Final answer:
Disordered eating includes a range of harmful dietary habits and can lead to serious health issues. A focus on 'clean' eating can be a sign of disordered eating, and very low calorie diets can result in electrolyte imbalances. All major eating disorders, not just anorexia, can potentially be life-threatening, and there are various behaviors associated with bulimia beyond vomiting.
Explanation:
Disordered eating can present itself through various behaviors and does have significant health implications if not addressed properly. Let’s clarify which statements about disordered eating are correct:
A person who obsesses over "clean" eating may have disordered eating habits, even if energy intake is adequate. This statement is correct because an obsession with eating foods that are considered 'clean' or 'pure' can be indicative of an underlying issue with food and self-control, regardless of the adequacy of energy intake.Anorexia is not the only eating disorder that is life-threatening. Bulimia nervosa, binge eating disorder, and other eating disorders can also have serious health consequences.Individuals with bulimia nervosa do not always purge by vomiting. They may use other methods such as excessive exercise, fasting, or the use of laxatives and diuretics.Very low calorie diets can lead to dangerous electrolyte imbalances. This is a correct statement as such diets can disrupt the balance of minerals necessary for vital organ functions.An individual can have disordered eating habits without having a clinically diagnosed eating disorder, which is also true. Disordered eating includes a range of irregular eating behaviors that may or may not warrant an eating disorder diagnosis.Discuss the biological importance of each of the following organic compounds in relation to cellular structure and function in plants and animals.
a) Carbohydrates
b) Proteins
c) Lipids
d) Nucleic acids
Explanation:
Carbohydrates: they are the main source of ATP inside the cell and in the membrane, they can participate in cell recognition, adhesion, signaling and also as a physical barrier. Proteins: they form channels or bombs to exchange different molecules into or out of the cell, they also conform most of the enzymes responsible for many metabolic tasks inside the cell. Lipids: they are the main content of the cell's membrane, they form a fluid lipidic bilayer that permeates the cell from its environment and allows the movement of proteins and carbohydrates in it. Nucleic acids: they contain the genetic information of the cellI hope you find this information useful and interesting! Good luck!
The biological importance components:
The carbohydrates are the main ATP and participate in cell recognition. Proteins are forms of channels that exchange different molecules. Lipids are the main content of the cell membrane. Nucleic acids consist of genetic information from cellsWhat are Carbohydrates ?These are the main source of ATP in the cells and membrane, they can participate in cell recognition, adhesion, signaling, and also as a physical barrier.
While Proteins are channels or bombs to exchange different molecules into or out of the cells. Lipids: they are the main content of the cell's membrane, fluid layers. Nucleic acids consist of the genetic matter of the cell.
Find out more information about the compounds.
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DNA codes for the sequence of amino acids in the primary structure of a protein, but not for sugars or lipids. This is because?A. only proteins are involved in living metabolic reactions. B. sugars and lipids code for their own replication. C. sugars and lipids are ever present in the living environment and are not used in living structures. D. other hereditary molecules code for sugars and lipids. E. proteins are the main structural and functional components of cells.
Answer:
E. proteins are the main structural and functional components of cells.
Explanation:
DNA sequence codes for proteins since proteins are the main structural and functional components of cells. Proteins are structural components of cell membranes. The DNA packaging proteins pack the DNA into chromatin. The immune system distinguishes the self from non-self by identification of specific peptides present on the cell surfaces.
Hemoglobin is the oxygen carrier protein of RBCs. Immunoglobulins are one of the most abundant proteins of blood serum and are responsible for adaptive immune response. Enzymes are proteins with catalytic functions. Myosin and actin are the major structural and functional proteins of muscle cells. Proteins with different primary structures perform these diverse functions.
Sympathetic nerves may leave the spinal cord at which vertebra?
a. first coccyx
b. first thoracic
c. third lumbar
d. second cervical
Answer:
b. first thoracic
Explanation:
Sympathetic nervous system is one of the divisions of the autonomic nervous system. The other divisions are parasympathetic nervous system and enteric nervous system.
Autonomic nervous system is composed of preganglionic neurons and postganglionic neurons.
Similarly, the sympathetic nervous system is composed of postganglionic and preganglionic neurons.
The cell bodies of preganglionic neurons of sympathetic nervous system is found in intermediolateral cell columns of spinal cord segments T-T12 and L1-2 or 3. Therefore, sympathetic nervous system is also called thoracolumbar division.
The cell bodies of parasympathetic nervous system are found in the brainstem nuclei which exit via cranial nerves 3, 7, 9, 10 and also in the sacral spinal cord segements S2-4, therefore parasympathetic nervous system is also known as craniosacral division.
QUESTION:
b. first thoracic
This is the correct answer choice as sympathetic nervous system cell bodies are found in intermediolateral cell column of T1-12 and L1-2 or 3 spinal cord segments. Hence, a sympathetic nerve is most likely to leave via first thoracic vertebra.
Third lumbar is less likely as sympathetic nervous system does not extend to this level in every individual.
Sympathetic nerves leave the spinal cord at the first thoracic vertebra, playing a critical role in the regulation of the body's involuntary mechanisms.
Explanation:The sympathetic nerves, which are a key part of the autonomic nervous system that regulates body's unconscious actions, typically leave the spinal cord at the first thoracic vertebra. Therefore, the correct answer to your question is b: first thoracic. It's important to note that while they start at this vertebra, sympathetic nerves extend throughout the body to help regulate various involuntary mechanisms, such as heart rate, blood pressure, and digestion.
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In an ocean ecosystem, seabirds are predators of fish. What would happen to the balance of an ocean ecosystem if the number of seabirds decreased? A. The population of fish in the ecosystem would increase. B. The population of fish in the ecosystem would decrease. C. The population of fish in the ecosystem would stay the same. D. The population of fish would disappear completely.
Answer: The correct option is A.
Explanation: Seabirds are predators of fish in an ocean ecosystem. When the population of sea birds decreases, the population of fish in the ocean increases. This is because sea birds feed on the fish and fish population is controlled by sea birds whenever seabirds decrease in number, then very low amount of fish is required to feed the sea birds so automatically fish population is increased.
What is aquaculture?a. Farm production of fish or seafood by raising animals in tanks, ponds, or ocean net pens.b. A fishery that ensures that fish stocks are maintained at healthy levels, the ecosystem is fully functional, and fishing activity does not threaten biological diversity.c. Fishing in the open ocean without the use of trawl nets and thus without bycatch.
Answer: Option A - Farm production of fish or seafood by raising animals in tanks, ponds, or ocean net pens.
Explanation:
Aquaculture is the cultivation of aquatic produce such as aquatic plants, fish, and other aquatic animals.
Answer:
The answer is A
Explanation:
Farm production of fish or seafood by raising animals in a controlled aquatic environment like tanks, ponds, or ocean net pens
Which bacterial modification allowing the bacteria to survive harsh times, like dessication?
Answer:
The correct answer is endospore and capsule.
Explanation:
Bacteria can do modification in them during harsh conditions which protect them from the harsh time. These modifications are the formation of endospores or capsules.
Endospore is the hard and non-reproductive structure formed outer to the cell wall by mainly bacteria of phylum firmicutes in the absence of nutrients. Endospore helps the bacteria to remain dormant for several years until the right condition comes.
Capsule is a gelatinous layer secreted by bacterial cell which is made up of polysaccharides. It forms outer to cell wall and protects the bacteria from desiccation and engulfment by phagocytes.
Final answer:
Bacteria can survive harsh conditions like desiccation by forming endospores, which are highly resistant structures that protect the bacterial DNA in a dormant state until favorable conditions return.
Explanation:
The bacterial modification that allows bacteria to survive harsh times, such as desiccation, is the formation of endospores. Formed under unfavorable conditions, endospores encapsulate the genetic material of the bacterium and become dormant, allowing the bacterium to withstand extreme conditions including heat, lack of moisture, and exposure to chemicals. They remain in this resistant state until conditions become favorable again, at which point they can re-animate and resume normal cellular functions. Bacterial genera such as Bacillus and Clostridium are well-known for their ability to form endospores.
Other adaptions include changes in cell wall structure, accumulation of energy reserves like starch or oils, and alterations in membrane and protein structure to reduce metabolic activity during harsh conditions. Some bacteria, like Halobacterium, can survive in environments with extremely high salt concentrations, which would be lethal to most other organisms. This extreme halotolerance is an example of how microbial life can adapt to survive and even thrive under conditions that would normally be considered inhospitable.
(a) Addition of malate to an extract of muscle cells stimulates oxygen consumption by the extract. Explain why malate stimulates oxygen consumption.
(b) Malonate is an inhibitor of succinate dehydrogenase. If malonate is added to a suspension of bacteria that are catabolizing glucose then oxygen consumption is inhibited. Explain why malonate inhibits oxygen consumption.
Answer:
Oxygen consumption is a measure of the activity of the first two stages of cellular respiration: glycolysis and the Kreb's cycle. The addition of malate stimulates the citric acid cycle and thus stimulate respiration.
Explanation:
The added malate serves a catalytic role, because it is regenerated in the later part of the citric acid cycle. With higher concentrations of oxaloacetate or malate, higher flux of acetyl CoA will be utilized into the citric acid cycle increasing the oxygen consumption much greater levels.
The inhibition of succinate oxidation by malonate is a well known phe-
nomenon. Since the oxidation of succinate to fumarate is an integral
part of the Krebs cycle of oxidations, it has been generally assumed that
the inhibitory effect of malonate upon the oxidation of any member of
the cycle is the result of the inhibition of the succinate to fumarate step. malonate inhibits oxidations in the cycle by at least two mechanisms: in addition to the inhibition resulting from a block of succinate oxidation, malonate inhibits oxidation by another mechanism that is believed to involve combination with magnesium ions.
Calculate the difference in blood pressure between the feet and top of the head.
Answer: this is an incomplete question but pressure is said to decrease with height and increase with depth.
This means pressure at the top of the head is lower than pressure at the feet.
It is taken that blood pressure at the arms is roughly 10% lower than blood pressure at the legs.
Explanation:
If Avery had observed transformation using only the extracts containing degraded DNA, degraded RNA, and degraded protein, but NOT the extract containing degraded polysaccharides, he would have concluded that
Answer:
Polysaccharides are the genetic material.
Explanation:
Avery did not observe transformation using the extracts containing degraded DNA. On the other hand, extracts with degraded RNA, proteins, and polysaccharides exhibited transformation. Therefore, he concluded that DNA is the genetic material responsible for transformation. If he would have observed the process of transformation using extracts containing degraded DNA but not with degraded polysaccharides, he might have concluded that "polysaccharides were the genetic material responsible for the process of transformation."
Reducing equivalents are key to the process of cellular respiration and ultimately ATP production. Illustrate the direct inter-linkage of the glycolytic pathway to cellular respiration, concluding with the general illustrative mechanism and description to the production of ATP. (Hint: draw the reason that you sometimes get 30 ATP out of glycolysis instead of 32.
→The correct question should be 'draw the reason that you sometimes get 30 ATPs out of cellular respiration instead of 32'
Answer :Glucose is high energy-rich ccompound. Therefore to initiate the breakdown down of bonds in glucose for phosphorylation of glucose; to obtain Pyruvate for ATPs synthesis 2ATPs must be supplied...
The 2 ATPs were ‘borrowed’ – to convert
1. Glucose to fructose phosphate,
2. And later to fructose bisphosphate. This is phosphorylation of glucose.
The end product of this reaction is PYRUVATE. THE NET ATPs produced is 4ATPs per molecule of glucose, but the total gained ATPs is 2, because the 2ATPs 'borrowed' from the phosphorylation must be paid back. This is substrate level phosphorylation and it takes place in the cytoplasm without OXYGEN.
Link reaction is the process of conversion of pyruvate from glycolysis to Acetyl CoA .It is the reaction pathway which linked glycolysis with citric acid cycle, in cellular respiration in the presence of oxygen. If no oxygen is available, link reaction does not take place; rather the pyruvate moves to fermentation
It involves decarboxylation, and dehydrogenation (with NADH) of 3Cpyruvate, and reaction with Coenzyme A to form 2C-acetyl CoA,
The 2C-acetyl Coenzyme A reacted with 4c oxaloacetate in the mitochondrial matrix to form 6C citrate. This involves decarboxylation and dehydrogenation with NADH.
The citric cycle continues on till 2ATPs are produces with each glucose molecule; until the 4 molecule oxaloacetate is reproduced. 4CO2, 6NADH, 2FADH2 and 2ATPs molecules per molecule of glucose are produced.
The NADH and FADH2 fetched Hydrogen into the mitochondrial matrix where it is split to
Protons and electrons.
The electrons are transferred by protein carries in the matrix. As these electrons moved through proteins carriers of different energy levels ENEGY is released;and used to pump protons from the matrix across the intramembrane to the cytosol. The proton gradient drives protons down the concentration gradient through the hydrogen channel of ATPase synthase. The enzyme tapped the energy from the protons to synthesise ATP from ADP.
This is oxidative phosphorylation.it produces 28ATPs
The balanced sheet of ATP per molecule of glucose =
→4ATPS from glycolysis ( 2ATPs used ).
→2ATPs from Citric Acid
→28 ATPs from Oxidative phosphorylation.
⇒Total =34 - (2ATPs) used =30ATPs.
Explanation:
Vascular bundles neatly and uniformly arranged around the stem is _________.
Answer: Xylem
Explanation:
Xylem is a vascular tissue that helps in distributing water and minerals taken up by the roots.
It is located in the innermost portion of the stem, and is neatly arranged in a ring form
How many of the following statements concerning the loss of hind limbs during whale evolution are true? 1. It is well documented by a series of transitional fossils. 2. It explains why modern whales have vestigial pelvic girdles. 3. It involved changes in the sequence or expression of Hox genes. 4. It is an example of macroevolution. 5. It, and the loss of limbs by snakes, are an example of similar adaptations to a similar environment. A) Only one statement is true. B) Two statements are true. C) Three statements are true. D) Four statements are true. E) All five statements are true.
Answer:
D) Four statements are true.
Explanation:
True statements are:
1. It is well documented by a series of transitional fossils.
2. It explains why modern whales have vestigial pelvic girdles.
3.It involved changes in the sequence or expression of Hox genes.
4. It is an example of macroevolution.
Which is a uniquely sympathetic function?
a. regulation of body temperature
b. regulation of pupil size
c. regulation of respiratory rate
d. regulation of cardiac rate
Answer:
The correct answer will be option-A
Explanation:
Homeostasis refers to maintaining the internal temperature of the body irrespective of the external condition. However, the signals from the outer environment directly influence the mechanism which maintains the homeostatic state.
The homeostatic mechanisms maintain the body temperature and pH of the blood.
The sympathetic nervous system regulates the body temperature by producing hormones and activation of the parasympathetic nervous system. All other responses of the sympathetic nervous system are not unique.
Thus, option-A is the correct answer.
Regulation of body temperature is a uniquely sympathetic function. The correct option is A.
Thus, Controlling cardiac rate, also known as heart rate, is a solely sympathetic function. The "fight-or-flight" response, which primes the body for strenuous exercise or stress, is brought on by the sympathetic nervous system.
Norepinephrine, which binds to beta-adrenergic receptors in the heart and causes an increase in heart rate, is released by sympathetic neurons in this reaction.
In times of increased activity or stress, this aids in supplying the body's tissues with more oxygen and nutrients.
Thus, Regulation of body temperature is a uniquely sympathetic function. The correct option is A.
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The researchers predict that for any particular concentration of substrate the c aurantiacus enolase catalyzed reaction is more rapid at 55c than at 37c
Answer:
because increase in temperature will increase the enzyme catalyzed reaction.
Explanation:
As the temperature increases the kinetic energy of molecules increases means the collision between substrate and enzyme increases that is substrate attached to its appropriate active site of enzyme which results in increase in reaction. But if temperature is too high it can denature the enzymatic activity.
In catalyzed reaction increase in temperature will loosen the bond between molecules which results in catalyzation.
Enzyme activity is influenced by environmental conditions such as temperature and substrate concentration. Hence, the enolase enzyme from c aurantiacus would catalyze reactions faster at 55°C than at 37°C, although extremely high temperatures can denature enzymes. Enzyme activity also increases with substrate concentration until a saturation point.
Explanation:The question is about the effect of temperature on the activity of the enzyme enolase from the microorganism c aurantiacus. Enzymes, including enolase, speed up chemical reactions. They work best under the environmental conditions in which the organisms that produce them live.
One factor that affects enzyme activity is temperature. Increasing the environmental temperature generally increases reaction rates. Thus, the rate of the c aurantiacus enolase-catalyzed reaction would be faster at 55°C than at 37°C. However, if the temperature is too high it can cause enzymes to denature, i.e., lose their three-dimensional structure and function.
Another factor that affects enzyme activity is substrate concentration: Enzyme activity is increased at higher concentrations of substrate until it reaches a saturation point at which the enzyme can bind no additional substrate.
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With independent assortment, the ratio of phenotypes in the F2 generation of a cross between true-breeding strains (AA bb x aa BB) can be described as 9:3:3:1 when A and B are dominant over a and b. To what phenotype does the "9" in the ratio refer?
a. recessive for both traits
b. dominant for one trait and recessive for the other
c.dominant for the A trait and recessive for the B trait
d. dominant for both traits
Final answer:
The "9" in the 9:3:3:1 ratio refers to offspring that are dominant for both traits, which occurs with a probability of 9/16 when considering two traits with independent assortment.
Explanation:
The "9" in the 9:3:3:1 phenotypic ratio refers to the phenotype that is dominant for both traits. This ratio emerges from the cross between true-breeding strains (AA bb x aa BB) when A and B are dominant alleles over a and b. Independent assortment ensures that alleles for different traits are transmitted to offspring independently of each other. Combining the probabilities for the dominant phenotypes of each trait, both texture and color in this context, we apply the product rule. Since each trait has a 3:1 dominant to recessive ratio, the probability of an offspring being dominant for both traits is (3/4) × (3/4) = 9/16.
The correct option is d. dominant for both traits. The 9 therefore represents the phenotype that is dominant for both traits.
The 9 in the 9:3:3:1 ratio refers to the phenotype that is dominant for both traits. This ratio is derived from the principles of Mendelian genetics, where each trait is determined by a pair of alleles, with one allele inherited from each parent.
In the given cross (AA bb x aa BB), the parents are true-breeding for different traits: one parent is homozygous dominant for trait A (AA) and homozygous recessive for trait B (bb), while the other parent is homozygous recessive for trait A (aa) and homozygous dominant for trait B (BB).
The F1 generation from this cross would all be heterozygous for both traits (Aa Bb), showing the dominant phenotype for both traits due to the presence of at least one dominant allele for each trait.
When the F1 individuals are crossed to produce the F2 generation, the alleles for each trait assort independently. The possible gametes for the F1 individuals are AB, Ab, aB, and ab, each with an equal probability of 1/4. The Punnett square for the F2 generation would look like this:
AB Ab aB ab
-------------------------
AB | AA BB | AA Bb | Aa BB | Aa Bb
-------------------------
Ab | AA Bb | AA bb | Aa Bb | Aa bb
-------------------------
aB | Aa BB | Aa Bb | aa BB | aa Bb
-------------------------
ab | Aa Bb | Aa bb | aa Bb | aa bb
The resulting phenotypic ratio is calculated by counting the number of each phenotypic class:
- Dominant for both traits (AA BB, AA Bb, Aa BB): 4 offspring with the dominant phenotype for both traits.
- Dominant for trait A and recessive for trait B (AA bb, Aa bb): 2 offspring with the dominant phenotype for trait A and recessive for trait B.
- Recessive for trait A and dominant for trait B (aa BB, aa Bb): 2 offspring with the recessive phenotype for trait A and dominant for trait B.
- Recessive for both traits (aa bb): 1 offspring with the recessive phenotype for both traits.
The ratio of these phenotypes is 4:2:2:1, which simplifies to 9:3:3:1 when multiplied by 2 to avoid fractions.
Many muscular/skeletal problems and injuries especially in adults can be attributed to a lack of ___
Answer:
flexibility
Explanation:
Stretching may help prevent injury according to many doctors. Stretching improves your flexibility. "A sports medicine physician will determine your level of flexibility and form an activity or exercise prescription with specific exercises and stretches to improve your flexibility." This shows that your level of flexibility contributes to what activities you can do without injury.
Muscular/skeletal problems and injuries can be attributed to a lack of exercise or physical activity. Regular physical activity and exercise help to strengthen and maintain the health of skeletal muscles.
Explanation:Health: Understanding Musculoskeletal Problems and Injuries
Muscular/skeletal problems and injuries can be attributed to a lack of exercise or physical activity. When a person leads a sedentary life, their skeletal muscles can atrophy due to lack of use, while smooth muscles, which are involuntary muscles found in organs, do not typically experience the same level of atrophy.
Regular physical activity and exercise help to strengthen and maintain the health of skeletal muscles. When muscles are not regularly used and stimulated, they can weaken and lose mass, making them more susceptible to injuries and problems.
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Synthesis of glucose from pyruvate during the Cori cycle occurs primarily in the__________.
Synthesis of glucose from pyruvate during the Cori cycle occurs primarily in the Liver.
What is the Cori cycle?The Cori cycle refers to the metabolic pathway in which lactate is produced by the fermentation in the muscles and moves to the liver and is converted to glucose which then returned to the muscles and is converted back to lactate.
According to the context of this question, in animals, lactate formed in the muscles is successfully recycled to glucose in the liver. Lactate produced in the muscles is transported from the muscles to the liver where it is reoxidized by liver LDH to pyruvate.
Through the process of gluconeogenesis, pyruvate is finally converted into glucose in the liver.
Therefore, the synthesis of glucose from pyruvate during the Cori cycle occurs primarily in the Liver.
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Final answer:
Glucose synthesis from pyruvate in the Cori cycle primarily occurs in the liver, where pyruvate undergoes gluconeogenesis to form glucose.
Explanation:
Synthesis of glucose from pyruvate during the Cori cycle occurs primarily in the liver. The liver plays a crucial role in this metabolic cycle by converting lactate, which is produced by anaerobic glycolysis in the muscles during periods of intense activity, back to pyruvate. This pyruvate is then synthesized into glucose through a process called gluconeogenesis. The newly formed glucose is released into the bloodstream, providing energy for muscle cells and other tissues, thereby completing the Cori cycle.
Read the following information and then answer follwoing questions:
Heartburn is an esophageal symptom characterized by a burning sensation behind the breastbone. It may be related to gastric hyperacidity. There is little doubt that emotional disturbance, excitement, and nervous tension result in hyperacidity, which in turn may lead to heartburn. A gesture that usually accompanies heartburn is movement of the open hand up and down the breastbone. This is in contrast to the stationary tightly clenched fist of angina pectoris. Another important aspect of heartburn is that it is usually relieved, if only temporarily, by taking antacids. Antacids are used in other instances as well, particularly in cases of peptic ulcer. Pain is the outstanding symptom of peptic ulcer, a disorder that is chronic, periodic, almost always characterized by gnawing or burning, and coinciding with digestion. Nausea, emesis, and anorexia are uncommon.
1. Where does the burning sensation occur in heartburn?
2. Which factors lead to heatburn?
3. What is the use of antiacids?
4. What are the symptoms of peptic ulcer?
Answer:
Following are the answers for the given questions.
Heart burn is a burning sensation behind the Breast bone. Breast bone is another name for Sternum.Gastric hyper acidity is the main cause of heart burn. The other factors that would likely lead to hyper acidity of stomach are emotional disturbance, excitement and nervous tension that ultimately leads to heart burn.Antacids are medicines that are mainly used for symptomatic treatment of heart burn and peptic ulcers.Pain is the main and outstanding symptom of peptic ulcers. In chronic cases, peptic ulcer presents with gnawing and burning sensation after food ingestion. Uncommon presentations of chronic peptic ulcers are Nausea, Emesis (vomiting) and Anorexia (loss of appetite)Explanation:
The answers had been taken from the given paragraph in the asked question.
During intramembranous ossification, the developing bone grows outward from the ossification center in small struts called _
Answer:
During intramembranous ossification, the developing bone grows outward from the ossification center in small struts called spicules.
Explanation:
The intramembrane ossification is the one that preferentially produces flat bones and, as the name implies, takes place within a connective tissue membrane. In this process, some of the mesenchymal cells that form the connective tissue membranes are transformed into osteoblasts constituting an ossification center around which bone is formed. Consequently, a certain amount of flat bones develop that are characterized by the presence of bone spicules. These spicules radiate from the primary ossification centers to the periphery. As growth continues during fetal and postnatal life, these bones increase in volume by apposition of layers on their outer surface and by osteoclastic resorption from the inside of each bone.
Populations evolve for many reasons. Suppose there is a population of plants that have either purple flowers or white flowers, and the allele for purple flowers is dominant. This means that plants with two purple alleles have purple flowers. Plants with one purple allele and one white allele also have purple flowers. Only plants with two white alleles have white flowers. For each event or condition described below, answer the following questions.
a. Which mechanism of evolution is at work?b. How does this event affect the populations gene pool? Do the frequencies of the two alleles change, and if so, how?
Answer:
Natural selection by directional selection, was the driving force of evolution at work. The purple flowers parents were able to withstand the selective pressure and therefore emerged from completions with other flowers from previous generation, thus their dominance follow directional selection. This was the trait inherited by their offspring, the dominant purple
The gene pool of purple flower will be large; because of dominance in expression with each successive generation in both homozygote and heterozygote state, and therefore higher frequency of expression compared to the white flowers.
The frequency of allele depends on the rate expression of the allele in the population. Thus, the dominant purple flowers have high frequency of expression compared to the recessive white flowers which can only express itself in homozygote recessive state
Hydrothermal processes are most likely involved in the formation of which set of resources?
Answer:
mineral resources like the Zinc, copper, gold and silver.
Explanation:
The hydrothermal process usually includes the movement s of the subsurface hot waters and the upwelling of the magma from the earth mantel. This process leads to the formation of deposits such as those of the iron, zinc and gold and silver.After the first exposure to an antigen, a ________ stimulates growth and multiplication of antigen-reaction cells. After the first exposure to an antigen, a ________ stimulates growth and multiplication of antigen-reaction cells. phagocytic immune response secondary innate immune response hyperactive cytotoxic response primary adaptive immune response
Answer:
Primary adaptive response
Explanation:
After the first exposure to an antigen, a primary adaptive response stimulates growth and multiplication of antigen-reaction cells. phagocytic immune response secondary innate immune response hyperactive cytotoxic response primary adaptive immune response
What would be the consequences of the following?(A) A mutation of a promoter sequence such that the promoter is deleted.(B) A mutation of the gene that encodes RNA polymerase such that the polymerase is not made.(C) Deletion of intron consensus sequences from a gene.
Answer A)
The gene would not be transcribed. This is because the gene transcribing machinery would not be activated if the promoter sequence is not recognized. Transcription will not start.
Answer B)
The gene would not be transcribed. This will result in no RNA polymerase being formed. Hence, transcription and translation will not occur.
Answer C)
The mRNA would contain intron sequences, which would be translated into extra amino acids in the protein. A non- functional protein would be made.
Mutations in the promoter sequence can prevent RNA polymerase from attaching, reduce or cease transcription. Deletion of intron consensus sequences can cause improper splicing during transcription.
Explanation:Mutations in the promoter sequence can have various consequences on gene expression.
If a mutation in the promoter sequence deletes the promoter, it will result in RNA polymerase not being able to attach, leading to a significant decrease or complete cessation of transcription. This will prevent the gene from being expressed into a protein.
On the other hand, if a mutation occurs in the gene that encodes RNA polymerase and prevents its production, it will have a similar effect as the promoter deletion, as without RNA polymerase, transcription cannot occur.
Deletion of intron consensus sequences from a gene can lead to improper splicing during transcription. Intron consensus sequences are important for the proper removal of introns and the splicing of exons. Deletion of these sequences can result in incorrect splicing and the production of abnormal mRNA and proteins.
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Let’s suppose that weight in a species of mammal is polygenic, and each gene exists as a heavy and light allele. If the allele frequencies in the population are equal for both types of alleles (i.e., 50% heavy alleles and 50% light alleles), what percentage of individuals will be homozygous for the light alleles in all of the genes affecting this trait, if the trait was determined by the following number of genes?
Two
Three
Four
Final answer:
Using the Hardy-Weinberg principle, the percentage of individuals homozygous for the light alleles in a population with equal allele frequencies is 6.25% for two genes, 1.56% for three genes, and 0.39% for four genes.
Explanation:
When assessing the population's genetic structure, biologists focus on the frequencies of the resultant genotypes to infer phenotype distribution. Assuming equal frequencies of heavy and light alleles for a polygenic trait (50% each), the Hardy-Weinberg principle can be applied. Using this principle, the probability of an individual being homozygous for the light alleles can be calculated using p² + 2pq + q² = 1, where p and q represent the frequencies of the heavy and light alleles, respectively. In this case, the frequency of homozygous light alleles (qq) will be q².
For two genes, the probability an individual will be homozygous for the light alleles at both loci is q² for one gene times q² for the other gene:
q² × q² = q⁴ = (0.5)⁴ = 0.0625 or 6.25%.
For three genes, the calculation becomes:
q² × q² × q² = q⁶ = (0.5)⁶ = 0.015625 or 1.56%.
For four genes, the calculation would be:
q² × q² × q² × q² = q⁸ = (0.5)⁸ = 0.00390625 or 0.39%.
The percentage of individuals homozygous for light alleles are approximately 6.25% for two genes, 1.56% for three genes, and 0.39% for four genes.
In this scenario, the frequency of alleles for the mammal's weight is equally divided between heavy and light alleles, each at 50%. If we denote the frequency of the light allele as q (where q = 0.5), we can use the formula for homozygous recessive genotype frequency (q²) to determine the frequency of individuals homozygous for the light alleles in each gene.
Calculation for Two Genes:
For two genes, the probability of an individual being homozygous for light alleles in both genes is calculated by:
Calculate q² for one gene: (0.5)² = 0.25Since each gene is independent, the frequency of homozygous light alleles for both genes is: (0.25)² = 0.0625 or 6.25%Calculation for Three Genes:
For three genes, follow the same steps:
Calculate q² for one gene: (0.5)² = 0.25
For three genes, the frequency is: (0.25)³ = 0.015625 or 1.56%Calculation for Four Genes:
Similarly, for four genes:
Calculate q² for one gene: (0.5)² = 0.25For four genes, the frequency is: (0.25)⁴ = 0.00390625 or 0.39%Uniden's condition is an X-linked recessive trait in the Redialer population. The trait allele F has a frequency of 0.005 in this population; that is. the frequency is the same in the male and female populations. Individuals with this trait have a strong desire to use land-line phones. Note, the trait gene is in Hardy Weinberg Equilibrium (HWE) for the female Redialer population. An man with the trait and an woman without the trait have a son. What is the probability that the son has Uniden's condition? (Use the HWE equation to determine the most likely genotype in the woman). 0.0 0.25 0.005. 0.5 1E-05.
The probability that the son has Uniden's condition is 0.5 or 50%.
The trait allele F has a frequency of 0.005 in the Redialer population.
The trait is X-linked recessive.
The trait gene is in HWE for the female Redialer population.
A man with the trait and a woman without the trait have a son.
Step 1: Determine the genotype of the woman.
Since the trait is X-linked recessive, the woman without the trait must be homozygous for the dominant allele (XX).
Genotype of the woman: XX
Step 2: Determine the genotype of the man.
Since the man has the trait, he must be hemizygous for the recessive allele (XF).
Genotype of the man: XF
Step 3: Calculate the probability of the son having Uniden's condition.
The son inherits one X chromosome from the mother (X) and one X chromosome from the father (XF).
The probability of the son having Uniden's condition (XF) is 0.5.
A 35-year-old male, Mr. NX, presents to your clinic today with complaints of back pain and "just not feeling good." Regarding his back, he states that his back pain is a chronic condition that he has suffered with for about the last 10 years. He has not suffered any specific injury to his back. He denies weakness of the lower extremities, denies bowel or bladder changes or dysfunction, and denies radiation of pain to the lower extremities and no numbness or tingling of the lower extremities. He describes the pain as a constant dull ache and tightness across the low back.He states he started a workout program about 3 weeks ago. He states he is working out with a friend who is a body builder. He states his friend suggested taking Creatine to help build muscle and Coenzyme Q10 as an antioxidant so he started those medications at the same time he began working out. He states he also takes Kava Kava for his anxiety and garlic to help lower his blood pressure.1. His historical diagnoses, currently under control, are:A. Type II diabetes since age 27B. High blood pressureC. Recurrent DVTs2. His prescribed medications include:A. Glyburide 3 mg daily with breakfastB. Lisinopril 20 mg dailyC. Coumadin 5 mg daily
Answer:
Recurrent DVT
Coumadin 5mg daily
Explanation:
DVT is called Deep venous thrombosis.. It is characterized by blood clots in the legs which can dislodge to any part of the body and cause blockage of blood vessels causing recurrent pain..
Coumadin is a drug used in treating and preventing DVT
A researcher follows a protocol to test the activity of a mitochondrial extract containing all of the soluble enzymes of the matrix. Because the mitochondrial extract was dialyzed, the protocol lists low molecular weight cofactors that must be added to the extract in order to catalyze the oxidation of acetyl-CoA to CO2. The list does not include lipoic acid, a known cofactor of the citric acid cycle. Why is lipoic acid omitted from the list of cofactors to add back to the extract? O O O O O The added TPP can substitute for lipoic acid in the pyruvate dehydrogenase complex. Lipoic acid is covalently attached to the pyruvate dehydrogenase complex. The disulfide bond in lipoic acid prevents diffusion through the dialysis membrane. The Kd of lipoic acid binding to pyruvate dehydrogenase is extremely low. The oxidation of acetyl-CoA to CO2 does not require lipoic acid in vitro.
Answer:
Lipoic acid is covalently attached to the pyruvate dehydrogenase complex.
Explanation:
The pyruvate dehydrogenase complex has lipoic acid bound covalently to it.
This explain why dialysis is not effective in separating the lipoic acid from the enzyme
Answer:
The answer is: Lipoic acid is covalently attached to the pyruvate dehydrogenase complex
Explanation:
It can be said that via a convalent amide bond lipoic acid binds to the terminal lysine residue found in the lipophilic domains of the enzyme. The importance of lipoic acid is that it acts as a cofactor in the pyruvate dehydrogenase enzyme complex.
What is the genus of plant that is easily identified by having no true leaves, dichotomous branching as well as lobed sporangia (usually yellow in color)
-None of these
-Psilotum
-Cardiospermum
-Ginkgo
-Equisetum -Marchantia
Answer:
Psilotum.
Explanation:
Psilotum is also known as whisk fern, it is a fern like vascular plants genus. This plant is found in tropical areas, and more temperate areas. Plants of this genus lack true leaves, and roots (they are anchored by creeping rhizomes).
The stem of Psilotum contains many branches with paired enations (like small leaves) but they do not have vascular tissues. By the fusion of three sporangia, above the enations synangia formed, and produced the spores.
Reproduction by mitosis duplicates:A. chromosomesB. genesC. DNAD. all of the above
I do not understand why this was deleted, but Reproduction by mitosis duplicates all of the above, because all of these are in the gene being duplicated, the chromosomes, the gene itself, and the DNA.
Hope this helps, if not, comment below please!!!!
Reproduction by mitosis duplicates by chromosomes, genes and DNA which makes the answer option D
Reproduction by mitosis duplicates?Reproduction by mitosis results in the duplication of chromosomes, genes, and DNA. Mitosis is a type of cell division that ensures each daughter cell receives an identical set of genetic information to the parent cell. During mitosis, the chromosomes are replicated, and the genetic material, including genes and DNA, is duplicated to be equally distributed between the two daughter cells. This process allows for the growth, development, and tissue repair in multicellular organisms and ensures the genetic continuity of the cell's genome.
The answer to this problem is D. All of the above
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In humans, the AMY1 gene produces the enzyme amylase in cells of the salivary glands. Amylase breaks down starch (a polysaccharide) into the sugar maltose (a disaccharide). People from cultures with diets high in starch produce more amylase than people from cultures with diets low in starch because of a mutation in the AMY1 gene. Explain in two to three sentences why the frequency of this AMY1 mutation would have increased in frequency in populations with a high starch diet.
Answer:
Natural selection
Explanation:
AMY1 produces amylase, a protein that breaks down starch from food into maltose, which is the first step in properly digesting starch.
A mutation in the AMY1 gene that causes it to produce more amylase, would mean that individuals carrying that mutation would have more amylase in their saliva, and would be better at breaking down the starch.
If you can more efficiently break down the starch in your diet, it will be a better source of energy for you. This would have been particularly important thousands of years ago, when food security was poorer. If you lived in a region or were part of a culture where starch was an important part of your diet, you might have been better nourished if you could properly break down these starches.
Over time, natural selection would be at work: the individuals carrying the AMY1 mutation would have improved fitness, because of their ability to digest starch and get all the nutrition and energy from it (i.e. maybe they were less likely to get sick, to starve etc.). Therefore, very slowly, this mutation would become more frequent in those regions where it is beneficial.
The AMY1 mutation likely became more prevalent in high-starch diet cultures because it offers a survival advantage, allowing for more efficient starch digestion and nutrient absorption.
Explanation:The frequency of the AMY1 mutation that increases amylase production would have increased in populations with high-starch diets due to the principles of natural selection and evolution. Essentially, individuals with this mutation could more efficiently digest starch resulting in greater nutrient absorption and, consequently, a potential survival and reproductive advantage. Over many generations, these benefits would increase frequency of this AMY1 mutation within the population, reflecting the adage 'survival of the fittest'.
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