When is conflict said to be sexual? In what way is genomic imprinting an outcome of sexual conflict?

Answers

Answer 1

Conflict is said to be sexual when it involves sexual traits that may benefit one sex while harming the other. In this case, the conflict is usually between males and females, as they have different reproductive strategies.

One example of sexual conflict is mate choice, where males may want to mate with as many females as possible, while females may be selective and only mate with the best males.Genomic imprinting is an outcome of sexual conflict as it results from the differing interests of the maternal and paternal genomes in offspring development. Genomic imprinting occurs when only one allele from either the mother or the father is expressed, leading to differences in gene expression depending on the parent of origin. This process is thought to result from the evolutionary battle between the sexes, where females may benefit from limiting the resources invested in male offspring, while males may benefit from overproducing sperm and mating with as many females as possible. Thus, genomic imprinting can be seen as a way of resolving sexual conflict and ensuring that offspring receive the optimal combination of genes from their parents.

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Related Questions

A drug is noted to cause a change in the resting membrane potential of renal epithelial cells from -60 mV to -50 mV. Which of the following mechanisms is most likely to be employed by the drug?
A. Decreased rate of diffusion of potassium into the cells
B. Increased rate of diffusion of potassium into the cells
C. Decreased rate of diffusion of sodium into the cells
D. Increased rate of diffusion of sodium into the cells
E. Decreased rate of diffusion of calcium into the cells

Answers

The mechanism most likely to be employed by the drug that causes a change in the resting membrane potential of renal epithelial cells from -60 mV to -50 mV is "Increased rate of diffusion of sodium into the cells".Sodium ions play a crucial role in determining the membrane potential of cells.

Their concentration gradient across the plasma membrane generates a potential difference (or voltage), which is maintained by the ATP-dependent Na+/K+ pump. As a result, any substance that alters the rate of Na+ entry or exit from cells will impact the membrane potential, either by depolarization (i.e., making the potential less negative) or hyperpolarization (i.e., making the potential more negative).

Here, we are given that a drug is noted to cause a change in the resting membrane potential of renal epithelial cells from -60 mV to -50 mV. This means that the drug is increasing the membrane potential of the cells (i.e., depolarizing them) by allowing more positive ions (e.g., sodium) to enter the cells.

Therefore, the most likely mechanism employed by the drug is "Increased rate of diffusion of sodium into the cells". Hence, the correct answer is option D.

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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.

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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).

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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.

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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.

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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.

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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.

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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.

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Question 37
Which of the following is NOT produced by TFH?
A. TGF beta
B. BIL-4
C. IL-21
Question 38
A woman was seen in a rheumatology clinic with fatigue, a low-grade fever, weight loss, and a nonspecific rash on her face and chest. What do you think is the cause of the systemic symptoms (fever, weight loss)?
A. immune complexes
B complement activation
C.Inflammatory cytokines
D. ADCC

Answers

The correct answer to the given question is option A.

TGF beta.TFH (T follicular helper) cells are a particular kind of T cell that plays a crucial role in the immune system. These cells are essential in helping B cells create immunological memory and antibodies. TFH cells secrete various cytokines to help B cells develop into antibody-secreting cells.

However, TGF-beta is not produced by TFH cells.

The correct option is C. Inflammatory cytokines.

The cause of these systemic symptoms is due to inflammatory cytokines. Inflammatory cytokines are a kind of signalling molecule released by immune cells. They cause inflammation and fever by stimulating the body's immune cells to attack pathogens.

Inflammation causes the immune system to move towards the affected area, causing redness, swelling, and warmth. This can be seen in the woman with fatigue, a low-grade fever, weight loss, and a nonspecific rash on her face and chest. Therefore, the correct option is C. Inflammatory cytokines.

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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.

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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.

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Tryptic soy agar is an example of (select all that apply) General Purpose Media Semi-Solid Media Selective Media Solid Media Enriched Media Liquid Broth Media

Answers

Tryptic soy agar is an example of General Purpose Media, Solid Media, and Enriched Media.

General Purpose Media:

This media supports the growth of most non-fastidious bacteria, including gram-positive and gram-negative bacteria.

Solid Media: Solid agar is used in a variety of lab applications.

It aids in the isolation and analysis of bacteria in microbiology labs.

Solid media, unlike liquid media, provides a solid surface for bacteria to grow on and allows for colony-forming units (CFUs) to be counted.

Enriched Media:

This is a type of media that has been formulated to supply microorganisms with all of the nutrients that they need to thrive.

Enriched media typically contains added nutrients that promote the growth of fastidious bacteria or support the growth of bacteria with unique nutritional requirements.

So, the correct options are General Purpose Media, Solid Media, and Enriched Media.

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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.

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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.

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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.

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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.

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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.

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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

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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.

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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.

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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.

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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.

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1. Semen travels through the male reproductive tract in this order: a. ejaculatory duct, vas deferens, epididymis, urethra b. epididymis, vas deferens, ejaculatory duct, urethra c. urethra, ejaculator

Answers

Semen is produced in the testicles and travels through the male reproductive system in the following order:

The testes produce sperm, which are stored and matured in the epididymis.

When sperm are needed, they travel through the vas deferens and into the ejaculatory duct.

Seminal fluid is added to the sperm in the seminal vesicles and prostate gland, which is then mixed and expelled through the urethra during ejaculation.

The correct order in which semen travels through the male reproductive tract is:

The epididymis is a long, coiled tube that sits on top of each testicle and serves as a site of sperm maturation and storage.

The vas deferens is a muscular tube that connects the epididymis to the urethra.

The ejaculatory duct is formed by the union of the vas deferens and seminal vesicles, and it passes through the prostate gland to empty into the urethra.

Understanding the anatomy and function of the male reproductive system is important for overall health and wellness.

Semen is composed of fluid and sperm.

It is ejaculated from the male reproductive system during orgasm.

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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.

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What
have been the impact of widespread destruction of California's
Tidal Marshes/Estuaries?

Answers

The widespread destruction of California's tidal marshes/estuaries has had significant ecological and socio-economic impacts.

The destruction of California's tidal marshes and estuaries has resulted in profound ecological consequences. These habitats serve as vital breeding, nesting, and feeding grounds for numerous species, including fish, birds, and mammals. With their destruction, the loss of critical habitat has led to declines in biodiversity, negatively impacting the overall health of ecosystems. Additionally, tidal marshes and estuaries play a crucial role in water filtration and nutrient cycling, helping to maintain water quality and support healthy fisheries. The destruction of these habitats disrupts these processes, leading to imbalances in the ecosystem.

The destruction of California's tidal marshes and estuaries also has socio-economic implications. These habitats provide essential services such as coastal protection by acting as natural buffers against storms and reducing the risk of coastal erosion. Without them, coastal communities are more vulnerable to the impacts of storms, leading to increased property damage and potential loss of life. Tidal marshes and estuaries also contribute to the economy through recreational activities like birdwatching, fishing, and boating, attracting tourists and supporting local businesses. Their destruction not only impacts the livelihoods of those directly dependent on these activities but also affects the broader coastal economy.

In conclusion, the widespread destruction of California's tidal marshes and estuaries has had far-reaching impacts on both ecological systems and human communities. Conservation and restoration efforts are crucial to mitigate these effects, protect biodiversity, and ensure the resilience and sustainability of California's coastal ecosystems.

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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.

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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."

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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.

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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.

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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.

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Other Questions
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. 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