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.
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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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.
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
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%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 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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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.
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.
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
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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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.
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.
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.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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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.
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.
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: