Answers
Answer:
Simplified diagram of pyruvate oxidation. Pyruvate—three carbons—is converted to acetyl CoA, a two-carbon molecule attached to coenzyme A. A molecule of coenzyme A is a necessary reactant for this reaction, which releases a molecule of carbon dioxide and reduces a NAD+ to NADH.
Simplified diagram of pyruvate oxidation. Pyruvate—three carbons—is converted to acetyl CoA, a two-carbon molecule attached to coenzyme A. A molecule of coenzyme A is a necessary reactant for this reaction, which releases a molecule of carbon dioxide and reduces a NAD+ to NADH.
In eukaryotes, this step takes place in the matrix, the innermost compartment of mitochondria. In prokaryotes, it happens in the cytoplasm. Overall, pyruvate oxidation converts pyruvate—a three-carbon molecule—into acetyl two-carbon molecule attached to Coenzyme A—producing an t, N, A, D, H, end text and releasing one carbon dioxide molecule in the process. Acetyl C, o, A, end text acts as fuel for the citric acid cycle in the next stage of cellular respiration.
Pyruvate oxidation steps
Pyruvate is produced by glycolysis in the cytoplasm, but pyruvate oxidation takes place in the mitochondrial matrix (in eukaryotes). So, before the chemical reactions can begin, pyruvate must enter the mitochondrion, crossing its inner membrane and arriving at the matrix.
In the matrix, pyruvate is modified in a series of steps:
More detailed diagram of the mechanism of pyruvate oxidation.
1. A carboxyl group is removed from pyruvate and released as carbon dioxide.
2. The two-carbon molecule from the first step is oxidized, and NAD+ accepts the electrons to form NADH.
3. The oxidized two-carbon molecule, an acetyl group, is attached to Coenzyme A to form acetyl CoA.
More detailed diagram of the mechanism of pyruvate oxidation.
A carboxyl group is removed from pyruvate and released as carbon dioxide.
The two-carbon molecule from the first step is oxidized, and NAD+ accepts the electrons to form NADH.
The oxidized two-carbon molecule, an acetyl group, is attached to Coenzyme A to form acetyl CoA.
Image credit: "Oxidation of pyruvate and the citric acid cycle: Figure 1" by OpenStax College, Biology, CC BY 3.0
Step 1. A carboxyl group is snipped off of pyruvate and released as a molecule of carbon dioxide, leaving behind a two-carbon molecule.
Step 2. The two-carbon molecule from step 1 is oxidized, and the electrons lost in the oxidation are picked up 2 \text{NADH}NADHstart text, N, A, D, H, end text are generated from \text{NAD}^+NAD
Step 3. The oxidized two-carbon molecule—an acetyl group, highlighted in green—is attached to Coenzyme A (\text{CoA}CoAstart text, C, o, A, end text), an organic molecule derived from vitamin B5, to form acetyl \text{CoA}CoAstart text, C, o, A, end text. Acetyl \text{CoA}CoAstart text, C, o, A, end text is sometimes called a carrier molecule, and its job here is to carry the acetyl group to the citric acid cycle.
The steps above are carried out by a large enzyme complex called the pyruvate dehydrogenase complex, which consists of three interconnected enzymes and includes over 60 subunits. At a couple of stages, the reaction intermediates actually form covalent bonds to the enzyme complex—or, more specifically, to its cofactors. The pyruvate dehydrogenase complex is an important target for regulation, as it controls the amount of acetyl \text{CoA}CoAstart text, C, o, A, end text fed into the citric acid cycle^{1,2,3}
1,2,3
start superscript, 1, comma, 2, comma, 3, end superscript.
If we consider the two pyruvates that enter from glycolysis (for each glucose molecule), we can summarize pyruvate oxidation as follows:
Two molecules of pyruvate are converted into two molecules of acetyl \text{CoA}CoAstart text, C, o, A, end text.
Two carbons are released as carbon dioxide—out of the six originally present in glucose.
2 \text{NADH}NADHstart text, N, A, D, H, end text are generated from \text{NAD}^+NAD
+
start text, N, A, D, end text, start superscript, plus, end superscript.
Why make acetyl \text{CoA}CoAstart text, C, o, A, end text? Acetyl \text{CoA}CoAstart text, C, o, A, end text serves as fuel for the citric acid cycle in the next stage of cellular respiration. The addition of \text{CoA}CoAstart text, C, o, A, end text helps activate the acetyl group, preparing it to undergo the necessary reactions to enter the citric acid cycle.
Explanation:
Related Questions
This perspective would study the pro's and con's of cramming for a test. *
Psychodynamic
Biolgical
Behavioral
Humanistic
Cognitive
Social Cultural
Evolutionary
Answers
Cognitive perspective would study the pro's and con's of cramming for a test, 4th option.
What is Cognitive perspective?In psychology, the cognitive perspective focuses on how the interactions of thinking, emotion, creativity, and problem-solving abilities influence how and why you think the way you do. Cognitive perspective would examine how people process, perceive, and think about information.
It would study how cramming affects the cognitive processes such as attention, memory and organization of information, and the impact of cramming on academic performance. This perspective would also consider the individual's beliefs and attitudes towards cramming, and how those influence the effectiveness of this study technique.
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A. A study has concluded that the toxic chemicals do not affect the urchins.
B. Urchin spines have no effect on their movement.
C. Urchins with short spines are extremely vulnerable to predators.
D. Urchins with long spines need to eat less kelp in order to survive.
Answers
A premise that could be used to back up the conclusion that urchins are undergoing natural selection would be that urchins with short spines are extremely vulnerable to predators. Option C.
Natural selectionUrchins with short spines are extremely vulnerable to predators.
This is because natural selection is a process in which organisms with certain traits are better able to survive and reproduce than those without those traits, and those traits are then passed on to future generations.
In this case, the vulnerability of urchins with short spines to predators means that they are less likely to survive and reproduce, while those with longer spines are more likely to do so.
Therefore, the trait of long spines is being naturally selected for in the urchin population. The other options do not provide evidence for natural selection.
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A 56-year-old man with an insufficient aortic valve is scheduled for surgery. Identify the chamber of the heart that the regurgitated blood most likely enter. Will the blood regurgitated primarily during systole or diastole?
Answers
Answerand Explanation:
The regurgitated blood will enter the left chamber of the heart. This is because as the patient's aotic valve is not working properly, it does not close properly, which allows a backflow of blood to occur towards the left chamber of the heart, as this chamber relaxes to be filled with blood pumped by the right camera.
As the regurgitated blood is a pumped blood, we can consider that it will happen, mainly during systole, which is the moment when the heart contracts to pump blood. This contraction will be made with the malfunction of the aortic valve which will cause blood regurgitation.
-6x-4x+2 need more than one answer
Answers
-6x-4x+2
=-6x+-4x+2
Combine Like Terms:
=-6x+-4x+4
=(-6x+-4x)+(2)
=-10x+2=Answer
Factor way answer for factor would be 2(-5x+1)
Hope this helps pls give me BRAINLIEST
State the relationship of living things and non living things.Living things:frog,water bug,beetle,water weeds,fish,insects.
Non living things: sunlight,water,soil.
Answers
Living things, including frogs, water bugs, beetles, water weeds, fish, and insects, exist in a dynamic relationship with non-living things such as sunlight, water, and soil.
This interplay between the living and non-living components is fundamental to the functioning of ecosystems. Sunlight, as a source of energy, drives photosynthesis in plants, enabling them to convert light into chemical energy. This energy is transferred through the food chain as organisms consume one another.
Water, a vital resource, sustains life by providing hydration, serving as a habitat, and facilitating nutrient transport. It influences the distribution and abundance of living organisms in aquatic ecosystems. Soil, an essential medium, supports plant growth by providing nutrients, anchoring roots, and housing a diverse community of microorganisms. Plants, in turn, provide food and shelter for animals, contributing to the intricate web of life.
Therefore, living things depend on non-living things for resources, while non-living things are influenced by and interact with living organisms, creating a delicate balance in ecosystems.
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Which of the following is an example of terrestrial plant
a.talahib
b.sea lettuce
c.duckweeds
d.algae
Answers
Answer:
algale is the right response
what describes an ion
Answers
29. Brain has one son, two grandchildren, and three great-grandchildren. This is example of
O exponential growth.
Odynamic growth.
O linear growth.
Obinary growth.
Answers
This is an example of exponential growth.
Exponential growth refers to a pattern of growth where the quantity or size of something increases at an accelerating rate over time.
In the given scenario, Brain has one son, two grandchildren, and three great-grandchildren.
Each generation is producing more offspring, resulting in a rapidly increasing number of descendants.
The progression from one generation to the next demonstrates exponential growth because each subsequent generation adds more individuals than the previous one.
As the generations continue, the number of descendants grows exponentially larger.
Exponential growth can be observed in various natural and human-made systems, such as population growth, compound interest, and the spread of infectious diseases.
In this case, the number of descendants in Brain's family tree follows an exponential pattern, as each new generation multiplies the number of offspring, leading to a significant growth rate.
Therefore, the example given aligns with the concept of exponential growth.
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Write a question to match this answer:
6 boxes of 12
Answers
Answer:
The chocolate lover bought 6 boxes of chocolate, each containing 12 truffles
Explanation:
mark me brainly
Answer:
The chocolate lover bought 6 boxes of chocolate, each containing 12 truffles. How many truffles did she buy?
a year after cats were let loose with the lizards, most of the surviving lizards should be which of the following?
Answers
The majority of the lizards who are still alive one year after cats were allowed to roam free with them should be little.
How does the fact that the lizards survived prove that natural selection is at work?The characteristic of having longer horns is advantageous for survival and will probably grow more prevalent in the population over time as the short-horned lizards die off. Longer horned lizards have a higher chance of surviving than shorter horned lizards do.
Which of the following is necessary for the process of natural selection-based evolution to take place?Reproduction, inheritance, variation in organism fitness, and variation in individual attributes across population members are the four conditions necessary for natural selection to occur. Whenever they are satisfied, natural selection takes place on its own.
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Question:-
A year after cats were let loose with the lizards, most of the surviving lizards should be which of the following?
If a 160lb individual has 28% body fat how much of their body weight is fat in lbs.?
Answers
Answer:
44.8 lb
Explanation:
160 lb * 0.28= 44.8 lb
Answer:
44.8 lb
Explanation:
160 x 0.28 = 44.8
What difference would it make if the southern sea otter (Core Case Study) became extinct primarily because of human activities?
Answers
Which of the following phenotypic classes reflect offspring that were generated as a result of a crossover event? Select all that apply. Select all that apply. wild type miniature wings garnet eyes miniature wings, garnet eyes SubmitRequest Answer Part B If 800 offspring were produced from the cross, in what numbers would you expect the following phenotypes? __wild type : __ miniature wings : __ garnet eyes : __ miniature wings, garnet eyes Enter your answer as the number of flies of each phenotype separated by a colon (example: 100:300:100:300).
Answers
Complete question:
The genes for miniature wings (m) and garnet eyes (g) are approximately 8 map units apart on chromosome 1 in Drosophila. Phenotypically wild-type females (m + g / mg +) were mated to miniature-winged males with garnet eyes.
Part A
Which of the following phenotypic classes reflect offspring that were generated as a result of a crossover event?
Select all that apply.
garnet eyes wild type miniature wings miniature wings, garnet eyesPart B
If 800 offspring were produced from the cross, in what numbers would you expect the following phenotypes?
__wild type : __ miniature wings : __ garnet eyes : __ miniature wings, garnet eyes
Enter your answer as the number of flies of each phenotype separated by a colon (example: 100:300:100:300).
Answer:
Part A:
2. wild type (m+g+/mg)
4. miniature wings, garnet eyes (mg/mg)
Part B:
32:368:368:32
Explanation:
When calculating the recombination frequency, we need to know that 1% of recombinations = 1 map unit = 1cm. And that the maximum recombination frequency is always 50%.
The map unit is the distance between the pair of genes for which every 100 meiotic products, only one results in a recombinant one.
So, en the exposed example:
PART A:
First, we need to Identify the gametes for the individual with the genotype m + g / mg + and label parental and recombinants
Gametes:
m+ g parental type m g+ parental type m+ g+ recombinant typem g recombinant typeKnowing that the only possible gamete type for the other parental is mg, then we can identify the phenotypic classes of the offspring that were generated as a result of a crossover event. These would be
2. wild type (m+g+/mg)
4. miniature wings, garnet eyes (mg/mg)
PART B:
We know that the total number of individuals equals 800.We also know that the distance between genes equals 8 MU.To calculate the number of the phenotypes among the offsprings,
8 map units = 8% of recombination in total
= 8/2% m+g+/mg + 8/2% mg/mg
= 4% + 4%
The recombination frequency of m+g+/mg equals 0.04
The recombination frequency of mg/mg equals 0.04.
100% - 8% = 92% of parental in total =
= 92/2% of m+ g/mg + 92/2% m g+/mg
= 46% + 46%
The recombination frequency of m+g/mg equals 0.46
The recombination frequency of mg+/mg equals 0.46.
The frequency for each ggenotype is:
F (m+g+/mg) = 0.04 F (mg/mg) = 0.04. F (m+g/mg) = 0.46 F (mg+/mg) = 0.46.The number of individuals with each genotype is:
m+g+/mg = 0.04 x 800 = 32mg/mg = 0.04 x 800 = 32m+g/mg = 0.46 x 800 = 368m g+/mg = 0.46 x 800 = 36832 wild type : 368 miniature wings : 368 garnet eyes : 32 miniature wings, garnet eyes
Think about the ecosystem that you live in. Write a response below describing it in complete sentences. Use each of these terms correctly in your description: community . population • ecosystem . individual .biotic abiotic
Answers
Answer:
Within an ecosystem, we can identify different levels of organization, such as individuals, populations, communities, and the overall ecosystem.
Explanation:
An individual refers to a single organism, such as a plant or an animal, living within the ecosystem. For example, a squirrel or a tree in your ecosystem would be considered individuals.
Populations are groups of individuals of the same species that live in the same area and interact with one another. For instance, if there is a group of squirrels living in your ecosystem, we can refer to them as a population of squirrels.
Communities include all the populations of different species living together in a specific area. In your ecosystem, there might be various populations, such as squirrels, birds, insects, and plants, interacting and coexisting with one another. Together, they form a community of different organisms.
The ecosystem encompasses both the living organisms (biotic factors) and the non-living components (abiotic factors) in a given area. The abiotic factors include things like sunlight, temperature, water, soil, and air. They influence the distribution and behavior of the organisms within the ecosystem. For example, the availability of sunlight and the temperature range in your ecosystem will affect the growth of plants and the activities of animals.
An ecosystem consists of the interactions between abiotic and biotic components in a specific area. Abiotic components include non-living factors like water, air, sunlight, temperature, and soil, while biotic components include living organisms like plants, animals, and microbes.
Populations of different species of plants, animals, and microbes make up the biotic components of an ecosystem. Each individual organism, whether a plant, animal, or microbe, interacts with its biotic and abiotic surroundings. The community of different species of organisms in an ecosystem all interact with each other to obtain food, shelter, or other necessary resources.
Some species form mutualistic relationships, while others may be in competition with each other for resources. Therefore, the ecosystem that we live in is the local community of living organisms interacting with the abiotic environment. These interactions between abiotic and biotic components of an ecosystem result in complex and dynamic ecosystems that can change over time.
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6.
Perform transcription by using the following piece of DNA.
Introns are underlined. Write the final strand of mRNA created.
TACCGIA CA ATG G GATC CouTube
Answers
Answer:
A-T-G-G-T G-T T-A-C C C-T-A-G
Explanation:
TACCGIA CA ATG G GATC
A-T-G-G-T G-T T-A-C C C-T-A-G
The "I" in the first few letters shouldn't probably be there since there are only 4 bases: A, T, C, and G.
Students have expressed a protein according to the BASIL protocols. They collected a pre-induction sample, a post-induction sample, and a sample following an affinity column purification. The samples were analyzed by SDS-PAGE followed by Coommassie Blue staining. Which lane corresponds to which sample?
Answers
Answer:
The first lane with the protein band is the post-induction sample, i.e., the lane with the sample of interest.
Explanation:
Antibody binding is specific and will only bind to whichever protein it has been produced for particularly, which in this case would be the post induction sample.
Can you help me with this question?
Answers
Answer:
Explanation:
I DONT KNOW THE ANSWER
Which organism is likely to carry out the process shown in the diagram?
CO, + H,O
Heat
www
Chloroplast
Mitochondrion
MASS
Respiration
OA. Barn owl
OB. Mushroom
C. Alligator
-0₂
Sugar
ATP
Answers
Answer:
C
Explanation:
took the test
This disorder is characterized by severe pain or throbbing on one side of the head and
can last up to a few days.
O Migraines
Epilepsy
O Dementia
O Quadriplegia
Answers
None of the others have that
in step 1 of protein synthesis describe two of the differences between the DNA template and its complementary RNA molecule
Answers
Answer:Proteins are synthesised (made) by the ribosomes using messenger RNA transcribed from DNA in the nucleus.
Explanation:
In every eukaryotic cell, such as plants and animals, there is DNA in the nucleus. The nucleus is an area surrounded by a membrane (nuclear membrane) which has some pores for materials to enter and exit, but these are controlled. The DNA is the cell's instructions, it is a series of code that can be transcribed and translated to make proteins. As in the image below DNA is a series of bases attached to a sugar phosphate backbone and bonded to another base pair to made a ladder, this ladder is twisted making the shaped called a double helix . There are four base pairs: Adenine (A), Thymine (T), Guanine (G), and Cytosine (C). The bases have bases they bond to making complementary base pairs , A and T bond together while the C and G bond together. on each strand of DNA one end is labelled 5' (read as 5 prime) and other end is the 3'. The 5' end base pairs of one strand are always attached to the 3' end of the complementary strand. Labelling the ends helps us to describe in which direction DNA is read.
E1 Biol 2281 usage of internet to analyze nuclei acid
Are these right?
Answers
We can see here that going through the answers provided in the image, we can deduce that they are correct though it will be proper to cross-check before submitting.
What is nuclei acid?Nucleic acids are biomolecules that play a crucial role in storing, transmitting, and expressing genetic information in living organisms. They are essential for the functioning and inheritance of genetic traits.
Both DNA and RNA are polymers made up of repeating units called nucleotides. Each nucleotide consists of three components: a sugar molecule (deoxyribose in DNA and ribose in RNA), a phosphate group, and a nitrogenous base.
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One out of every hundred people of the Hopi Indian tribe inherits the autosomal recessive disease Glucose-6-phosphate deficiency. How many people in the population of six million are carriers of this disease?
Answers
Following the Hardy-Weinberg equilibrium theory, in a population of six million people, 1,080,000 individuals are healthy and carriers (heterozygous) for the trait.
What is the Hardy Weinberg equilibrium?The Hardy-Weinberg equilibrium is a theory that states that populations in equilibrium get the same allelic and genotypic frequencies generation after generation.
These populations do not evolve. No evolutive forces are acting on them.
Assuming a diallelic gene, frequencies are
Allelic frequencies
p is the dominant allele frequencyq is the recessive allele frequencyGenotypic frequencies
p² is the frequency of the homozygous dominant individualsq² is the frequency of the homozygous recessive individuals2pq is the frequency of heterozygous individuals
In a population in equilibrium,
p + q = 1p² + 2pq + q² = 1We will consider these equations to solve the problem.
Glucose-6-phosphate deficiency is an autosomal recessive disease.One out of every hundred people of the affection. So this is the homozygous recessive genotypic frequency.Because we have the frequency of the homozygous recessive genotype, we can get the allelic frequency.
q² = 1/100 = 0.01q = √q² = √0.01 = 0.1Now, by clearing the equation p + q = 1, we can get the dominant allele frequency. And then, the frequency of the homozygous dominant genotype.
p = 1 - 0.1 = 0.9p² = 0.9² = 0.81Finally, we can get the frequency of the heterozygous genotype
2pq = 2 * 0.9 * 0.1 = 0.18The number of individuals with each phenotype in the population is,
0.81 * 6,000,000 = 4,860,000 individuals are healthy not carrier0.18 * 6,000,000 = 1,080,000 individuals are healthy carriers. 0.01 * 6,000,000 = 60,000 individuals have the disease.In a population of six million people, 1,080,000 individuals are healthy and carriers (heterozygous) for the trait.
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Persistence length for a cytoskeletal filament is the minimum filament length at which random thermal fluctuations are likely to cause it to bend. Which of the following comparisons are true, considering the persistence lengths of (a) an actin filament, (b) a bundle of cross-linked actin filaments, and (c) a microtubule
Answers
The provided question is incomplete as the group of choices were not given, however, the choices as follows:
A. (a > b) and (a > c)
B. (a > b) and (a < c)
C. (a < b) and (a > c)
D. (a < b) and (a < c)
E. (a > b) and (b > c)
Answer:
The correct answer is - D. (a < b) and (a < c)
Explanation:
The persistent length of actin filaments, a bundle of cross-linked actin filaments and microtubules can be compared on their typical persistent length.
The persistent length of actin filaments is 15 µm. And due to the crosslink with one another to make a bundle their length reduced As the actin filaments cross-link each other their persistent length will be reduced. however, it is still longer than single actin filament around 20 nm.
The persistent length of microtubules is 6mm.
So b > a, a < b and a < c.
So fourth option a < c and b > a is the correct answer.
#5 In phagocytosis, the membrane
vesicle eventually fuses with the cell's
A. lysosome
B. nucleus
C. Golgi bodies
D. mitochondria
Answers
Answer: A. Lysosome
Explanation:
Phagocytosis is a process of endocytosis. Basically bringing stuff into the cell. Lysosomes are the organelles that break stuff down in a cell.
The material is entrapped/engulfed by the cell with its plasma membrane and inwardly cleaves to create a vacuole in the cell. Lysosomes seek out and break down this material so that its constituents can be utilized.
A hetero gous ail yellow plant is crossed with a homorygous short green. Show the genoype od phenotypes of the Fi offspring and the probability of each.
Answers
The genotype of the F1 offspring is YyTt, and the phenotype is yellow and tall. The probability of each phenotype in the F1 offspring is as follows yellow and tall is 1/2 or 50%, green and short is 1/2 or 50%.
The genotype and phenotype of the F1 offspringGenotype refers to the genetic makeup or combination of alleles present in an organism, while phenotype refers to the observable traits or characteristics expressed by an organism. Genotype represents the genes an organism carries, while phenotype represents the physical or observable features resulting from the interaction between genotype and the environment.
When a heterozygous ail yellow plant (genotype Yy) is crossed with a homozygous short green plant (genotype tt), the F1 offspring will have the genotype YyTt and the phenotype of yellow and tall. The probability of this phenotype occurring in the F1 generation is 50%. Additionally, there is a 50% probability of the F1 offspring having the phenotype of green and short.
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Briefly explain why DNA replication is semiconservative and discontinuous. Make sure to name the proteins involved in the synthesis of DNA, synthesis of RNA, the removal of RNA and the sealing of the nicks.
Answers
Answer:
DNA replication is semiconservative: each new strand has a strand of the parent DNA and another strand of new synthesis. DNA synthesis is discontinuous, since one of the strands, the lagging strand, replicates intermittently within the replication bubble, while the leading strand replicates continuously. For replication, apart from DNA polymerases, requires the action of enzymes that synthesize the primer, DNA ligases, enzymes that are responsible for joining pieces formed of chains, etc. RNA is synthesized through a process called transcription in making a complementary copy of a piece of DNA.
Explanation:
Replication is discontinuous since one of the strands, the lagging strand, replicates intermittently within the replication bubble, while the leading strand replicates continuously. It is bi-directional, because the strand is copied in both directions, and depending on the point of origin and the way it is viewed, a strand can be either discontinuous or continuous. Finally, replication is said to be semi-conservative since the original strands are part of each of the double strands resulting from replication.The unwinding of the DNA and the maintenance of the two separate strands in the replication fork so that the two new strands can be synthesized requires the action of proteins: Gyrase allows the release of the torque tension of the DNA double strand while the helicase separates the strands, allowing the action of DNA polymerases, which are responsible for the inclusion of new nucleotides as the DNA model strand is copied. Finally, the primases allow the incorporation of leader sequences at the beginning of the replication forks, allowing the action of DNA polymerases. DNA ligases seal cuts or nicks (broken phosphodiester bonds). DNA ligase also splices the ends of DNA into duplex regions. Transcription consists of the synthesis of RNA taking DNA as a template and means the passage of the information contained in DNA to RNA. The transfer of information from DNA to RNA is carried out following the complementary rules of the nitrogenous bases. When the entire strand has been copied, at the end of the process, the RNA chain is free and the DNA is closed again, by pairing of its complementary chains.
The border between two tectonic plates lies below the state of California. The interaction of these two plates results in which of the following?
a.) Mild Climate
b.) Rich, fertile soil
c.) Frequent earthquakes
d.) Occasional wildfires
Answers
Answer:
Frequent Earthquakes
Explanation:
californication
The interaction of these two plates results in frequent earthquakes.
What is an earthquake?An earthquake is a sudden and intense shaking of the ground caused by the movement of tectonic plates in the Earth's crust. Earthquakes can result in damage to buildings, roads, and other structures, as well as cause landslides, tsunamis, and other natural disasters.
They are caused by the release of energy from tectonic plates as they shift and interact with one another. Earthquakes are measured on the Richter scale, which is used to determine their magnitude and potential for causing damage. The frequency and intensity of earthquakes can vary depending on location, with some areas experiencing more frequent or severe earthquakes than others.
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In the ATP Cycle energy is transferred from sunlight or glucose to build _____ from ADP and _____________.
Answers
Answer:
1. ATP
2. Inorganic phosphate (Pi)
Explanation:
Adenosine triphosphate (ATP) is the usable form of energy by the cells of living organisms. During the light reaction of photosynthesis, energy from sunlight (solar energy) is converted to ATP in order to power the Calvin cycle stage. Also during cellular respiration, energy stored in glucose molecule is transferred to ATP.
The processes by which this synthesis of ATP from ADP (Adenosine diphosphate) occurs is called oxidative phosphorylation and chemiosmosis. Energy is stored in the ATP molecule when ADP combines with inorganic phosphate (Pi).
Small, nonpolar hydrophobic molecules such as fatty acids
a. pass readily through a
membrane's lipid bilayer
b. diffuse very slowly through the lipid bilayer
c require special channels to enter a cell
d. are actively transported across cell membranes
c. must enter the cell via endocytosis
Answers
Answer:
a. pass readily through a membrane's lipid bilayer
Explanation:
Small, nonpolar hydrophobic molecules such as fatty acids pass readily through a membrane's lipid bilayer.
Small, nonpolar hydrophobic molecules such as fatty acids pass readily through a membrane's lipid bilayer. Therefore, the correct option is option A.
The hydrophilic (water-loving) heads of the two layers of phospholipids in the lipid bilayer face outward towards the aquatic surroundings, while the hydrophobic (water-hating) tails point inward towards one another. The hydrophobic tails of phospholipids share characteristics with nonpolar hydrophobic compounds like fatty acids. These tiny molecules can easily dissolve in the hydrophobic area of the lipid bilayer because of their hydrophobic nature.
Therefore, the correct option is option A.
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Which of the following is an example of the Doppler effect? (2 points)
a
A water bug on the surface of a pond is producing small ripples in the water. The bug is not moving.
b
A police car with its siren on is driving towards you, and you perceive the pitch of the siren to increase.
c
Scientists measure the speed of light and find that it is constant.
d
The gravitational pull of two objects on each other
Answers
Answer:
this would be b
Explanation:
Answer:
AMONG
Explanation:
US
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