Answer:
The correct answer is the third statement.
Explanation:
The epidermal growth factor refers to a ligand that stimulates the epidermal growth factor receptor or EGFR. The EGFR refers to a receptor tyrosine kinase. With the binding of a ligand to a receptor, the phenomenon of dimerization takes place, which makes the receptor to cross-phosphorylate each other within the intracellular domain, this eventually results in the phosphorylation of the downstream signaling molecules.
This transduction of signals eventually results in cell differentiation. However, when the intracellular domain is not present, the signal transduction cannot occur within the cell. Thus, if the mutated EGFR, that is, without intracellular domain gets overexpressed, the ligand combines with the mutated receptor, its dimerization takes place with the help of the extracellular domain, but it will not be able to perform signal transduction, due to the absence of intracellular domain. Therefore, dimerization of EGFR will take place, without performing cross phosphorylation, which will eventually prevent cell growth.
You fill a shallow pan with water and place a drop of red ink in one end of the pan and a drop of green ink in the other end. Which of the following is true at equilibrium? a. The red ink is uniformly distributed in one-half of the pan, and the green ink is uniformly distributed in the other half of the pan. b. The red and green inks are uniformly distributed throughout the pan. c. Each ink is moving down its concentration gradient. d. The concentration of each ink is higher at one end of the pan than at the other end.
Explanation:
When we fill a shallow pan with water and place a drop of red ink in one end of the pan and a drop of green ink in the other end we see that the red nd green inks are uniformly distributed throughout the pan. This is due to the Diffusion of red ink and green ink in the water. Diffusion is the process where the movement of particles takes place from the region of higher concentrations to the region of lower concentartions. Until the concentration of substances is uniform throughout, this phenomena continues.
At equilibrium, the red ink is uniformly distributed in one-half of the pan, and the green ink is uniformly distributed in the other half of the pan.
Explanation:At equilibrium, the correct statement is:
a. The red ink is uniformly distributed in one-half of the pan, and the green ink is uniformly distributed in the other half of the pan.
Equilibrium is reached when the concentration of each ink becomes uniform throughout the pan. This happens because the molecules of the red and green ink are constantly moving due to Brownian motion. Over time, they will mix and spread evenly, resulting in a uniform distribution of the inks in the pan.
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In avocets, curved beaks result from a Z-linked recessives gene. A curve-beak male is mated with a flat-beaked (normal) female. The F1 are interbred to produce the F2. What is the probability of producing a curve-beak female in the F2
Answer:
50%
Explanation:
Z O
z Zz zo
z Zz zo
The males will have two copies of the alleles of the genes present on the Z chromosomes. As females have only one Z chromosome, they will have only one allele for the genes present on the Z chromosome.
The results from the punnet square show that there will be chance that the females of the F2 progeny will have curved-beaks.
Final answer:
The probability of producing a curve-beak female in the F2 generation when a curve-beak male is mated with a flat-beaked female, and their F1 offspring are interbred, is 25%. This calculation assumes the trait for a curved beak is Z-linked recessive, and it uses the basic principles of genetics and the specific sex chromosome composition of birds.
Explanation:
The question asks about the inheritance of a Z-linked recessive trait in avocets, specifically the probability of producing a curve-beak female in the F2 generation when a curve-beak male is mated with a flat-beaked female. In birds, males have two Z chromosomes (ZZ) and females have one Z and one W chromosome (ZW). A Z-linked recessive trait, like the curved beak in avocets, will manifest in females (ZW) if the Z chromosome carries the recessive allele since there is no corresponding allele on the W chromosome to mask it.
First, consider the parental generation (P): The male is curve-beaked and therefore has the genotype ZZc where c indicates the recessive allele for the curved beak. The flat-beaked female, assuming she does not carry the recessive allele (as it's not expressed and she's normal), would have the genotype ZW.
The F1 generation, resulting from their mating, would consist of males (ZcZ) and females (ZcW), all of which would have flat beaks because the curved beak is recessive.
When these F1 individuals interbreed, the F2 generation could have the following genotypes:
ZZ (flat-beaked males)
ZcZ (flat-beaked males carrying the recessive gene)
ZW (flat-beaked females)
ZcW (curve-beaked females)
The only way to produce a curve-beaked female is from an F1 male (ZcZ) mating with an F1 female (ZcW), resulting in the ZcW genotype. Therefore, the probability of this outcome is 1/4 or 25% since there are four possible combinations (ZZ, ZcZ, ZW, ZcW) and only one produces a curve-beaked female.
The cornea is the transparent outer layer of the human eye. Because it must be transparent to light, it does not normally contain blood vessels. Therefore, it must receive its nutrients via diffusion. Oxygen from the surrounding air diffuses to the cornea through the surface tears whereas other nutrients diffuse to the cornea from the inner parts of the eye, such as the vitreous humor and lens.
During operation, the cornea produces waste in the form of CO2 gas that must be expelled to keep the eye healthy and functioning. This is accomplished by the simultaneous diffusion of CO2 from the cornea to the surrounding atmosphere, which generally features a low CO2 concentration.
It is therefore critical that modern contact lens materials allow sufficient diffusion rates of oxygen and carbon dioxide. Without oxygen, the cornea will warp, loose transparency, and become susceptible to scarring. The body may also react by growing additional blood vessels into the eye, which can damage the cornea.
If an increased steady-state flow rate of O2 (oxygen molecules per second) to the cornea is desired, which of the following contact lens / ambient condition modifications is not likely to be useful?
Note: the flow rate is equal to product of the diffusion flux and an area of interest through which diffusion occurs.
(a) Increase the contact lens thickness
(b) Increase the diffusivity of oxygen gas by increasing the contact lens porosity
(c) Increase the ambient temperature
(d) Increase the ambient partial pressure of oxygen gas
(e) All of the suggestions (a-d) are useful for increasing the flow rate of oxygen
the answer is not c
Increasing the thickness of a contact lens would decrease the flow rate of oxygen to the cornea, not increase it. Thus, this modification would not be useful if the goal is to increase the steady-state flow rate of oxygen to the cornea.
Explanation:The best answer is (a) Increase the contact lens thickness. In the context of diffusion, the flow rate of a molecule is inversely proportional to the thickness of the layer it has to go through. Increasing the contact lens thickness would effectively reduce the rate at which oxygen reaches the cornea, thereby reducing the flow rate, not increasing it as desired. On the other hand, solutions (b), (c), and (d) could potentially increase the flow rate of oxygen to the cornea. Increasing the diffusivity, ambient temperature, or ambient partial pressure of oxygen gas would all potentially increase the rate of oxygen diffusion to the cornea.
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In the bacteria that carry this out, dissimilatory nitrate reduction can lead directly to the production of ___________ in these cells. A. amino acids and nucleotides. B. NO2- C. N2 D. NH3 E. None of the above.
Answer: D. NH3
Explanation:
The dissimilatory nitrate reduction is a process which is conducted by the anaerobic bacteria. In this process the nitrate is to reduce to the ammonia. This process is conducted by the bacteria or chemoorganoheterotrophic microbes. These organisms use nitrate as an electron acceptor for the respiration. These organisms organisms oxidize the organic matter and utilize the nitrate as an electron acceptor. The nitrate is further reduced to nitrite and then to ammonia. The ammonia can be found in the cells of these organisms.
Imagine a cell whose membrane is permeable to water, but impermeable to solute X. If solute X is more concentrated outside the cell than inside, which way will water move by osmosis?
A. Water will move equally in both directions, in and out of the cell.
B. Water will enter the cell.
C. Water will not move in either direction, in or out of the cell.
D. Water will leave the cell.
Answer: D
Explanation:osmosis is the movement of molecules from a region of lower concentration to a region of higher concentration through a semi permeable membrane.
Final answer:
Water will move from inside the cell to the outside by osmosis because the concentration of solute X is higher outside the cell, making the outside a region of lower water concentration.
Explanation:
If solute X is more concentrated outside the cell than inside, and the cell's membrane is permeable to water but impermeable to solute X, water will move by osmosis from inside the cell to outside. This is because osmosis is the diffusion of water across a membrane from an area of higher water concentration (or lower solute concentration) to an area of lower water concentration (or higher solute concentration). Therefore, the correct answer is D. Water will leave the cell.
Which term is not correctly matched with its proper definition? Prototroph: possesses the same metabolic/growth capabilities as the parent, wild type strain. The prefixes chemo- and photo-: define the electron source Macronutrient: phosphorous and sulfur are in this category Chemoorganotroph: a heterotroph could be considered this Growth factor: components, either defined or undefined, that promote/support growth
Answer:
The prefixes chemo- and photo-: define the electron source
Explanation:
The prefix 'chemo' indicates something that is a “chemical” or “chemically induced” while the prefix "photo" relates to light. An electron source is that which produces electrons. The prefixx chemo and photo do not relate to sources of electron production.
The term 'Prototroph' is incorrectly matched with its description. A Prototroph is a microorganism that can synthesize all compounds needed for its growth and does not specifically relate to maintaining the same metabolic/growth capabilities as the parent, wild type strain. Other terms are correctly matched with their definitions.
Explanation:In the list of terms and their corresponding definitions, the term that isn't aptly matched is Prototroph. The definition of Prototroph usually refers to a microorganism that, unlike an auxotroph, does not require any additional nutrient other than a source of organic carbon for growth. It does not specifically relate to maintaining the same metabolic/growth capabilities as the parent, wild type strain.
Specifically, a Prototroph is a strain of microorganism that possesses a complete set of genes, enabling it to synthesize all compounds needed for its growth, thereby not requiring any growth factors. Similarly, a wild strain usually refers to the 'normal' or 'natural' condition of an organism, but it may or may not always have the same growth capabilities as its parent strain.
On the other hand, the terms and definitions of Macronutrient, Chemoorganotroph, Growth factor, and terms with the prefixes chemo- and photo- are correctly matched. These terms are standard in the field of Microbiology.
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How does hemoglobin function as a pH buffer?a.Hemoglobin binds hydrogen ions when carbon dioxide exits the red blood cell. b.Hemoglobin releases hydrogen ions after carbon dioxide enters the red blood cell. c.Hemoglobin binds hydrogen ions after carbon dioxide enters the red blood cell. d.Hemoglobin releases hydrogen ions when oxygen exits the red blood cell.
Answer: Option C.
Haemoglobin binds Hydrogen ion after carbondioxide enters red blood cells.
Explanation:
Haemoglobin is the protein in the red blood cells that help to transport oxygen in the blood. It is an iron compound. Haemoglobin acct as buffer by binding to acid or hydrogen ion in the blood when carbondioxide enters the blood, to remove the acid in the blood before it changes the blood pH.
Final answer:
Hemoglobin acts as a pH buffer by binding free hydrogen ions when carbon dioxide is converted into bicarbonate ions within red blood cells, preventing significant pH changes. So, the correct option is c : Hemoglobin binds hydrogen ions after carbon dioxide enters the red blood cell.
Explanation:
Hemoglobin functions as a pH buffer in the bicarbonate buffer system, which plays a crucial role in the body's regulation of blood pH. Carbon dioxide (CO2) enters red blood cells and is converted into carbonic acid (H2CO3) by the enzyme carbonic anhydrase (CA). This unstable intermediate quickly dissociates into bicarbonate ions (HCO3-) and hydrogen ions (H+). The key to buffering is that hemoglobin within the red blood cells binds to these free H+ ions, thereby preventing a drastic shift in pH levels. This buffering capability keeps the blood's pH within a narrow and healthy range. When carbon dioxide enters the red blood cell and is converted into bicarbonate ions, hemoglobin binds hydrogen ions. Therefore, the correct answer to the original question is that hemoglobin functions as a pH buffer by binding hydrogen ions after carbon dioxide enters the red blood cell (option c).
Make a hypothesis about how you think the two species of Paramecium sp. will grow alone and how they will grow when they are grown together.
I hypothesize that Paramecium species will grow faster alone due to reduced competition. When together, resource competition might slow growth.
I expect that when each species of Paramecium is cultured alone, their growth rates will be relatively higher due to the absence of interspecific competition for resources. In a mixed culture, where both species coexist, I anticipate a potential decrease in growth rates compared to individual cultures.
This could be attributed to the sharing of limited resources, leading to heightened competition between the two species. Consequently, the presence of interspecific interactions might result in a more constrained growth environment, affecting the growth trajectories of both Paramecium species when compared to their growth in isolation.
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When grown separately, both Paramecium aurelia and Paramecium caudatum will probably proliferate successfully. However, when they share the same environment and compete for resources, Paramecium aurelia is likely to outcompete Paramecium caudatum. This hypothesis relies on previously observed behaviors, but must still be validated by experimental testing.
Explanation:A hypothesis about the growth of two species of Paramecium, specifically Paramecium aurelia and Paramecium caudatum, could be: When each species is grown separately, they will both proliferate successfully. However, if they are grown together and forced to compete for the same resources, Paramecium aurelia is likely to outcompete Paramecium caudatum. This hypothesis is based on the observations shared in figures 45.24, 36.25, and 19.19 that reveal this competitive dynamic.
Beyond competition, these species might engage in symbiosis, which represents different types of close, long-term interactions between species. Depending on the conditions and specific interactions, these could be commensal (one species benefits, the other is unaffected), mutualistic (both species benefit), or parasitic (one species benefits at the expense of the other).
The interaction between these Paramecium species could potentially mirror other observed species interactions, such as between different groups of minnows in a shared environment or different mussel cultures in a lab. It's important to note however that any hypothesis must be tested experimentally to verify its accuracy and be subjected to further refinement based on findings.
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Frank has Klinefelter syndrome (47, XXY). His mother has normal skin, but his father has anhidrotic ectodermal dysplasia, an X-linked condition where the skin does not contain sweat glands. Frank has patches of normal skin and patches of skin without sweat glands. Which of the following options is correct? a. Frank received the mutant chromosome from his mother. Nondisjunction occurred in his mother during the second meiotic division b. Frank received the mutant chromosome from his father. Nondisjunction occurred in his father during the first meiotic division. c. Frank received the mutant chromosome from his father. Nondisjunction occurred in his father during the second meiotic division d. Frank received the mutant chromosome from his mother. Nondisjunction occurred in his mother during the first meiotic division. e. Frank received the mutant chromosome from his mother. Nondisjunction occurred in his father during the second meiotic division f. Frank received the mutant chromosome from his father. Nondisjunction occurred in his mother during the first meiotic division.
Answer:
b. Frank received the mutant chromosome from his father. Nondisjunction occurred in his father during the first meiotic division.
Explanation:
As you can see in the question above, Frank has Klinefelter syndrome which causes him to have normal skin patches and skin patches without sweat glands. Her mother has completely normal hair, which may indicate that the defective gene was not supplied by her. In addition, Frank's father has anhydrotic ectodermal dysplasia, an X-linked condition where the skin does not contain sweat glands.
Although Frank's father's defective gene is linked to the X chromosome, it is likely that Frank inherited the defective gene from his country. This may have occurred because during meiosis I, his father's genes did not show disjunction. As a result, Frank presents a mosaic of his phenotype, because an inactivation of the X chromosome occurred.
One of Mendel’s Laws says that the probability of inheriting a maternal allele for a gene does not affect the probability of inheriting the maternal allele for another gene. If each gene has two alleles, what is the probability that one individual inherited the paternal allele at three different genes?
Answer:
There's 2 alleles for every quality and there are 3 qualities. The likelihood of acquiring the fatherly allele of those 3 qualities would be 1/2 x 1/2 x 1/2. This is on the grounds that you have a 50/50 possibility of getting the allele from your mom or your dad for three unique alleles. Since you are making sense of the likelihood of acquiring 3 qualities from your dad you duplicate at that point to get 1/8.
What is the function of rooting hormone?
A Encouraging a cutting to develop roots
B Encouraging successful budding
C Encouraging successful grafting
D Encouraging bulbs to grow
Answer:
Option a
Explanation:
For cuttings used in propagation to grow root faster, rooting hormones are usually used.due to the fact that the plant root hormones have not been produced, the rooting hormones added facilities the growth of root till the plant root hormones are produced.this lead to strong root and faster growth of plants. Rooting hormones which has auxis present inside of it helps in cell length growth. Rooting hormones which also help increase auxin concentration in plants leads to faster and stronger plant growth.using when it is placed or applied into the cuttings, it will help produce root cell via(through) the replacing of stem cell in plant with undifferentiated cell ( that form the root cell)
For each of the following characteristics, indicate whether the trait is common to Phylum Arthropoda or specific to certain classes of arthropods: wings, chewing mouthparts, jointed appendances, number of legs, segmented bodies, exoskeleton.
Answer: wings: Insecta class
Chewing mouthparts: some orders in the insects class;
Jointed appendages: common to all athropods
Number of legs: not all the same in some class and even orders
Segmented bodies: common to all
Exoskeleton: common to all
Explanation:
Memeberss in the phylum athropoda all have exoskeleton with itsain component being chitin. They all have segmented bodies; some head, thorax and abdomen like insects, some head, cephalothorax and abdomen like in the crustaceans. All have jointed appendages. Some class possess wings eg insecta and some possess chewing mouthparts egg grasshopers in the class insecta
Wings and chewing mouthparts are traits specific to certain classes within Phylum Arthropoda (e.g., Insecta). Jointed appendages, segmented bodies, exoskeleton, and a fixed number of legs (typically eight) are characteristics common to all members of Phylum Arthropoda.
The characteristics common to Phylum Arthropoda and those specific to certain classes of arthropods are as follows:
1. Wings: Wings are not common to all members of Phylum Arthropoda. They are specific to certain classes within the phylum, such as the Insecta (insects) and Pterygota (winged insects). Therefore, this trait is specific to certain classes of arthropods.
2. Chewing mouthparts: Chewing mouthparts are a trait found in many arthropods, but not all. For example, insects like grasshoppers have chewing mouthparts, while other arthropods like mosquitoes have piercing-sucking mouthparts. This trait is more common in certain orders and families within the class Insecta, making it specific to certain classes of arthropods.
3. Jointed appendages: Jointed appendages are a defining characteristic of the Phylum Arthropoda. All arthropods, regardless of their class, possess jointed appendages. This trait is common to the entire phylum.
4. Number of legs: The number of legs can vary widely among arthropods. While insects typically have six legs, arachnids have eight, and myriapods can have many more. Therefore, the number of legs is specific to certain classes of arthropods.
5. Segmented bodies: A segmented body is a common trait among all arthropods. The degree of segmentation and the function of those segments may vary, but the presence of a segmented body is a characteristic of the Phylum Arthropoda as a whole.
6. Exoskeleton: An exoskeleton is a trait common to all members of Phylum Arthropoda. The exoskeleton provides support and protection and is a key feature that defines the phylum.
Mutations in which of the following genes lead to transformations in the identity of entire body parts?A) morphogensB) segmentation genesC) egg-polarity genesD) homeotic genesE) inducers
Answer: Option A
Explanation:
A morphogen gradient is an important concept in case of the developmental biology. It produces the signals from the part of embryo which decides the fate of development of the parts.
The differentiation of the cell takes place by the help of the signals produced by the embryo in order to decide where will which part of the embryo grow.
The surrounding cells receive the signals and the development of entire body parts takes place.
So, the mutation in this gene will transform the entire body part.
Answer: Option A.
Explanation:
Morphogens are signaling molecules that originate from a specific location that spread from their origin via diffussion to form a concentration gradients acting at long distances to induce responses in cells base on the gradients the cells interact with. It determines the pattern of tissue development and morphogens gradients cause the differentiation of unspecialized stem cells into different cell types.
Saturated fatty acids are degraded by the stepwise reactions of beta-oxidation, producing acetyl-CoA. Under aerobic conditions, how many ATP molecules would be produced as a consequence of removal of each acetyl-CoA?
Answer:
4 ATP molecules
Explanation:
Normally, about 11 ATP molecules are generated as the result of Beta-Oxidation of saturated fatty acid in Kreb's Cycle. But the total removal of acetyl-CoA under certain aerobic condition decreases the overall yield and approximately 4 ATP molecules comes out from each removal of acetyl-CoA.
Final answer:
Each acetyl-CoA molecule produced from the stepwise degradation of saturated fatty acids through beta-oxidation can yield around 10 ATP molecules.
Explanation:
Saturated fatty acids are degraded through beta-oxidation, which produces acetyl-CoA. Each acetyl-CoA molecule can yield a certain number of ATP molecules when it enters the citric acid cycle and oxidative phosphorylation. One mole of acetyl-CoA metabolized by the citric acid cycle yields about 10 ATP molecules. Therefore, the energy produced by the removal of each acetyl-CoA molecule from the beta-oxidation of saturated fatty acids can be estimated to be around 10 ATP molecules.
An article in the Los Angeles Times (Dec. 3, 1993) reports that 1 in 200 people carry the defective gene that causes inherited colon cancer. In a sample of 1000 individuals, what is the approximate distribution of the number who carry this gene
Answer: 5
Explanation:
Since 1 out of 200 people carry the defective gene that causes inherited colon cancer.
In a sample of 1000 individuals, we say
1 = 200
y = 1000
cross multiply
y x 200 = 1000 x 1
200y = 1000
y = 1000/200
y = 5
Thus, in a sample of 1000 individuals, approximately 5 will be carriers
An example of anabolism is A. glucose oxidation to pyruvate. B. breakdown of starch or glycogen to glucose. C. oxidation of fats to carbon dioxide and water. D. NONE of A-D are examples of anabolism E. DNA replication.
Answer: Option E. DNA replication
Explanation:
DNA replication is described as anabolism because it involves the synthesis of newly synthesized strands of DNA from the double stranded parental DNA units.
Thus, since complex molecules are formed from simpler unit, DNA replication is an example of anabolism.
The genetic condition Xeroderma pigmentosum, which can lead to skin cancer, results from A. inability to correct UV induced lesions B. inability to process phenylalanine. C. inability to produce functional hemoglobin. D. inability to correct transitions E. breaks in the X chromosome
Answer:
Option A
Explanation:
Xeroderma pigmentosum arises as a result of the cell being unable to correct lesions induced by UV. This can be as a result of mutations in the enzymes which include XP A-E needed for correction of the lesions. Failure to correct these lesions leads to their accumulation and then damage to the cell.
Suppose that a scientist performed an mRNA mixed copolymer experiment to decode the nonstandard genetic code of a recently discovered species of archaea. The mixture used to synthesize the mRNA contained four parts uracil (U) and one part cytosine (C) nucleotides. The polypeptide translated from this mRNA contained only phenylalanine, serine, leucine, and proline. From previous experiments, the scientist already knew that the organism's genetic code is a triplet code, like the genetic code of humans. Furthermore, the scientist knew that all codons starting with UU code for phenylalanine, UC code for serine, CC code for proline, and CU code for leucine. Using this information, calculate the percentage of amino acids in the polypeptide expected to be composed of proline.
Answer:
The percentage of amino acids in the polypeptide expected to be composed of proline is 4%
Explanation:
Uracil: 80%= 4/5
Cytosine: 20%= 1/5
Proline: CC: 1/5 x 1/5: 1/25= 4%
When _______________ occurs between nonsister chromatids genetic exchange between chromosomes provides new combination of genes that are different from either parent.
Answer:
crossing over
Explanation:
Crossing over occur during meiosis and this involves an exchange of genetic materials between non-sister chromatids in a tetrad. This exchange of genetic materials ensures that the daughter cells produced at the end of meiosis are genetically different from either of the parental cells.
When crossing over occurs between non-sister chromatids genetic exchange between chromosomes provides a new combination of genes that are different from either parent.
Crossing Over:
It is the process by which non-sister chromatids of homologous chromosomes, exchange their genetic material.
Crossing over occurs during prophase I.Crossing over results in the genetic diversity between gametes.Therefore, when crossing over occurs between non-sister chromatids genetic exchange between chromosomes provides a new combination of genes that are different from either parent.
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50.Botox, which is used to reduce wrinkles for a limited time interval, inhibits membrane fusion by targeting a protein that allows overcoming the repulsive force of the membranes. Based on this information, what is the protein that is inhibited by Botox?
Answer: Botox functions by inhibiting the function of SNARE protein.
Explanation:
The primary function of the SNARE protein is to mediate the vesicle fusion, which shows the fusion of the target membrane bound compartment with the vesicle.
The botox functions by inhibiting the function of this protein. This is a drug which weakens or paralyzes the muscles.
in small doses it is used to reduce the wrinkles on the face. This drug is made of bolulinum toxin which acts on the SNARE protein.
Catecholamines mediate their effects throughI. α-adrenoreceptors.II. β-adrenoreceptors. III. receptor tyrosine kinases.IV. monomeric G protein receptors
Explanation:
Catecholamines are hormones made by your adrenal organs, which are situated on your kidneys. Models incorporate dopamine; norepinephrine; and epinephrine (this used to be called adrenalin or adrenaline). Our adrenal organs send catecholamines into our blood when you're genuinely or sincerely focused. They cause you to inhale quicker, raise your pulse, and send more blood to significant organs, similar to your cerebrum, heart, and kidneys.Chromatin structure can increase or decrease transcription of a gene according to the:
A. combination of amino acid modifications in the histone tails.
B. coding sequences in the messenger RNAs for histone proteins.
C. combination of histone proteins found within the nucleosome.
D. proportion of arginine and lysine amino acids in the histone proteins.
Answer:
D. proportion of arginine and lysine amino acids in the histone proteins.
Explanation:
Epigenetic modifications are one of the main reasons of methylation and de-methylation, acylation and de-acylation of histone proteins that trigger the transcriptional process. All of these mentioned modification causes the change in the ration of lysine and arginine residues and these are the main reasons that affect the structure of chromatin as required.
At the neuromuscular junction, an electrical signal from the motor neuron is translated to a chemical signal and then back to an electrical signal in the muscle cell. What proteins interact with the released chemical to cause the electrical signal in the muscle cell?
a.Sodium voltage-gated channels
b.Potassium ligand-gated channels
c.Cation ligand-gated channels
d.Calcium mechanical-gated channels
Answer:
Cation ligand-gated channels
Explanation:
At the neuromuscular junction, the arrival of nerve impulse triggers the release of chemical signals called neurotransmitters in the cleft. Binding of these chemical signals to the receptor present on the motor endplate results in the opening of a ligand-gated cation ion channel. These channels are present in the receptor. Opening of ion channels allow cations, specifically Na+, to flow across the membrane. The entry of cations makes the inside of the muscle fiber more positively charged and triggers a muscle action potential.
Cation ligand-gated channels on the muscle cells interact with the neurotransmitter (chemical signal) released from the motor neuron to generate an electrical signal in the muscle cell at the neuromuscular junction.
Explanation:The proteins that interact with the released chemical to cause the electrical signal in the muscle cell at the neuromuscular junction are the cation ligand-gated channels. Here's a brief step-by-step of what happens: An electrical signal or action potential travels down the motor neuron. When this signal reaches the end of the neuron, it causes the release of a neurotransmitter called acetylcholine. The acetylcholine diffuses across the neuromuscular junction and binds to the acetylcholine receptors - which are a type of cation ligand-gated channels - on the muscle cell. This binding causes these channels to open and allow positively charged ions to flow into the muscle cell, creating an electrical impulse in the muscle cell that ultimately leads to muscle contraction.
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The evolutionary ancestor of fruit flies had two pairs of wings. Modern fruit flies have one pair of wings and one pair of halteres, an organ involved in balance. A mutation in the modern fruit fly results in conversion of the halteres to wings. The mutant fly has two pairs of wings, so it resembles its evolutionary ancestor. What is the most likely reason the mutation converts one organ to another?
Answer:
Genetic silencing.
Explanation:
The program that commands the expression and development of an organ can be silenced by interruption of a molecular switch, for instance a blocking by a protein of a promoting region. This way it is possible to convert one organ initially silenced to a functional organ.
What is the expected Hardy-Weinberg frequency of heterozygotes in a population of green anoles, where the locus under investigation has two alleles and the frequency of one allele is 0.005 (give your answer to five decimal places)?
Answer:
0.00995
Explanation:
If the population is in Hardy Weinberg Equilibrium, the sum total of frequencies of the alleles of a locus would be one. The frequency of one allele for a locus is 0.005. So, the frequency of the other allele for the same locus would be=1-0.005= 0.995
According to the given information, there are only two alleles for the locus in the population. So, the frequency of heterozygous genotype in a population= 2 x frequency of one allele x frequency of other allele= 2 x 0.005 x 0.995= 0.00995
Consider a locus with two alleles - A and a. These alleles are codominant, meaning that the fitness of the heterozygote is halfway between either homozygote. Consider further a population of randomly mating green frogs where the genotype counts are AA = 500, Aa = 250, and aa = 250. In this population the relative fitnesses of each genotype are AA = 1.00, Aa = 0.80, and aa = 0.60. What is the mean realtive fitness within this population? Please give your answer to two decimal places.
Consider a locus with two alleles - A and a. These alleles are codominant, meaning that the fitness of the heterozygote is halfway between either homozygote. Consider further a population of randomly mating green frogs where the genotype counts are AA = 500, Aa = 250, and aa = 250. In this population the relative fitnesses of each genotype are AA = 1.00, Aa = 0.80, and aa = 0.60. What is the expected allele frequency change for A after one generation with selection? Please give your answer to two decimal places.
Final answer:
The mean relative fitness of the population is 0.85. The expected allele frequency change for allele A after one generation with selection is a decrease to approximately 0.62 due to the relative fitness differences among the genotypes.
Explanation:
The question at hand requires an understanding of genetic structure, allele frequency, and the Hardy-Weinberg principle to calculate both the mean relative fitness within a population and the expected allele frequency change for allele A after one generation with selection.
Mean Relative Fitness Calculation:
To calculate the mean relative fitness (w), we multiply the relative fitness of each genotype by its proportional representation in the population and sum the results:
w(AA) = 1.00 × (500/1000) = 0.50,w(Aa) = 0.80 × (250/1000) = 0.20,w(aa) = 0.60 × (250/1000) = 0.15.The sum is the mean relative fitness: wμ = 0.50 + 0.20 + 0.15 = 0.85.
Expected Allele Frequency Change:
The next step is to calculate the expected change in allele frequency using the Hardy-Weinberg equation. First, determine the frequency of the alleles (p for A and q for a). The current frequency of A (p) = (2×500 + 250) / (2×1000) = 0.625. The frequency of a (q) = 1 - p = 0.375. To calculate the expected frequency of allele A after one generation of selection, we must account for the relative fitness of each genotype:
p' = (p² × w(AA) + pq × w(Aa)) / wμ,p' = (0.625² × 1.00 + 0.625 × 0.375 × 0.80) / 0.85 ≈ 0.62.Hence, the expected frequency of A after one generation is approximately 0.62, indicating a slight decrease due to selection.
The mean relative fitness within the population is 0.85, rounded to two decimal places.
To find the mean relative fitness within the population, we need to calculate the weighted average of the relative fitnesses of each genotype, considering their frequencies.
Given:
- AA genotype count [tex](n_AA)[/tex] = 500
- Aa genotype count [tex](n_Aa)[/tex] = 250
- aa genotype count [tex](n_aa)[/tex] = 250
- Relative fitness of AA genotype [tex](w_AA)[/tex] = 1.00
- Relative fitness of Aa genotype [tex](w_Aa)[/tex] = 0.80
- Relative fitness of aa genotype [tex](w_aa)[/tex] = 0.60
First, calculate the total number of individuals in the population (N):
[tex]\[ N = n_{AA} + n_{Aa} + n_{aa} \][/tex]
[tex]\[ N = 500 + 250 + 250 \][/tex]
[tex]\[ N = 1000 \][/tex]
Next, calculate the contribution of each genotype to the total fitness:
[tex]\[ Contribution_{AA} = n_{AA} \times w_{AA} = 500 \times 1.00 = 500 \][/tex]
[tex]\[ Contribution_{Aa} = n_{Aa} \times w_{Aa} = 250 \times 0.80 = 200 \][/tex]
[tex]\[ Contribution_{aa} = n_{aa} \times w_{aa} = 250 \times 0.60 = 150 \][/tex]
Now, sum the contributions of all genotypes:
[tex]\[ Total \, Contribution = Contribution_{AA} + Contribution_{Aa} + Contribution_{aa} \][/tex]
[tex]\[ Total \, Contribution = 500 + 200 + 150 = 850 \][/tex]
Finally, calculate the mean relative fitness (w_mean):
[tex]\[ w_{mean} = \frac{Total \, Contribution}{N} \][/tex]
[tex]\[ w_{mean} = \frac{850}{1000} \][/tex]
[tex]\[ w_{mean} = 0.85 \][/tex]
So, the mean relative fitness within this population is 0.85, rounded to two decimal places.
Information transfer is fundamental to all living organisms. For the following examples, explain in detail how the transfer of information is accomplished.
1. The genetic material from one bacterial cell enters another via transformation, transduction, OR conjugation?
2. The genetic material in one eukaryotic cell is copied and distributed to two identical daughter cells.
3. A gene in a eukaryotic cell is transcribed and translated to produce a protein.
Answer:
Explanation:
We are to explain in details how
genetic material from one bacterial cell enters another via transformation, transduction, OR conjugation.
For transformation, here, there is genetic alteration of a cell. It is a process of horizontal gene transfer where bacteria take up foreign genetic material from the environment.
Transduction is genetic recombination in bacteria where genes from a host cell are carried into the genome of a bacterial virus and then carried to another host during infection cycle.
Conjugation. In this, DNA plasmid is transferred from one bacterium to another through pilus.
2. We explain also in details how genetic material in one eukaryotic is copied and distributed.
Genetic material is copied during DNA replication and genetic material is distributed during mitosis.
The followings take place during DNA replication.
• DNA is copied during interphase, S phase of cell cycle.
• Site origin of DNA is known.
• Unwinding of DNA at the origin
• Synthesis of new strands with enzyme called helicase.
• Formation of replication forks.
• Proteins associated with replication forks help in the initiation and continuation.
• DNA polymerase synthesizes new strands by adding nucleotides that complement each strand.
Mitosis.
• Chromosomes are doubled
• Prophase: phase of condensation of chromosomes, formation of spindle.
• Metaphase: Alignment of chromosomes
• Anaphase: Chromatids are separated
• Telophase: reformation of nuclear membrane, division of cells, cell cycle control.
3. We also explain how transcription and translation of gene in eukaryotic.
Transcription
• Here we have DNA sequence to RNA sequence.
• Occurs in the nucleus
• DNA is carried into nucleosomes.
• RNA polymerase: add complementary RNA nucleotides to DNA template.
• Growth of new strands
• Leads to mRNA processing.
Translation
• Here we have mRNA base sequences to amino acids initiation
• Sequence of events such as complexes, small unit of ribosomes.
• Also we have the first tRNA
• Structure of ribosomes formed.
• Complete description of two subunits, 2 action sites, rRNA and proteins.
A transcription factor that binds to a gene first and facilitates binding of other transcription factors is called a(n) ________. A transcription factor that binds to a gene first and facilitates binding of other transcription factors is called a(n) ________. pioneer factor repressor transcription factor transcription factor II D activator transcription factor
Answer:
A transcription factor that binds to a gene first and facilitates binding of other transcription factors is called an activator transcription factor.
Explanation:
Transcription factors are proteins that regulate the transcription of genes.Transcription is the process where a gene's DNA sequence is transcribed into an RNA molecule. Transcription is a key step in using information from a gene to make a protein.
The enzyme RNA polymerase, which makes a new RNA molecule from a DNA template, must attach to the DNA of the gene. It attaches at a spot called the promoter.In eukaryotes, RNA polymerase can attach to the promoter only with the help of basal (general) transcription factors. They are part of the cell's core transcription toolkit, needed for the transcription of any gene.
A typical transcription factor binds to DNA at a certain target sequence. Once it's bound, the transcription factor makes it either harder or easier for RNA polymerase to bind to the promoter of the gene.
Some transcription factors activate transcription, other transcription factors repress transcription.
The transcription factor that binds to a gene first and facilitates binding of other transcription factors is called a Transcription factor II D (TFIID)
To start transcription, a transcription factor (TFIID) is known to be the first to bind to the TATA box. The Binding of TFIID bring up also other transcription factors.
Transcription factors are referred too as proteins that help turn specific genes "on" or "off" by binding to close DNA. They act as activators that boost a gene's transcription.
Transcription factor II D (TFIID) is one of different general transcription factors that set up the RNA polymerase II preinitiation complex.
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A young couple went to see a genetic counselor because each had a sibling affected with cystic fibrosis. (recessive disease, neither member of the couple is affected. Their four parents are also not affected))(a) What is the probability that the female of this couple is a carrier?(b) What are the chances that their child will be affected with cystic fibrosis?(c) What is the probability that their child will be a carrier of the cystic fibrosis mutation?
Answer: A: I would say a 50% chance that she's the carrier SF
B: 25%
C. The child will most likely be a carrier
Explanation: I'm not really sure if i'm correct but how I figured it out was. Mapping out the family tree and shading the circles if the had SF and if they did not have SF and was a carrier I only shaded half of it. After I finished shading everything out I made a punnet square to predict the probability of the child.
Describe one structure or metabolic process that purple bacteria or cyanobacteria might have dispensed with once they became endosymbionts in a eukaryotic cell. Describe one structure or metabolic process that purple bacteria or cyanobacteria might have dispensed with once they became endosymbionts in a eukaryotic cell. One of the most important features the bacteria might have dispensed with is the cell wall. One of the most important features the bacteria might have dispensed with is the gas vacuole. One of the most important features the bacteria might have dispensed with is the flagellum. One of the most important features the bacteria might have dispensed with is cytoplasmic membrane.
Answer:
The answer is photosynthesis or thylakoid region of plasma membrane.
Explanation:
Cyanobacteria are photosynthetic, prokaryotic organisms which are green-blue in colour. They possess a simple structure at sub-cellular level and lack a nucleus. Whereas, eukaryotic cells are more complex, multicellular and have membrane-bound organelles; this includes a nucleus that holds their DNA.
During endosymbyosis with bacteria, the eukaryotic cell develops an organelle called chloroplast, which is then responsible for photosynthesis.
Hence on metabolic process the bacteria might dispense would be photosynthesis, or more specifically its thylakoid region (infolded regions of the plasma membrane responsible for photosynthesis).