Answer:
Cells divide twice
Explanation:
Meiosis may be defined as the process of cell division in which a single cell divides to give four daughter cells and each daughter cells have half number of chromosomes.
Meiosis is also known as reduction division as the chromosome number reduces upto half in the progeny cells as compared with the parent cell. The cells divide twice in meiosis and give rise to four daughter cells.
Thus, the correct answer is option (4).
Meiosis is a cellular division that results in four daughter cells with half the number of chromosomes of the parent cell. Specifically, cells in meiosis divide twice across two phases: Meiosis I and Meiosis II.
Explanation:During meiosis, cells divide twice. Meiosis is a type of cellular division that results in four daughter cells, each with half the number of chromosomes of the parent cell. It consists of two successive divisions: Meiosis I and Meiosis II. During Meiosis I, homologous chromosomes pair up and then separate, resulting in two cells. In Meiosis II, these cells divide again, resulting in four cells with half the number of chromosomes as the original cell.
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In some specialized cells of eukaryotes, a very high level of gene expression can be achieved when DNA is copied repeatedly without cell division for example, some insect larvae maximize production of salivary proteins by means of this process:
Answer:
In some specialized cells of eukaryotes, a very high level of gene expression can be achieved when DNA is copied repeatedly without cell division for example, some insect larvae maximize production of salivary proteins by means of this process: polytene chromosomes.
Which kind of psychology seeks to identify and promote those qualities that lead to happy, fulfilled, and contented lives?
a.self-psychology
b.positive psychology
c.eustress psychology
d.psychology of joy
Answer:
B - Positive psychology
Explanation:
By definition, positive psychology is the study of a healthy mind and it is a applied to achieve peak operational functioning of the mind. It does this by identifying and promoting qualities that lead to happy, fulfilled and contented lives. Positive psychology is based on nurturing what is best withing oneself, thereby improving both individuals and communities.
Describe how the schwann cells form the myelin sheath
Explanation:
The myelin sheath, lipoprotein structure deposited around axons selected in inner nodules, interrupted periodically by Ranvier nodules, allows saltatory, rapid and effective conduction in the nervous system of vertebrates. The cells that construct myelin are the oligodendrocyte in the nervous central nervous system (CNS) and the Schwann cells in the peripheral nervous system. An oligonucleotide myelinates one or more axons, while the Schwann always forms a single myelin internodule. The process of myelination begins when a cell-sheath projection involves the axon and a loose spiral shape around it. Over time the formed layers are compacted by the expulsion of the cytoplasm and the formation of a lamellar structure with thick electrodes bands - derived from the apposition of the cytoplasmic phases of the membranes - and fewer electrodes - derived of the external phases of the membranes.
Schwann cells form the myelin sheath by wrapping their membrane around an axon segment in the PNS, creating a lipid-rich layer that facilitates rapid transmission of electrical signals. This insulation ensures efficient nerve impulse conduction with nodes of Ranvier allowing impulses to travel quickly by 'jumping' from gap to gap.
Explanation:Schwann cells are a type of glial cell responsible for the production of the myelin sheath around axons in the peripheral nervous system (PNS). During the process of myelination, a Schwann cell envelops an axon segment by wrapping its cell membrane around the axon multiple times. This wrapping forms a lipid-rich layer with very little cytoplasm between the layers, effectively insulating the axon and aiding in the rapid transmission of nerve impulses.
The myelin sheath's appearance has been compared to a pastry wrapped around a hot dog, with the sausage representing the axon and the pastry the myelin layers. Unlike oligodendrocytes in the central nervous system (CNS), which can myelinate several axons at once, a single Schwann cell myelinates just one segment of a peripheral nerve. Myelin is not only comprised of the phospholipids from the Schwann cell membrane but also includes proteins that help maintain the structure of the sheath, supporting fast electrical signaling along the nerve fiber.
The role of Schwann cells and myelin is crucial as they contribute to the formation of nodes of Ranvier—regularly spaced gaps in the myelin sheath that allow for saltatory conduction, wherein nerve impulses 'jump' from node to node, drastically increasing the speed at which they travel down the axon.
Two parents consult a genetic counselor. They do not understand why all their sons are colorblind (an X-linked trait) but none of their daughters are colorblind. Both of the parents appear normal. How would you explain this to the parents?
Answer:
There is 50% probability that a son of carrier mother (X^cX) and normal father (XY) is affected with colorblindness. Since father is normal, all the daughters will be normal.
Explanation:
Being X linked recessive trait, males can be affected with colorblindness if their mother is either colorblind or carrier for the disease. Since both the parents are normal, the mother is carrier for the disease and is heterozygous dominant with genotype X^c X. She would deliver the "X^C" to 50% of her sons making them colorblind.
Daughters can get affected with X linked disorder such as colorblindness if their father is affected with the disease. In the given question, father is normal (XY). So, none of the daughters can be colorblind.
Final answer:
Color blindness is an X-linked recessive trait. Boys can be colorblind if they inherit the colorblind gene from their mother, as they only have one X chromosome. Girls can be carriers of the colorblind gene but do not express color blindness due to the presence of a normal gene on their second X chromosome.
Explanation:
Color blindness is an X-linked recessive trait, which means it is associated with the X chromosome. Since boys only have one X chromosome, if it carries the colorblind gene, they will be colorblind. On the other hand, girls have two X chromosomes, so even if one X chromosome carries the colorblind gene, the other X chromosome with a normal gene for color vision will mask the recessive colorblind gene. Thus, girls with one colorblind allele and one normal allele are carriers and do not express color blindness.
What kind of channels are located in the input region?
a. Voltage gated channels opened by neurotransmitter molecules
b. ligand gated channels opened by neurotransmitter molecules
c. voltage gated channels opened by a change in charge across the membrane
d. ligand gated channels opened by a change in charge across the membrane
Answer:
The correct answer is option B ( ligand gated channels opened by neurotransmitter molecules).
Explanation:
Input region or postsynaptic region or dendrite is the site of neuron cell which receives the impulse from the pre-synaptic neuron at the synaptic junction.
The cell membrane of the dendrite is embedded with ligand-gated channels which opens up in response to the ligand (neurotransmitters) produced by the neurotransmitter vesicle of axon or output region of neuron.
The neuromuscular junction is such a case where acetylcholine receptors present in the dendrite opens in the presence of acetylcholine.
Thus, option B ( ligand-gated channels opened by neurotransmitter molecules) is the correct option.
Read this excerpt from a persuasive essay about animal testing. Which two sentences form a hook to draw the reader’s attention? Somewhere in a claustrophobic laboratory a dog is forced to breathe tobacco smoke. In another lab in another part of the country, a rat is squeezed into a narrow glass tube and forced to breathe in tobacco smoke. Scenes from a horror novel? No, all of this is real. Modern technology has advanced to new heights. Yet scientists continue to perform cruel and painful experiments on innocent animals all in the name of progress. Moreover, most of these experiments on animals have ended up being worthless—scientists don’t even have a good reason to continue with them. We must put an end to it before more animals are harmed.It is time we show humanity to our fellow creatures.
The first two sentences of the given essay extract function as the hook, drawing readers into the contentious issue of animal testing via vivid and emotionally charged descriptions of laboratory practices.
Explanation:The first two sentences of this excerpt form the hook that is designed to draw the reader's attention. These sentences are: 'Somewhere in a claustrophobic laboratory a dog is forced to breathe tobacco smoke. In another lab in another part of the country, a rat is squeezed into a narrow glass tube and forced to breathe tobacco smoke.' A hook in an essay is the first one or two sentences that are intended to capture the attention of a reader and engage them in the topic. In this essay, the hook introduces the contentious issue of animal testing by painting a vivid, disturbing picture of what animals may undergo in the process.
By using the technique of vivid description and somewhat shocking imagery, the writer effectively brings the issue of animal testing to life for the reader. This hook catches the reader's attention and makes them want to read further to find out more about the situation, the ethical considerations, and what might be done to change it.
It is crucial to note that a strong hook not only draws in the reader but also sets the tone of the essay and introduces the main topic in a compelling and engaging way. The mentioned sentences accomplish all of these tasks effectively, setting the stage for a persuasive argument against animal testing.
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Know the special characteristics of epithelial cells.
Answer:
Functions of epithelial cells include secretion, selective absorption, protection, transcellular transport, and sensing. Epithelial layers contain no blood vessels, so they must receive nourishment via diffusion of substances from the underlying connective tissue, through the basement membrane
Which of the following statements is FALSE about muscles? a. toned muscles react to stimuli quickly and powerfully c. excessive use of muscles can cause hypertrophy b. atrophy of muscles can cause an increase in their size d. prolonged disuse of muscles can cause atrophy
Answer:
The correct answer is option b. "atrophy of muscles can cause an increase in their size".
Explanation:
Muscle atrophy is defined by the partial or total loss of muscles mass that results from disabling circumstances. Muscle atrophy results in weakness, since the ability of the muscle to exert force is closely related to its size. Therefore, is false to state that muscle atrophy can cause an increase in its size. Actually, is the opposite, an atrophy in a muscle results in a decrease in its size.
At the conclusion of experiments involving growing bacterial cultures, _____ should be added to the plates or tubes and everything should be soaked for _____.
Sterile broth should be added to the plates or tubes and everything should be soaked for incubation.
Explanation:At the conclusion of experiments involving growing bacterial cultures, sterile broth should be added to the plates or tubes and everything should be soaked for incubation.
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At the conclusion of experiments involving growing bacterial cultures, a sterilizing agent such as ethanol or bleach should be added to the plates or tubes and everything should be soaked for a sufficient amount of time to kill any remaining bacteria.
Explanation:At the conclusion of experiments involving growing bacterial cultures, a sterilizing agent such as ethanol or bleach should be added to the plates or tubes, and everything should be soaked for a sufficient amount of time to kill any remaining bacteria. This is done to prevent any contamination and to ensure the accuracy of the results.
For example, after completing an experiment to grow bacterial cultures, you would add ethanol or bleach to the plates or tubes and let them soak for at least 10 minutes. This ensures that any bacteria left on the surfaces are killed before handling or disposing of them.
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What are the two major components of extracellular matrix?
The two main components of the extracellular matrix are Elastin and Collagen.
The extracellular matrix is an intricate macromolecular network that is found in the extracellular space. The matrix is composed of polysaccharides and very diverse proteins, locally secreted and assembled forming a complex network that surrounds the cells. The matrix is highly developed in connective tissue and its derivatives. The extracellular matrix is formed mainly by proteins, glycosaminoglycans,proteoglycans and glycoproteins, organized in diverse networks that constitute the different tissues. The most abundant proteins are collagen and elastin.
Collagen is a family of very abundant proteins in the body of animals. Collagen molecules can represent 25 to 30 % of all body proteins. Its main mission in the tissues is to form a framework that supports the tissues and that resists the forces of mechanical tension.
The elastin molecules are very close to each other through links between the regions rich in the amino acid lysine. It is an abundant protein in may extracellular matrices and appears as a component of the so called elastic fibers, which are onsoluble aggregates of proteins.
The extracellular matrix, a structure provided by cells that lack cell walls, majorly consists of proteoglycans and fibrous proteins such as collagen. Proteoglycans form the bulky mass, while fibrous proteins provide strength.
Explanation:The two major components of the extracellular matrix are proteoglycans and fibrous proteins such as collagen. Proteoglycans form the bulky mass of the extracellular matrix. They are proteins that are heavily glycosylated, meaning they have several carbohydrate molecules attached to them. On the other hand, fibrous proteins such as collagen provide strength to the structure. They are long and stringy in structure and are known primarily for their tensile strength and resistance to stretching. Both of these components are attached to fibronectin proteins which, in turn, are connected to integrin proteins. These integrin proteins interact with transmembrane proteins in the plasma membranes of eukaryotic cells that lack cell walls.
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You are on a jury to decide the following case: The Jones family claims that baby Jane, given to them at Ucsc Medical Hospital, does not belong to them and that baby Sara, who was presented to the Smith family, really belongs to the Jones family. The Jones allege that the babies were exchanged on 4/20 by the medical staff. The staff deny the allegation Blood group determinations show the following results: Ms. Jones = AB Mr. Jones = O Mrs. Smith = A Mr. Smith = O Baby Jane = A Baby Sara = O A and B are dominant over O and A and B are codominant to one another. Which baby belongs to which family? Sara: Jane: Human cells normally have 23 pairs of chromosomes. For each of the following stages state the number of DNA molecules present in a human cell. Metaphase of meiosis II Metaphase of meiosis I Anaphase of meiosis II Egg cell
Question #1
To know which baby belongs to who you need to draw both families Punnett squares.
Let's start with the Jones family. Mr. Jones is AB this means, he has co-dominant alleles. Ms. Jones is 0 this means, she has two recessive alleles.
When you draw the Punnett square, you'll find that chances are 50% group A, and 50% group B.
Now, analyze the Smith family Mr. Smith is group A this means, he has either two A alleles or A0. Ms. Smith is group 0 this means, she has two recessive alleles.
If Mr. Smith is AA, 100% of children will be A, and if Mr. Smith is A0, 50% of children will be group A, and 50% will be group 0.
Baby Jane is group A, she could be either Smith or Jones, but baby Sara is group 0 the only chance is that her family was the Smith.
Question #2
Meiosis is a process in which gametes create. You have two stages: meiosis I, and II. At the end of meiosis I, you still have diploids cells (2n) this means, you have a pair of chromosomes in each cell. At the end of meiosis II, you have all haploid (n) cells, with half of the genetic charge.
Metaphase of meiosis II: Sister chromatids are in the equatorial plane to be pulled apart without being replicated. Here you have 46 chromosomes about to create two cells with 23 (n) chromosomes.
Metaphase of meiosis I: homologs chromatids are in the equatorial plane to be pulled apart after replication. Here you have 46 pairs of chromosomes about to create two cells with 46 chromosomes each (2n).
Anaphase of meiosis II: all singles chromatids (n) are in the poles and the plasmatic membrane begins to close to obtain two haploids (23 chromosomes each) cells.
Egg cell is the result of meiosis II and is the breeding cell. Its chromosomic charge is 23.
50Pts. Will give Brainliest if correct.
What percent of all the genes in E. coli are active all the time?
10%
20%
60%
80%
Answer:
60%
Explanation:
29. Approximately how many molecules of ATP are synthesized via cellular respiration for every glucose molecule used? (assume oxygenated environment)
a. 12
b. 24
c. 36
d. 48
Answer: c. 36
Explanation:
Cellular respiration oxidizes food molecules. The chemiosmotic model suggests that one ATP molecule is generated for each proton pump activated by the electron transport chain. Since the electrons from NADH activate three pumps and those from FADH2 activate two, we would expect each molecule of NADH and FADH2 to generate three and two ATP molecules, respectively.
However, because eukaryotic cells carry out glycolysis in their cytoplasm and the Krebs cycle within their mitochondria, they must transport the two molecules of NADH produced during glycolysis across the mitochondrial membranes, which requires one ATP per molecule of NADH. Thus, the net ATP production is decreased by two. Therefore, the overall ATP production resulting from aerobic respiration theoretically should be 4 (from substrate-level phosphorylation during glycolysis) + 30 (3 from each of 10 molecules of NADH) + 4 (2 from each of 2 molecules of FADH2) – 2 (for transport of glycolytic NADH) = 36 molecules of ATP
Which of the following pairs is mismatched? Helper T Cell : Killing of virus-infected cells T cell activation : proliferation Effector B cell : Plasma cell plasma cell : antibody secretion
Answer:
T cell activation : proliferation
Explanation:
The pairs that are correctly matched are:
Helper T cell: Killing of Virus infected cells
Helper T cells are also known as cytotoxic T cells which destroys cells with infected virus and tumor.
Plasma cell: Antibody Secretion
Plasma cell is also known as white blood cell. The cells are present in the bone marrow and secrete proteins called antibodies.
Effector B cell: Plasma cell
Plasma B cells are also known as Effector B cells.
Taking about,
T cell activation : proliferation
T cells are a type of lymphocytes that develops in the thymus gland. It is plays a central role in immune response. Thus, both T cell activation and proliferation are the different steps in the immune response by the T cell.
Thus, they are mismatched.
Chimpanzee attacks on members of other groups can be extremely brutal and result in severe injury or death. never result in injury. are definitely motivated by territoriality. are so rare that only one or two have been observed. always occur in the form of displays and nothing more.
Answer:
severe injury or death
You have isolated a newly discovered viral strain and sequenced its genome. The genome contains 30% adenine, 40% guanine, 25% cytosine, and 5% thymine. You conclude that the virus has a single-stranded DNA genome. Based on what evidence can you make this conclusion?
Answer:
It is a single stranded DNA since adenine percentage is not equal to that of thymine and the percentage of guanine is not equal to cytosine.
Explanation:
According to Chargaff rule, the amount of adenine in a double-stranded DNA is always equal to that of thymine. This is due to the fact that adenine pairs with thymine in a double-stranded DNA. Likewise, the amount of guanine is always equal to that of cytosine in a double-stranded DNA.
The percentage of four types of bases in the genome of the virus does not follow the Chargaff rule and hence, it is a single-stranded DNA.
What is the physiologic progression that occurs in anaphylactic shock?
Answer:
A life threatening drop in blood pressure due to vasodilation caused by rapid release of histamine as allergic response.
Explanation:
Anaphylactic shock is one of the rapid immune response towards the entry of an allergen into the blood stream. Presence of allergen in the blood triggers the rapid release of histamine. Histamine is a vasodilator and increases the permeability of capillaries. The relation of blood vessels all over the body results in a fatal drop in blood pressure.
Which is a difference between the synaptic cleft and the synapse?
Answer:
Synaptic cleft
Synaptic cleft may be defined as the space between two neuron and the gap between post synaptic and pre synaptic neuron. This is one of the component of synapse. The signals are transmitted in the form of chemical signal.
Synapse:
Synapse may be defined as the functional contact between two neurons and the gap between two consecutive neuron. This synapse consists of Presynaptic and postsynaptic membrane. The signals can be transmitted in form of electrical and chemical synapse.
If isolated myofibrils were placed in a flask containing a solution of calcium and AMPPNP (a nonhydrozyable analog of ATP) you would expect to find the myofibrils arrested in which stage of the myosin power stroke cycle?
a. just completed the power stroke
b. all locked up in cross bridges
c. all cross bridges broken
d. in the high energy cocked head conformation
Answer:
All cross bridges broken
Explanation:
Muscle contraction is explained by the most accepted sliding filament theory given by Huxley and Huxley. During the process of muscle contraction actin filament slide over the myosin filament by the formation of cross-bridges. In this process the length of I band reduces whereas the length of A band remains the same. Also we should note that the actin and myosin filament remains unchanged.
The release of acetylcholine at the neuro-muscular junction stimulates the release of calcium ions from endoplasmic reticulum into the sarcoplasm. During the relaxed condition troponin-tropomyosin complex always block the myosin binding site on the surface of actin.
Once calcium is released it binds to the TpC component of troponin causing the shift of troponn-tropomyosin complex. Now actin can attach and slide over myosin.
The myosin head binds to actin forming the cross bridge and the pulling of myosin head towards M-line along with actin is called the power stroke. But myosin head also has a site for binding of ATP. Binding of ATP causes the detachment of myosin head from actin filament. Now since its the analog of ATP i.e AMPPNP present in the solution, cross-bridges will be broken. And only if ATP undergoes hydrolysis, the angle of myosin head changes into high energy cocked head conformation. Since AMPPNP is non-hydrolyzable we can rule out the fourth option.
1. Where is the CFTR protein found?
in heart cells
in cells lining the lungs
in every cell in the body
in muscle cells
in skeletal cells
Answer:
CFTR protein is found in the cells lining the lungs
Tay-Sachs is an autosomal recessive disease. A couple (Jack and Jill) is worried about having a child who has Tay-Sachs, because Jack had a brother with the deadly disease. Since they wanted to have children, they sought the advice of a genetic counselor. During the course of the interview the counselor determined that Jack’s parents were not affected with Tay-Sachs, yet his brother had it. What is immediately known about the genotypes of Jack’s parents? Is the genotype of Jack immediately known? Under what circumstances could Jack and Jill have a child with Tay-Sachs? Assume that neither Jack nor Jill has Tay-Sachs.
Answer:
Jack parents' genotype = Both Heterozygous dominant (carrier)
Jack's genotype = could be homozygous dominant or heterozygous dominant
Jack and Jill can have a child with Tay-Sachs if both of them are heterozygous dominant for the disease.
Explanation:
Since Jack's brother died of the disease, his parents should be heterozygous dominant for the disease. Both the parents served as the carrier (Aa x Aa, assuming that allele "a" is responsible for Tay-Sachs disease) and had one affected child (Jack's bother= "aa").
Since Jack is not affected by the disease, he could be homozygous dominant (AA) or heterozygous dominant (Aa). But, his exact genotype could not be determined by the given information.
Child of Jack and Jill could be affected with the disease only if both of them are heterozygous dominant (Jack: Aa and Jill: Aa) and serve as the carrier for the disease.
Jack's parents are carriers of the Tay-Sachs gene, and Jack has a 50% chance of being a carrier as well. For Jack and Jill to have a child with Tay-Sachs, both of them would need to be carriers and pass on the faulty copies of the gene.
Explanation:The fact that Jack's parents are unaffected but his brother has Tay-Sachs indicates that they are both carriers of the Tay-Sachs gene. This means that they each have one normal copy of the gene and one faulty copy, which they passed on to Jack's brother. Jack himself has a 50% chance of being a carrier like his parents. In order for Jack and Jill to have a child with Tay-Sachs, both of them would need to be carriers and pass on their faulty copies of the gene, resulting in the child inheriting two faulty copies and developing the disease.
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This problem has been solved!See the answerIn sheep, coat color is influenced by two genes. Gene A influences pigment production, while gene B produces black or brown pigment. If two heterozygous white sheep were crossed and offspring were produced in a ratio of 12 white sheep to 3 black sheep to 1 brown sheep, what are the genotypes of the offspring?
Answer:
White: A---
Black: aaB-
Brown: aabb
Explanation:
As per given information, the genotype of parent white sheep is AaBb (heterozygous white).
The gene "A" is dominant over gene "B". Hence, all the genotypes with one or two copies of "A" would exhibit white phenotype.
All the genotypes that have "aa" and one or two "B" alleles would exhibit black phenotype.
All the genotypes with "aabb" allele combination would exhibit brown phenotype.
A broth medium has been inoculated, and microbial numbers will be counted periodically to generate a bacterial growth curve. At 2 hours after inoculation, the culture has progressed through lag phase and is now in log phase. At this point, the population size is 1 million cells. The generation time is 30 minutes. Assuming the continuation of log growth, how many cells would there be at 2 hours of growth in log phase?
Answer:
16 million
Explanation:
The population at the beginning of the log phase is 1 million cells.
The generation time (doubling time) is 30 min.
In 2 h, there will be four generations.
The population will have doubled four times (2^4 = 16).
There will be 16 million cells.
In the logarithmic phase of bacterial growth, the population size doubles at regular intervals. Given a generation time of 30 minutes, a bacterial culture with a starting population of 1 million cells would grow to 16 million cells after 2 hours. This assumes ideal conditions for exponential growth.
Explanation:The bacterial culture is currently in the log phase of growth, also known as the exponential growth phase. Given that the generation time, or doubling time, is 30 minutes, the bacterial population will double every half hour. Therefore, starting from a population of 1 million cells, after 2 hours of growth (or four 30-minute cycles), the population size would be 16 million cells.
The calculation is straightforward: Starting with 1 million cells (2^0), after one 30-minute cycle (2^1), we have 2 million cells. After two cycles (2^2), we have 4 million cells. After three cycles (2^3), we have 8 million cells. And after four cycles (2^4), we have 16 million cells.
It's important to understand that this idealized exponential growth assumes that the conditions remain constant and optimal for the bacteria, without food shortages, waste accumulation, or other inhibitions to growth. In environmental conditions, such growth rate might not be sustained over time and could be affected by various factors.
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Which of the following statements is FALSE? Which of the following statements is FALSE? Myoglobin is a single polypeptide chain folded about a heme prosthetic group. Hemoglobin is a tetramer, each of which binds a heme group. In both hemoglobin and myoglobin, iron is chelated by a tetrapyrole ring system. The iron in both hemoglobin and myoglobin has two coordination sites that bind to oxygen.
That would be D(the last one).
Iron in both hemoglobin and myoglobin has 1 coordination site that bind to O2.
Each iron chelated in a tetrapyrrole/porphyrin ring can bind to one oxygen molecule. That is why we say hemoglobin can carry a maximum number of 4 O2 molecules. (See attached pic)
Final answer:
The false statement is that the iron in both hemoglobin and myoglobin has two coordination sites that bind to oxygen. Myoglobin has one vacant coordination site for O₂, not two, and in hemoglobin, each of the four iron atoms also has one binding site for oxygen.
Explanation:
The statement "The iron in both hemoglobin and myoglobin has two coordination sites that bind to oxygen" is the FALSE statement. In myoglobin, the heme iron is five-coordinate, meaning it has a single histidine imidazole ligand from the protein and the four nitrogen atoms of the porphyrin to bind, leaving one vacant coordination site for O₂ to bind. In contrast, while hemoglobin is a tetramer and each subunit binds a heme group, the iron at the center of these heme groups does not have two oxygen binding sites, but rather one like in myoglobin.
Myoglobin is a single polypeptide chain that includes a heme prosthetic group, and hemoglobin consists of four polypeptide chains (two alpha and two beta units), each binding a heme group. Both proteins have iron chelated by a tetrapyrole ring system, which is a fundamental component of the heme group that allows these proteins to bind oxygen and function in oxygen transport or storage.
Which of the following is a NOT a function of liver?
a. release enzymes for chemical digestion of food
b. detoxification
c. carbohydrate metabolism
d. bile formation
d. cholesterol excretion
Answer:
release enzymes for chemical digestion of food
Explanation:
Liver is a large organ located on the right upper quadrant just below the diaphragm of an organism.
Liver detoxifies the drug and harmful substances of the body. Liver produces bile juice, regulate carbohydrate metabolism and regulates the cholesterol excretion. Liver doesn't produce any enzyme for the the digestion of food. The bile juice of liver contains salts and chemical that helps in the emulsification of food.
Thus, the correct answer is option (A).
16. Which type of plastid organelle contains a high amount of red/orange photopigments and is known to give fruits & flowers their bright coloration? a. Chloroplast c. Amyloplast b. Chromoplast d. Lypoplast
Answer: b. Chromoplast
Explanation:
Plastids only occur in plant-like protists and are organeles surrounded by a double membrane. There are many types of plastid in plant cells. Chromoplasts include pigments different than chlorophyll and are associated with vividly colored structures and is known to give fruits & flowers their bright coloration. Plastids contain their own DNA in a small ‘plastid genome’ containing genes for some chloroplast proteins.
From the right atrium through the tricuspid valve to the __(1)__ through the __(2)__ valve to the pulmonary trunk to the right and left __(3)__ to the capillary beds of the __(4)__ to the __(5)__ to the __(6)__ of the heart through the __(7)__ valve, to the __(8)__ through the __(9)__ semilunar valve, to the __(10)__, to the systemic arteries, to the __(11)__ of the body tissues, to the systemic veins, to the __(12)__ and __(13)__, which enter the right atrium of the heart.
Answer:
Circulatory system
Explanation:
From the right atrium through the tricuspid valve to the right ventricle through the pulmonary sigmoid valve to the pulmonary trunk to the right and left lungs to the capillary beds of the pulmonary veins to the left atrium to the left ventricle of the heart through the mitral valve, to the aorta through the aortic semilunar valve, to the whole body, to the systemic arteries, to the capillaries of the body tissues, to the systemic veins, to the superior cava vein and inferior cava vein, which enter the right atrium of the heart.
The statement describes the journey of blood through the cardiovascular system. It involves the right ventricle, pulmonary valve, pulmonary arteries, lungs, pulmonary veins, left atrium, mitral valve, left ventricle, aortic valve, aorta, capillary beds of the body, superior and inferior vena cava.
Explanation:The blood flow pathway you're describing is basically the journey of blood throughout the cardiovascular system, including its passage through the heart. The blanks in your statement can be filled in the following way: From the right atrium through the tricuspid valve to the (1) right ventricle through the (2) pulmonary valve to the pulmonary trunk to the right and left (3) pulmonary arteries to the capillary beds of the (4) lungs to the (5) pulmonary veins to the (6) left atrium of the heart through the (7) mitral valve, to the (8) left ventricle through the (9) aortic semilunar valve, to the (10) aorta, to the systemic arteries, to the (11) capillary beds of the body tissues, to the systemic veins, to the (12) superior and (13) inferior vena cava, which enter the right atrium of the heart.
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What does the sympathetic nervous system do
Which of the following statements about the Bohr effect are true? This is the effect of pH on the binding of O2 to Hb. As blood travels from lung to tissue, there is a drop in pH, causing Hb to unload more oxygen than if there were no change in pH. As blood travels from lung to tissue, there is an increase in pH, causing Hb to unload more oxygen than if there were no change in pH. Myoglobin and Hb exhibit similar pH profiles.
Answer:
Option 1 and 2.
Explanation:
Bohr's effect explained the relationship between the binding affinity of oxygen with acidity and carbon dioxide content. Bohr effect was given by the scientist Christian Bohr.
The decrease in pH and increase in the amount of carbon dioxide reducing the binding affinity of oxygen. When the blood travels in tissue from the lungs, pH decreases and oxygen get unload from the hemoglobin as compared with the condition when there is no change in pH.
Thus, the correct answer is option (1)and (2).
The Bohr effect refers to the influence of pH on the interaction between oxygen and hemoglobin. A drop in pH (more acidic conditions) enhances the dissociation of oxygen from hemoglobin while an increase in pH (more basic conditions) inhibits this dissociation.
Explanation:The Bohr effect describes the influence of pH on the binding of O2 to Hemoglobin (Hb). As the blood travels from the lung to the tissues, the pH drops due to the increase in carbon dioxide and byproducts of cell metabolism like lactic acid, carbonic acid, and carbon dioxide which makes the blood more acidic. This drop in pH promotes oxygen dissociation from Hb. Conversely, a higher, or more basic, pH inhibits oxygen dissociation from Hb. This is because there are fewer hydrogen ions available to contribute to the breakdown of bicarbonate ions, leading to less carbon dioxide.
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How many substrate-level (gross) ATPs are generated in glycolysis alone?
A- 2
B- 4
C- 6
D- 10
Answer:
The answer is A-2.
Explanation:
Even though four ATP molecules are produced in the second half. The net gain of glycolysis is only 2 ATP because 2 ATP molecules are used in the first half of the glycolysis.