes present in neurons, such as tyrosine hydroxylase and dopamine β-hydroxylase. Once synthesized, monoamines are stored in vesicles in nerve terminals.Amino acids are synthesized by neurons themselves. GABA, for example, is synthesized from glutamate, while glutamate is synthesized from α-ketoglutarate.

Once synthesized, amino acids are stored in vesicles in nerve terminals. The release of neurotransmitters occurs when an action potential reaches the terminal of a presynaptic neuron. This causes the depolarization of the nerve terminal, which in turn triggers the influx of calcium ions into the terminal. The increase in calcium ion concentration causes synaptic vesicles containing neurotransmitters to fuse with the membrane, releasing their contents into the synaptic cleft. Neurotransmitters bind to receptors on the postsynaptic neuron and trigger a response that allows for the propagation of the signal.

To know more about Neurotransmitters  visit

https://brainly.com/question/30336628

#SPJ11

B.
• Briefly explain how the structure and chemical properties of each of the four biologically important molecules affects and influences their function.
C.
• Briefly explain how DNA stores and transmits information
• Describe three forms of RNA and list one function of each form

Answers

The structure and chemical properties of biologically important molecules play a crucial role in determining their functions. Lipids, with their hydrophobic nature, are involved in energy storage, insulation, and the formation of cell membranes.

Nucleic acids, specifically DNA, store and transmit genetic information through their unique double-stranded helical structure and the complementary base pairing of nucleotides.

DNA (deoxyribonucleic acid) stores and transmits genetic information through its specific structure and chemical properties. The double-stranded helical structure of DNA allows for the stable storage of genetic information. The sequence of nucleotides along the DNA molecule contains the instructions for building and maintaining an organism. During DNA replication, the complementary base pairing of nucleotides allows for accurate transmission of genetic information from one generation to the next.

RNA (ribonucleic acid) has multiple forms, each with distinct functions. Messenger RNA (mRNA) carries the genetic information from DNA to the ribosomes, where it serves as a template for protein synthesis. Transfer RNA (tRNA) is responsible for delivering amino acids to the ribosomes during protein synthesis. It recognizes specific codons on the mRNA and ensures the accurate assembly of amino acids into a polypeptide chain. Ribosomal RNA (rRNA) is a major component of ribosomes, the cellular machinery responsible for protein synthesis. It provides the structural framework for the ribosome and catalyzes the formation of peptide bonds.

Learn more about ribosomes here:

https://brainly.com/question/9333250

#SPJ11

Factor X can be activated O Only if the is Factor VII O Only if both intrinsic and extrinsic pathways are activated. O Only if the intrinsic pathway is acticated. O Only if the extrinsic pathway is ac

Answers

Factor X can be activated B. only if both intrinsic and extrinsic pathways are activated.

Blood clotting or coagulation is a complex process that requires the participation of several factors. Factor X is one of the clotting factors that participate in the coagulation cascade, a series of steps that culminate in the formation of a blood clot. When the lining of a blood vessel is injured, two pathways, the intrinsic and the extrinsic, initiate the clotting process. The extrinsic pathway is triggered by the release of tissue factor from damaged cells outside the blood vessels.

On the other hand, the intrinsic pathway is activated by the exposure of subendothelial collagen to blood after vessel damage. Once activated, the two pathways converge to activate factor X, which is then converted to factor Xa by a series of proteolytic cleavages. Factor Xa, in turn, activates prothrombin to thrombin, which converts fibrinogen to fibrin, the main protein that forms a blood clot. So therefore the correct answer  is B. only if both intrinsic and extrinsic pathways are activated, Factor X can be activated.

Learn more about blood clotting  at:

https://brainly.com/question/29841695

#SPJ11

When you recognize the characteristics of living
things, do you recognize virus as living?
if yes why?
if not, why not?
(please in your own words)

Answers

Although viruses share some similarities with living organisms, such as the ability to evolve and adapt to their environment, they lack the basic properties and cellular organization of living things. Therefore, viruses are not typically regarded as living things.

When you recognize the characteristics of living things, you may not recognize a virus as living as it lacks several fundamental characteristics of living things. For example, viruses cannot reproduce on their own; they require a host cell to replicate. Additionally, they do not generate or utilize energy, which is a fundamental characteristic of all living things.Furthermore, viruses do not have cellular organization and are not composed of cells, which is another vital characteristic of all living things. They are simply a piece of nucleic acid, either DNA or RNA, surrounded by a protein coat.Although viruses share some similarities with living organisms, such as the ability to evolve and adapt to their environment, they lack the basic properties and cellular organization of living things. Therefore, viruses are not typically regarded as living things.

To know more about environment visit:

https://brainly.com/question/5511643

#SPJ11

Which of the following 3 letter codon sequences serve as stop codon(s)?
a. UAG
b. UAA
c. UAU
d. UGA
Based on your answer above, of the remaining codons, which amino acids are encoded?
Group of answer choices
a. Tyr
b. Thr
c. Asn
d. Trp
Given the following DNA coding sequence: 3’ TGACCGATA 5’. Which of the answers below represents the mRNA sequence in the correct direction for this sequence?
a. DNA; 5’ GACTTACGT 3’
b. DNA; 3’ ACTGGCTAT 5’
c. RNA; 5’ UGACCGAUA 3’
d. RNA; 5’ AUAGCCAGU 3’
Consider the DNA non-template strand: 5’ – CAC GAA TAT – 3’. What is the correct amino acid sequence?
a. His – Glu – Tyr
b. Pro – Cys – Gly
c. Arg – Thr – Pro
d. Arg – Cys – Ser
Correct order of transcription and translation steps
a. Initiation, elongation, termination
b. Hot start, amplification, ligation
c. Indication, extension, completion
d. denaturation, annealing, extension
Which protein is involved in eukaryotic transcription termination.
a. Ligase
b. Transcription terminase
c. mfd
d. Rho protein
e. None of the above
If the coding DNA triplet TGG for tryptophan in the middle of the gene sequence mutates to TGT what would you expect during translation?
a. Tryptophan would be substituted with Cysteine
b. This codon will be skipped
c. Translation won’t be initiated
d. Translation would stop prematurely

Answers

If the coding DNA triplet TGG for tryptophan in the middle of the gene sequence mutates to TGT, during translation, you would expect Tryptophan to be substituted with Cysteine.

The correct answer is: Stop codon(s): a. UAG and b. UAA. The remaining codons encode the following amino acids: a. Tyr (Tyrosine)

b. Thr (Threonine)

c. Asn (Asparagine)

The correct mRNA sequence for the given DNA coding sequence (3’ TGACCGATA 5’) in the correct direction is:

c. RNA; 5’ UGACCGAUA 3’

The correct amino acid sequence for the DNA non-template strand (5’ – CAC GAA TAT – 3’) is:

a. His – Glu – Tyr

The correct order of transcription and translation steps is:

a. Initiation, elongation, termination

The protein involved in eukaryotic transcription termination is:

d. Rho protein

If the coding DNA triplet TGG for tryptophan in the middle of the gene sequence mutates to TGT, you would expect the following during translation:

a. Tryptophan would be substituted with Cysteine

Translation would continue with the substitution of the amino acid Cysteine instead of Tryptophan due to the change in the codon.

To know more about tryptophan

https://brainly.com/question/776786

#SPJ11

Question 24 (1 point) Chronically elevated cortisol may cause all of the following EXCEPT: O a) promotes insulin resistance and obesity Ob) increases muscle mass O c) promotes telomere shortening O d) weakens the immune response

Answers

Chronically elevated cortisol may cause all of the following except: increases muscle mass (option B).

What is the effect of elevated cortisol?

Cortisol is a steroid hormone produced and released by the adrenal glands, the endocrine glands above the kidneys.

Cortisol is an essential hormone that affects almost every organ and tissue in the body, however, higher-than-normal or lower-than-normal cortisol levels can be harmful to one's health.

Effects of chronic elevated levels of cortisol includes the following;

promotes insulin resistance and obesitypromotes telomere shorteningweakens the immune response

Learn more about cortisol at: https://brainly.com/question/30730834

#SPJ4

2. State whether decreasing the amount of oxygen (02) in inhaled air increased, reduced or did not change arterial carbon dioxide partial pressure from ordinary. 3. State whether decreasing the amount of O, in inhaled air increased, decreased or did not change plasma pH from normal.

Answers

Decreasing the amount of oxygen in inhaled air increases the arterial carbon dioxide partial pressure from ordinary. While decreasing the amount of oxygen in inhaled air decreases the plasma pH from normal. Arterial carbon dioxide partial pressure refers to the measure of the carbon dioxide concentration in the blood plasma of arteries.

The normal range for arterial carbon dioxide partial pressure is 35-45 mm Hg (millimeters of mercury). However, in the case of a decrease in oxygen inhalation, the arterial carbon dioxide partial pressure will increase. Why does this happen? It's because when oxygen levels are low, the body tends to retain carbon dioxide rather than expel it.What is plasma pH?The pH level of the plasma is referred to as plasma pH.

The normal range for plasma pH is between 7.35 and 7.45. When there is a decrease in the amount of oxygen inhalation, plasma pH decreases as well. This is because carbon dioxide is retained, which creates an acidic environment in the plasma.

To know more about arterial visit:

https://brainly.com/question/32366318

#SPJ11

which of the following microorganism inhibit adherence with
phagocytes because of the presence of m proteins
1. mycobacterium tuberculosis steptococcus pyogenes leishmania
klesiella pneumoniae

Answers

The microorganism that inhibits adherence with phagocytes because of the presence of m proteins is Steptococcus pyogenes.

What are m proteins?

M proteins are the fibrous surface proteins found on Streptococcus pyogenes bacteria.

M proteins are important virulence factors of the bacteria, and they play a role in the development of rheumatic fever and acute glomerulonephritis.

They can also be used to classify Streptococcus pyogenes bacteria into different strains.

They are capable of masking the bacteria's surface antigens, rendering them immune to phagocytosis.

The Streptococcus pyogenes bacterium has m proteins on its surface.

These proteins help the bacterium avoid being detected by immune cells and phagocytes.

As a result, the bacterium is able to evade the immune system and spread throughout the body, causing a variety of infections.

To know more about proteins visit:

https://brainly.com/question/2193769

#SPJ11

Other Questions
4.1.10 There are a number of ways in which cancer can evade the immune response. Which of the following cell types is able to kill malignant cells that have stopped expressing class I MHC?a.macrophagesb.CD4 T cellsc.NK cellsd.CD8 T cells 7. How does insulin release cause an increased uptake of glucose in skeletal muscle? How is glucose uptake maintained during exercise? Maximum word limit is 200 words. Four kilograms of carbon dioxide (CO2) is contained in a piston-cylinder assembly with a constant pressure of 2 bar and initial volume of 1m. Energy is transferred by heat to the CO2 at a rate of 15 W for 2.5 hours. During this process, the specific internal energy increases by 10 kJ/kg. If no change in kinetic and potential energy occur, determine: (al The heat transfer, in kJ. (b) The final volume, in m. Enter the value for problem 8, part (a). Enter the value for problem 8, part (b). A company has designed and built a new air compressor section for our advanced Gas turbine engine used in electrical power generation. They state that their compressor operates adiabatically, and has a pressure ratio of 30. The inlet temperature is 35 deg C and the inlet pressure is 100 kPa. The mass flow rate is steady and is 50 kg/s The stated power to run the compressor is 24713 kW Cp = 1.005 kJ/kg K k=1.4 What is the actual temperature at the compressor outlet? O 800 KO 656 KO 815 KO 92.6 deg C Find the probability of exactly five successes in seven trials of a binomial experiment in which the probability of success is 70%. Round to the nearest tenth of a percent. You are evaluating a child who has tested positive for one of the C4A alleles associated with schizophrenia. You are aware of a new drug that can reduce synaptic pruning in patients who are susceptible to developing schizophrenia. Why must you proceed with caution before giving this patient the drug? Multiple Choice The disease has a significant environmental component, so not everyone who has these gene variants will develop the disease. Because of the timing of the onset of schizophrenia, it may be too late to give the child the drug Because there are so many variants of the C4 genes, it is impossible to be sure which ones are associated with schizophrenia. You will need to be sure both parents had schizophrenia before administering the drug Which of these cells produces the factors for humorimmunity?A.Plasma B cellsB.CD4 T cellsC.NK CellsD.Naive B cellsE.Macrophages You would expect most endospres tobe difficult to stainstain easily Question 54 Which of the following is true regarding leukocidins? O They are secreted outside a bacterial cell They destroy red blood cells O They are superantigens O They are a type of A-B toxin O Th The random variable X has a uniform distribution over 0 x 2. Find v(t), Rv'(t, t), and v(t) for the random process v(t) = 6 cos (xt) Suggest a promotional campaign to be carried out by PIZZA HUT to reflect the core change that will be incorporated to its marketing strategy (as suggested by your answer in question 6). Make sure to outline (a) the chosen channel (b) what kind of content will be presented to the customers using this channel, and (c) what segmentation strategy will be used for this marketing campaign biomechanics questionA patient presents to your office with a complaint of low back pain. Upon examination you detect a rotation restriction of L3 around the coronal axis. What's the most likely malposition? a.-02 Ob.-8x A smooth, flat plate, 3.0 meters wide and 0.6 meters long parallel to the flow, is immersed in 15C water (p = 999.1 kg/m, v = 1.139 x 106 m /s) flowing at an undisturbed velocity of 0.9 m/s. a) How thick is the boundary layer at the plate's center? b) Find the location and magnitude of the minimum surface shear stress experienced by the plate. c) Find the total friction drag on one side of the plate. Nonhealing wounds on the surface of the body are often extremely difficult to manage, in part because the microbial cause of the lack of healing is often extremely difficult to identify. Create a list of reasons this might be the case. The minimum pressure on an object moving horizontally in water (Ttemperatu at10 degree centrigrade) at (x + 5) mm/s (where x is the last two digits of your student ID) at a depth of 1 m is 80 kPa (absolute). Calculate the velocity that will initiate cavitation. Assume the atmospheric pressure as 100 kPa (absolute). Scan the solution and upload in vUWS before moving to the next question. the stages of change theory and social cognitive theories are the two most widely cited theories that relate to What is the tolerance assuming the third order surveying when the closed loop distance is 1821 ft? a) 2.13 ft b) 1.68 ft O c) 0.23 ft d) None of the given answers O e) 0.29 ft Of) 0.03 ft g) 0.02 ft The hydrolysis of ATP above pH 7 is entropically favoredbecausea.The electronic strain between the negative charges isreduced.b.The released phosphate group can exist in multiple resonanceformsc In a DNA bisulfite sequencing experiment, the following read count data for a given cytosine site in a genome were obtained:Converted Read Unconverted Read(Not methylated) (Methylated)Cytosine Site 1 40 17Other Sites 2130 3611a : Specify a binomial statistical model for the above data and compute the MLE (Maximum Likelihood Estimation) for the model parameter, which should be the probability of methylation. (Round your answer to 3 decimal places)1b: Assume that the true background un-conversion ratio = 0.04 is known, compute the one-sided p-value for the alternative hypothesis that the methylation proportion of cytosine site 1 is larger than the background. In your answer, use the R code `pbinom(q, size, prob)` to represent the outcome of the binomial CDF, i.e. the outcome of `pbinom(q, size, prob)` is ( q) , where ~om( = prob, = size). 1c : Given the supplemented total counts for the rest of the genome, perform a new one- sided test to determine whether the methylation level on cytosine site 1 is significant or not.Converted Read Unconverted Read(Not methylated) (Methylated)Cytosine Site 1 40 17Other Sites 2130 361 P.S. You should not use the background un-conversion ratio in the last question. In your answer, you may use one of the pseudo codes ` pbinom(q, size, prob) `, ` phyper(q, m, n, k) `, and `pchisq(q, df)` to represent the CDF of binomial distribution, hypergeometric distribution, and chi-squared distribution respectively. For hypergeometric distribution, q is the number of white balls drawn without replacement, m is the number of white balls in the urn, n is the number ofblack balls in the urn, k is the number of balls drawn from the urn.1d : Assume you have obtained the following p-values for 5 sites at a locus in the genome:p-valueSite 1 0.005Site 2 0.627Site 3 0.941Site 4 0.120Site 5 0.022Compute the adjusted p-value with Bonferroni correction (if the adjusted p > 1, return the value of 1), and filter the adjusted p-value with alpha = 0.05. Which site remains significant after the adjustment? Name another adjustment method that is less stringent but more powerful than the Bonferroni correcti Unpolarised light is incident on an air-glass interface from the air side. You are told that the glass has a refractive index of 1.45, explain what measurement, involving polarisation, that you could do to confirm this is correct.