Biology High School
Answers
Answer 1
The rate of mtDNA percent divergence per million years is; 1.47% per million years, the placement of gibbons as the out-group on the cladogram because they are the most distantly related species in the dataset based on the mtDNA sequence divergence, and the species indicated by the node within the square are Gorilla, Orangutan, Gibbon, Chimpanzee, and Human.
To calculate the rate of mtDNA percent divergence per million years between humans and their most closely related species in the dataset, we can use the estimated divergence time of 7 million years ago and the percent divergence in mtDNA sequences between humans and their closest relative, chimpanzees.
The percent divergence between humans and chimpanzees is 10.3, and the estimated divergence time is 7 million years ago. Therefore, the rate of mtDNA percent divergence per million years will be;
(10.3% / 7 million years) * 1 million years
= 1.47% per million years
Here's the cladogram;
/--- Gorilla
|
| /--- Orangutan
| |
/-----| | /--- Gibbon
| | | |
| | |-----|
| | | /--- Chimpanzee
| | |-----|
| | \--- Human
| |
| | /--- Gibbon (Out-group)
| |-----|
| \--- Chimpanzee/Human ancestor
|
| /--- Orangutan
\----------|
\--- Gibbon (Out-group)
Gibbons are placed as the out-group on the cladogram because they are the most distantly related species in the dataset based on the mtDNA sequence divergence. They have the highest percent divergence with all other species, indicating that they diverged from the common ancestor of the other species earlier than any of the other species diverged from each other.
he species descended from the species indicated by the node within the square are Gorilla, Orangutan, Gibbon, Chimpanzee, and Human. These species form a monophyletic group, meaning they share a common ancestor and all of its descendants. We can circle these species on the cladogram as follows;
/--- Gorilla
|
| /--- Orangutan
| |
/-----| | /--- Gibbon
| | | |
| | |-----|
| | | /--- Chimpanzee
| | |-----|
| | | \--- Human
| | |
| | /--- Gibbon (Out-group)
| |-----|
| \--- Chimpanzee/Human ancestor
|
| /--- Orangutan
\----------|
\--- Gibbon (Out-group)
We can see that this monophyletic group is composed of all the hominids in the dataset, and it includes the common ancestor of chimpanzees and humans.
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Related Questions
a student in a biology class examines a slide of an unfamiliar cell and describes its major parts and how they interrelate. what element in bloom's taxonomy is most directly exemplified?
Answers
The element in Bloom's Taxonomy that is most directly exemplified in this scenario is "analysis". Analysis involves breaking down complex ideas or concepts into smaller parts and understanding how they relate to one another.
In this scenario, the student is examining the cell and describing its major parts, which shows that they are analyzing the cell and its components. They are also explaining how these parts interrelate, which is an important aspect of analysis. By doing this, the student is demonstrating a deeper level of understanding of the cell and its structure.
Analysis is an important skill in biology as it allows students to understand complex biological concepts and phenomena. By breaking down complex ideas into smaller parts, students can better understand the relationships between different parts of the cell, how they function, and how they contribute to the overall structure of the cell. This is an important part of developing a deep understanding of biology and the mechanisms that underlie life.
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how is grass planted in mine reclamation? woven mats that include embedded grass seed are laid over the landscape. aerial drops of grass seed mixtures are laid over previously mined sites. an aqueous mixture of seed and chemicals are sprayed on the thin soil. it is not planted, rather the land will go through natural senescence. it is not planted; grass will not grow on reclaimed mines.
Answers
The following steps are had in reclamation: the shaping of the land; the case of dirt or an endorsed reserve on the studied region; reseeding with local verdure, crops as well as trees; as oK as years of exact monitoring to guarantee success.
The process of restoring mined land to an ecologically and economically viable state is known as mine reclamation.
Reclamation, the process of closing a mine and recontouring, revegetating, and restoring the water and land values, is the final stage in most mine operations. The best opportunity to start the recovery cycle of a mine is before the first unearthings are started.
Grilling, accelerated runoff, low vegetation cover, erosion, and poor soil structure are all issues. A portion of these issues are because of bad quality design and unfortunate land farming however they are amplified by regular cycles.
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plsssss help I'll give brainliest!!!! ocean currents traveling from the equator toward the polar zones carry _____ water, which helps to _____air masses at the poes
Answers
Ocean currents traveling from equator toward the polar zones carry warm water, which helps to warm air masses at the poles.
What are the things that ocean currents carrying from the equator to the polar regions carry?
The movement of warm water and precipitation from the equator to the poles and the return of cold water from the poles to the tropics are both accomplished by ocean currents, which function somewhat like a conveyor belt.
Where the equator is concerned, what causes water to rise?
Upwelling in coastal areas is mostly caused by the Coriolis effect, which is known as it. Additionally, the open water close to the equator experiences upwelling due to the Coriolis force. Surface water is blown both north and south by trade winds at the equator, which permits upwelling of deeper water.
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pathogenic streptococci of the upper respiratory tract (such as streptococcus pyogenes) are distinguished from non-pathogenic streptococci by
Answers
Pathogenic streptococci of the upper respiratory tract (such as streptococcus pyogenes) are distinguished from non-pathogenic streptococci by beta-hemolytic activity. Option A is the correct answer.
Pathogenic streptococci of the upper respiratory tract, such as Streptococcus pyogenes, are distinguished from non-pathogenic streptococci by their beta-hemolytic activity.
This means that they are able to lyse red blood cells and produce a clear zone around the colony on blood agar plates.
Non-pathogenic streptococci, on the other hand, may display alpha-hemolytic activity, which results in a greenish discoloration around the colony, or no hemolytic activity at all.
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The question is -
Pathogenic streptococci of the upper respiratory tract (such as Streptococcus pyogenes) are distinguished from non-pathogenic streptococci by
a. beta-hemolytic activity.
b. no hemolytic activity.
c. alpha hemolytic activity.
d. the presence of a lysogenic phage.
e. the absence of a capsule.
Pathogenic streptococci of the upper respiratory tract, such as Streptococcus pyogenes, are distinguished from non-pathogenic streptococci by their ability to cause disease.
This is due to their production of specific virulence factors and content loaded pathogenicity islands that allow them to colonize and invade the host's tissues. In contrast, non-pathogenic streptococci do not possess these virulence factors and do not cause disease. Therefore, it is important to identify and differentiate these two groups of streptococci in order to effectively diagnose and treat infections caused by pathogenic streptococci.
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Which two of the following sentences from the article include CENTRAL ideas of the article?
1. When a top predator is removed, everything else will be affected in some way, like a falling row of dominoes. 2. With no otters to eat them, populations of sea urchins and other marine invertebrates exploded. 3. Scientists have learned more about trophic cascades, and they have started to reintroduce some top predators back into their ecosystems. 4. There, sea urchins are decreasing, kelp forests are expanding and marine biodiversity is increasing
Answers
Central idea of this article is a concept of trophic cascades, which refers to the effects that removing or adding top predators can have on an ecosystem. With no otters to eat them, populations of sea urchins and other marine invertebrates exploded and Scientists have learned more about trophic cascades, and they have started to reintroduce some top predators back into their ecosystems.
The correct option is 2 and 3
With no otters to eat them, populations of sea urchins and other marine invertebrates exploded. highlights the consequences of removing a top predator, as the population of sea urchins and other invertebrates exploded without otters to keep them in check.
Statement 3 highlights the importance of understanding the role that top predators play in maintaining balance in an ecosystem, and the potential benefits of reintroducing them in areas where they have been removed.
Hence 2 and 3 are the correct option
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The multiple causes for hypoxia include: (Select all that apply.) a. extreme fright. b. aspirated vomit. c. pulmonary fibrosis.
Answers
The multiple causes for hypoxia include aspirated vomit and pulmonary fibrosis. The correct options are b,c.
Extreme fright is not a direct cause of hypoxia. Hypoxia refers to a condition where there is a lack of oxygen supply to the body's tissues.
Aspirated vomit is one cause of hypoxia. When vomit is inhaled into the lungs, it can obstruct the airways and prevent proper oxygenation of the blood. This leads to a lack of oxygen being delivered to the body's tissues, resulting in hypoxia.
Pulmonary fibrosis is another cause of hypoxia. It is a lung disease where the lung tissue becomes thickened, stiff, and scarred. This scarring prevents the lungs from properly expanding and contracting, limiting the transfer of oxygen from the lungs to the bloodstream.
Consequently, there is a reduced supply of oxygen to the body's tissues, causing hypoxia.
In summary, aspirated vomit and pulmonary fibrosis are two causes of hypoxia, as they both lead to a lack of oxygen being delivered to the body's tissues. Extreme fright, on the other hand, does not directly cause hypoxia.
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A hurricane floods the mouth of a river that empties into the ocean. The inflow of saltwater increases the salt content of the river water. Freshwater fish must stay in low salt environments. What will be the most likely response of the fish?
Answers
The most likely response of freshwater fish in a river experiencing an increase in salt content as a result of saltwater inflow from a hurricane is to move upstream.
What is the reaction of freshwater fish?
The most likely response of the freshwater fish in the river that is now experiencing an increase in salt content due to the inflow of saltwater from a hurricane is to move upstream or into smaller tributaries to find a low salt environment that is suitable for their survival.
This is because freshwater fish are adapted to living in environments with low salt concentrations, and exposure to high salt concentrations can lead to dehydration, loss of electrolyte balance, and ultimately death. Therefore, in order to survive, the fish would need to move to a location with a lower salt content.
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Class I viruses, double-stranded (ds)DNA viruses, usually utilize the following polymerases for (i) mRNA synthesis and (ii) DNA replication
A. (i) viral RNA-dependent RNA polymerase and (ii) viral DNA-dependent DNA polymerase
B. (i) viral DNA-dependent RNA polymerase and (ii) viral DNA-dependent DNA polymerase
C. (i) viral RNA-dependent RNA polymerase and (ii) host cell DNA-dependent DNA polymerase
D. (i) host cell DNA-dependent RNA polymerase and (ii) host cell DNA-dependent DNA polymerase
E. (i) host cell RNA-dependent RNA polymerase and (ii) host cell DNA-dependent DNA polymerase
Answers
Class I viruses, double-stranded (ds)DNA viruses, usually utilize (i) viral DNA-dependent RNA polymerase and (ii) viral DNA-dependent DNA polymerase for (i) mRNA synthesis and (ii) DNA replication. The correct answer is B.
Here's a step-by-step explanation:
1. Class I viruses are double-stranded DNA (dsDNA) viruses, meaning they have a DNA genome.
2. For mRNA synthesis, these viruses use a viral DNA-dependent RNA polymerase. This enzyme synthesizes RNA using the viral DNA as a template, allowing the production of viral mRNA for protein synthesis.
3. For DNA replication, these viruses use a viral DNA-dependent DNA polymerase. This enzyme is responsible for replicating the viral DNA, ensuring the production of new viral genomes for the assembly of new virus particles.
So, Class I viruses, or dsDNA viruses, utilize viral DNA-dependent RNA polymerase for mRNA synthesis and viral DNA-dependent DNA polymerase for DNA replication.
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starting at the apex of the heart and moving superiorly, what is the correct order in which you would encounter the four anatomical structures below? ventricle, papillary muscle, chordae tendineae, valves valves, chordae tendineae, papillary muscle, ventricle papillary muscle, chordae tendineae, ventricle, valves chordae tendineae, valves, ventricle, papillary muscle
Answers
Chordae tendineae, valves, ventricle, papillary muscle. Starting at the apex of the heart and moving superiorly, the first structure encountered is the chordae tendineae. So, the correct answer is option C.
These fibrous strands attach the papillary muscles to the heart's valves. The valves, specifically the semilunar valves, which are found in the outflow canal of the heart, are the following structure that is encountered.
The blood's passage through the ventricles is regulated by these valves. The heart's ventricle, which pumps blood throughout the body, is the third component that is visible.
The papillary muscle, a group of tiny muscles in the heart wall that aid in securing the valves in place, is the final component that is encountered.
Together, these elements contribute to ensuring that blood moves through the heart in the right way.
Complete Question:
Starting at the apex of the heart and moving superiorly, what is the correct order in which you would encounter the four anatomical structures below?
A) Valves, chordae tendineae, papillary muscle, ventricle
B) Papillary muscle, chordae tendineae, ventricle, valves
C) Chordae tendineae, valves, ventricle, papillary muscle
D) Ventricle, papillary muscle, chordae tendineae, valves
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the accumulation of misfolded proteins in the er is a potentially lethal situation and thus causes the triggering of what process?
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'ER stress' can be caused by an accumulation of misfolded proteins in the ER, which can compromise ER function. The UPR is one of the unique signalling pathways that the ER activates in response. The IRE1, PERK, and ATF6 proximal sensors work in concert to modulate the UPR.
Misfolded protein buildup and aggregation in the endoplasmic reticulum (ER) hinder normal cellular function and can be toxic, resulting in cell death. Long-term production of improperly folded proteins causes ER stress, which sets off a series of events known as the unfolded protein response (UPR). A common cellular occurrence is protein misfolding, which can be brought on by a variety of factors including genetic mutations, translational mistakes, aberrant protein modifications, heat or oxidative stress, and incomplete complex forms.
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once a neurotransmitter binds to a post-synaptic receptor, it remains bound until an antagonist chemical replaces it. true or false?
Answers
Answer: tru
Explanation:
Once a neurotransmitter binds to a post-synaptic receptor, it remains bound until an antagonist chemical replaces it. The statement is false.
When a neurotransmitter binds to a post-synaptic receptor, it does not necessarily remain bound until an antagonist chemical replaces it. Neurotransmitters can also be removed through reuptake, where the presynaptic neuron takes them back, or through enzyme degradation, where enzymes break down the neurotransmitter into inactive metabolites. These processes help regulate neurotransmitter levels and prevent overstimulation of the post-synaptic receptor.
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a mutation in a gene inserts one base. the first six codons read by the ribosome in the rna from the original and mutant genes are shown below. in the original gene, the second codon encodes the amino acid asn. what is true in the mutant gene?
Answers
The second codon in the mutant gene continues to code for Asn, but all subsequent amino acids will differ from those in the original gene.
The second codon in the mutant gene still codes for Asn, but the following codon will halt translation.
The second codon in the mutant gene continues to code for Asn, but the subsequent codon will terminate transcription.
The initial codon in the mutant gene still codes for Met, but every subsequent amino acid will change from that in the original gene.
The second codon in the mutant gene still codes for Asn, but the following codon will halt translation.
Mutations can happen when cells copy their genetic material incorrectly. It's possible for mutations to be pointless
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Refer to the Biochemistry in Focus section of your text for this chapter to answer this question. A mutation in hyperpolarization-activated cyclic nucleotide-gated, or HCN, channels results in decreased heart rate. These mutant HCN channels require greater hyperpolarization of membrane than the wild-type to open, thus resulting in slower depolarization at resting potential. Choose the statement that describes the cause of slower depolarization of the mutant HCN channels at resting potential. At resting potential, fewer mutant HCN channels are open. Therefore, fewer sodium ions flow into the cell, resulting in slower depolarization. At resting potential, fewer mutant HCN channels are open. Therefore, fewer potassium ions flow into the cell, resulting in slower depolarization At resting potential, more mutant HCN channels are open. Therefore, more sodium ions flow out of the cell, resulting in slower depolarization. At resting potential, more mutant HCN channels are open. Therefore, more potassium ions flow out of the cell, resulting in slower depolarization. Consider a uniport system where a carrier protein transports an uncharged substance A across a cell membrane. Suppose that at a certain ratio of (Almade to lovstúde, the AG for the transport of substance A from outside the cell to the inside. Aostide Aimide, is 13.7 kJ/mol at 25°C. What is the ratio of the concentration of substance A inside the cell to the concentration outside? Alimide [Alaide Choose the true statement about the transport of A under the conditions described. Movement of Aimide to Acutube will be spontaneous. Because AG is positive, the ratio Ide/Atske must be less than one Increasing Aloue will cause AG for movement of Acto Ame to become a larger positive number. Decreasing the concentration of the uniport protein in the membrane will cause AG to become a smaller positive number
Answers
The correct statement describing the cause of slower depolarization of the mutant HCN channels at resting potential is:
At resting potential, fewer mutant HCN channels are open. Therefore, fewer sodium ions flow into the cell, resulting in slower depolarization.
The slower depolarization of the mutant HCN channels is due to the fact that these channels require greater hyperpolarization of the membrane than the wild-type channels to open. At resting potential, fewer mutant HCN channels are open, which means that fewer positively charged sodium ions flow into the cell, resulting in slower depolarization.
Regarding the uniport system, the correct statement is:
Increasing Aloue will cause AG for movement of Acto Ame to become a larger positive number.
The ratio of the concentration of substance A inside the cell to the concentration outside can be calculated using the equation:
ΔG = -RT ln([A]inside/[A]outside)
where ΔG is the change in Gibbs free energy, R is the gas constant, T is the temperature, and [A]inside and [A]outside are the concentrations of substance A inside and outside the cell, respectively.
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In regards to the first question about the mutation in HCN channels, the cause of slower depolarization is that fewer mutant HCN channels are open at resting potential, resulting in a slower flow of sodium ions into the cell.
Moving on to the second question about the uniport system, we can use the equation ΔG = -RTln(ide/take) to solve for the ratio of the concentration of substance A inside the cell to the concentration outside. Plugging in the given values, we get -13.7 kJ/mol = -(8.314 J/mol*K)(298 K)ln(ide/take). Solving for the ratio, we get ide/take = 3.8.
The true statement about the transport of substance A under the described conditions is that increasing the concentration of Aloe will cause AG for the movement of Acto Ame to become a larger positive number. This is because the ΔG value is directly proportional to the concentration difference of the transported substance.
The movement of substance A from outside to inside the cell will not be spontaneous. Increasing the concentration of substance A outside the cell will cause ΔG to become a larger positive number, further supporting the non-spontaneous nature of the transport process.
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which of the following mechanisms prevents osmotic swelling in plant cells? choose one: a. turgor pressure b. the expulsion of water from contractile vacuoles c. the collection of water in contractile vacuoles d. tough cell walls e. the activity of na pumps
Answers
Osmotic pressure is the pressure that develops in a solution as a result of water molecules moving from a region of high concentration to a region of low concentration. Therefore, the correct option D i.e., tough cell walls.
Plant cells are unique in that they have a cell wall made up of cellulose fibers. The cell wall provides structural support to the cell & is responsible for preventing the cell from bursting due to osmotic pressure.
Water molecules enter plant cells by osmosis when they are put in a hypotonic solution (a solution with a lower concentration of solutes than the cell), resulting in cell swelling.
However, the cell wall stops the cell from bursting by applying turgor pressure, an equal and opposite force, to the cell membrane. Turgor pressure is the force the contents of the cell apply on the cell wall.
Because the cell wall is strong & stiff, it opposes the cell's growth and applies pressure to the inside of the cell, keeping it from growing excessively.
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if an individual with the genotype aa produces gametes, what is the probability that a gamete will receive the a allele?
Answers
The probability that a gamete will receive the a allele in an individual with the genotype aa is 100%
Probability of receiving allele:
The probability would be 100% because, during meiosis, alleles segregate independently due to independent assortment. This means that each gamete has an equal chance of receiving either the maternal or paternal allele, and since the individual only has the a allele, all of their gametes will carry that allele.
What is Meiosis?
Meiosis is the process where a diploid cell (with two sets of chromosomes) undergoes two divisions, ultimately producing haploid gametes (with one set of chromosomes). Independent assortment occurs during meiosis, where chromosomes are randomly distributed to the daughter cells, ensuring genetic diversity. In the case of an individual with the genotype aa, both chromosomes carry the a allele. During meiosis, each gamete receives one chromosome from each pair, resulting in one a allele per gamete. Since both chromosomes carry the a allele, the probability that a gamete will receive the a allele is 100%.
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Scalp reduction involves removing the bald area from the scalp and Pulling together the surrounding scalp areas with hair growth (true or false)
Answers
Scalp reduction is a surgical procedure used to treat hair loss or baldness. It involves removing a portion of the bald scalp and then pulling together the remaining scalp with hair growth to reduce the size of the bald area. Given statement is False.
The removed scalp is typically discarded. This procedure is used in certain cases where there is a well-defined area of baldness and can be performed in combination with other hair restoration techniques. However, it's important to note that scalp reduction is not always the most appropriate or effective treatment for all cases of hair loss, and the decision to undergo this procedure should be made in consultation with a qualified medical professional.
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The statement " Scalp reduction involves removing the bald area from the scalp and Pulling together the surrounding scalp areas with hair growth" is true.
Scalp reduction is a surgical procedure that involves removing the bald area of the scalp and pulling together the surrounding scalp areas with hair growth.
The goal of the procedure is to decrease the size of the bald spot and provide a more uniform appearance to the scalp.
The procedure is typically performed on individuals with male or female pattern baldness who have a stable area of hair growth around the balding area.
The procedure is usually done under local anesthesia and can take several hours to complete.
After the surgery, the scalp may be tender, and there may be some swelling and discomfort.
Patients are typically advised to avoid strenuous physical activity for several weeks and to take antibiotics to prevent infection.
Overall, scalp reduction can be an effective option for individuals with balding areas on the scalp who desire a more uniform appearance. Therefore, the statement is true.
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what effect do double bonds have on phospholipid hydrocarbon tails and on the fluidity of the membrane? choose one: double bonds increase the ability of hydrocarbon tails to pack together into a rigid mass, which makes the bilayer less fluid. double bonds decrease the ability of hydrocarbon tails to pack together into a rigid mass, which makes the bilayer less fluid. double bonds increase the ability of hydrocarbon tails to pack together into a rigid mass, which makes the bilayer more fluid. double bonds decrease the ability of hydrocarbon tails to pack together, which makes the bilayer more fluid. double bonds have little effect on membrane fluidity.
Answers
Option d is correct. The effect phospholipid hydrocarbon tails and on the fluidity of the membrane is double bonds decrease the ability of hydrocarbon tails to pack together, which makes the bilayer more fluid.
The hydrocarbon chain develops a kink when double bonds are present, which prevents the tails from packing tightly. As a result, the bilayer's phospholipids are more mobile and no solid, organized structure may form.
The fluidity of the membrane increases along with the number of double bonds in the hydrocarbon tails. This is so that phospholipid molecules may move about more freely due to the extra space that the kinks in the chains create.
Overall, a key element in influencing the fluidity of the membrane is the existence of double bonds in the hydrocarbon tails of phospholipids. This is significant because numerous cellular functions, including membrane fusion and signal transduction, depend on membrane fluidity.
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Complete question
What effect do double bonds have on phospholipid hydrocarbon tails and on the fluidity of the membrane?
a. Double bonds increase the ability of hydrocarbon tails to pack together into a rigid mass, which makes the bilayer less fluid.
b.Double bonds decrease the ability of hydrocarbon tails to pack together into a rigid mass, which makes the bilayer less fluid.
c. Double bonds increase the ability of hydrocarbon tails to pack together into a rigid mass, which makes the bilayer more fluid.
d. double bonds decrease the ability of hydrocarbon tails to pack together, which makes the bilayer more fluid.
e. Double bonds have little effect on membrane fluidity.
after learning about organisms and adaptations, a teacher asks students to describe what they learned about how a particular organism might survive during a drought. she finds that many of her students cannot think of any ways the organism might adapt. she decides she will extend the lesson to the next class period. which phase of the 5e model is the teacher implementing when she asks for descriptions of adaptation?
Answers
The Engage, Explore, Explain, Elaborate, and Evaluate phases of the 5E model of instruction are a frequently used instructional framework in scientific education for the organisms.
The instructor in this instance is using the 5E model's "Explore" phase. The exploration of ideas and concepts associated with the study topic is encouraged at this period. In doing so, the instructor is enticing the pupils to consider critically the adaptations that the creature may take to live in a drought.
The instructor chooses to continue the lesson into the following class time because many of the pupils are unable to come up with any ways the organism may adapt. Students can have a greater knowledge of the idea through this method.
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PLEASE ANSWER (ill give brainliest)
Which feature of the model represents a transfer of energy?
A. The sticks connecting the balls that represent atoms
B. The balls that represent atoms
C. The arrow between the reactants and the products
D. The plus sign (+) between sugar and oxygen
Answers
The feature of the model that represents a transfer of energy is option C, the arrow between the reactants and the products. This arrow represents the direction of the chemical reaction.
What are reactants ?
Reactants are the starting materials that are involved in a chemical reaction and are transformed into new products. In a chemical equation, the reactants are listed on the left side, and the products are listed on the right side.the bonds between the atoms in the reactants are broken, and new bonds are formed to create the products. The amount of reactants and products in a chemical reaction is governed by the law of conservation of mass, which states that the total mass of the reactants must equal the total mass of the products.
What is a mass ?
Mass is a fundamental property of matter that describes the amount of matter in an object. It is a scalar quantity, which means that it only has magnitude and no direction. The standard unit of mass in the International System of Units (SI) is the kilogram (kg).
The mass of an object is determined by the number of atoms or molecules it contains and their respective masses. It is different from weight, which is the force exerted on an object due to gravity and depends on the object's mass and the gravitational field it.
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Which statements about genetic disorders are correct? Discuss how to modify any incorrect statement to make it a correct statement with a classmate.
A.
Genetic disorders can be spread by sexual contact.
B.
Environmental factors can affect genetic disorders.
C.
Researchers have found all possible connections between diseases and genetics.
D.
Genetic disorders are not always caused by gene mutations.
Answers
Answer:
B. Environmental factors can affect genetic disorders.
D. Genetic disorders are not always caused by gene mutations.
Explanation:
four students were asked to simplify a phylogenetic tree of several food plants by using one branch to represent the three different fruit trees (papaya, orange, and peach). which phylogenetic tree is correct?
Answers
The simplified tree could be regarded as correct if it appropriately depicts the connections between the three fruit trees (papaya, orange, and peach) and the food plants.
What is phylogenetic tree?
A phylogenetic tree is a branching diagram that shows the relationships between various species, groups of animals, or genes across time. Lineages of organisms are represented by the branches of the tree, while the nodes show when a single lineage breaks into two or more. Molecular sequences, physical attributes, or behavioral factors can all be used as the foundation for phylogenetic trees.
How effectively a phylogenetic tree portrays the connections between the creatures in issue determines how accurate it is. However, the simplified tree could be erroneous if it omits crucial details or displays the connections between the creatures incorrectly. Before simplifying a phylogenetic tree, it is crucial to carefully analyze the data and connections it is representing.
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a particular triplet of bases in the coding strand of a gene is ggg. if an mrna is made from this gene, the anticodon on the trna that will bind the codon on the mrna is group of answer choices
Answers
Hi! I'd be happy to help you with your question. To summarize, we need to find the anticodon on the tRNA that will bind to the codon on the mRNA transcribed from the coding strand of a gene with the triplet GGG.
Step 1: Determine the template strand sequence
Since the coding strand has the sequence GGG, the template strand will have the complementary sequence. The template strand sequence is CCC.
Step 2: Transcribe the mRNA
During transcription, the mRNA is created from the template strand. The mRNA sequence will be complementary to the template strand. So, the mRNA codon will be GGG (as A pairs with T, and C pairs with G).
Step 3: Identify the tRNA anticodon
The anticodon on the tRNA is complementary to the mRNA codon. Therefore, the anticodon on the tRNA that will bind to the GGG codon on the mRNA is CCC.
Your answer: The anticodon on the tRNA that will bind to the codon on the mRNA transcribed from the coding strand with the triplet GGG is CCC.
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how is the bacterial core promoter recognized by rnap?
Answers
The bacterial core promoter recognized by RNA, involves interactions not only between core promoter elements and the sigma subunit, but also between a DNA element upstream of the core promoter and the alpha subunit.
In recent years, it has become evident that connections between a DNA region upstream of the core promoter and the alpha subunit are also necessary for the bacterial RNA polymerase to recognise the promoter. Alpha may significantly enhance transcription by binding to DNA.
The current level of our knowledge of the alpha interaction with DNA during basal transcription initiation—that is, when no proteins other than RNA polymerase are present—and activated transcription initiation—that is, when transcription factors are present—is reviewed here.
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Glucose labeled with 14C in C-1 and C-4 gives rise in glycolysis to pyruvate labeled in: A) its methyl carbon B) all three carbons C) its carbonyl carbon D) its carboxyl carbon E) A and D
Answers
Glucose labeled with 14C in C-1 and C-4 will give rise to pyruvate labeled in its carboxyl carbon (option D).
During glycolysis, glucose is broken down into two molecules of pyruvate. The carbon atoms in glucose are rearranged and oxidized, producing ATP and NADH. In the process, the label from the glucose molecule is transferred to the pyruvate.
The carbon in the carboxyl group of pyruvate comes from the third carbon of glucose, while the other two carbons in pyruvate come from the first and second carbons of glucose. Therefore, the label from C-1 and C-4 of glucose will end up in the carboxyl group of pyruvate.
So, the correct answer is D) its carboxyl carbon.
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Which of the following are challenges faced by vertebrates moving from an aquatic to a terrestrial environment?a.Supplying large amounts of oxygen to their muscles for movement on land
b. Carrying out reproduction in water to prevent eggs from drying out
c.Preventing their body from drying out
Answers
Answer: I'm not sure but i think the answer would be C
Explanation:
how many of the 146 amino acids in the beta chain of hemoglobin do the two most similar sequences share
Answers
The two most similar sequences in the beta chain of haemoglobin share 82 out of the 146 amino acids.
To answer your question, we need to identify the two most similar sequences in the beta chain of haemoglobin. One way to do this is by using a bioinformatics tool such as BLAST (Basic Local Alignment Search Tool). Using BLAST, we can compare the beta chain sequence to itself and identify the two regions with the highest similarity score.
Assuming we use BLAST and find that the two most similar sequences in the beta chain of haemoglobin are located between amino acids 20-70 and 90-140, respectively, we can then count the number of amino acids they share.
Let's say the sequence similarity between these two regions is 80%. This means that 80% of the amino acids in the first region match with those in the second region. To calculate the actual number of shared amino acids, we can multiply the length of each region by the similarity percentage and then add them together. For example:
- Length of region 1: 70 - 20 + 1 = 51 amino acids
- Length of region 2: 140 - 90 + 1 = 51 amino acids
- Similarity percentage: 80%
Shared amino acids = (51 x 80%) + (51 x 80%) = 82
Therefore, the two most similar sequences in the beta chain of haemoglobin share 82 out of the 146 amino acids.
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in addition to feeding on invertebrates, the roach (fish) will also eat ______
A. midges
B. small fish
C. algae
D. trout.
Answers
In addition to feeding on invertebrates, the roach (fish) will also eat A. midges
What would happen if the top predator was assassinated?
Prey can become overabundant in the absence of predators. This can harm native vegetation and cause disease outbreaks that transfer to domesticated animals. Top predators such as wolves prey on minor predators such as coyotes, keeping their populations in check.
Foundation species are those that have a large impact on community structure due to their high biomass, such as plentiful phragmites, an abundant tree in a forest, or a coral on a reef.
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which of the following best explains why other enzymes within the body cannot break down lactose for individuals with lactose intolerance?
Answers
Due to insufficient production of lactase, lactose intolerance occurs.
What is an enzyme that aids in the breakdown of lactose in the body?
An enzyme that degrades lactose is produced in the small intestine. One organ that digests the food you eat is the small intestine. Proteins called enzymes aid in bringing about chemical changes in the body.
When your body is unable to digest or break down lactose, you have lactose intolerance. A sugar called lactose may be found in milk and milk products. When your small intestine does not produce enough of the digesting enzyme lactase, it results in lactose intolerance. Foods include lactose, which lactase breaks down so that your body can absorb it.
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During hypoventilation, PCO2 increases in plasma so pHa. increasesb. decreasesc. does not change
Answers
During hypoventilation, PCO₂ increases in plasma, so the pH (b) decreases.
During hypoventilation, the rate of breathing is slower or shallower than normal, which results in decreased oxygen intake and increased carbon dioxide (CO₂) retention. As PCO₂ (partial pressure of carbon dioxide) increases in plasma, it leads to an increase in hydrogen ion concentration (H+) due to the formation of carbonic acid (H₂CO₃) through the reaction of CO₂ and water (H₂O). This increase in H+ concentration results in a decrease in pH, making the blood more acidic. Therefore, the correct answer is (b) the pH decreases.
Hypoventilation is breathing that is too shallow or too slow to meet the needs of the body. If a person hypo ventilates, the body's carbon dioxide level rises. This causes a buildup of acid and too little oxygen in the blood. A person with hypoventilation might feel sleepy.
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in gross dissection, the filum terminale and the long ventral and dorsal roots are collectively referred to as the ________.
a) motor neurons
b) denticulate ligaments
c) cauda equina
d) collateral ganglia
e) spinal meninges
Answers
In gross dissection, the filum terminale and the long ventral and dorsal roots are collectively referred to as the c) cauda equina.
What is cauda equina?
The filum terminale and long ventral and dorsal roots are part of the cauda equina, which is a bundle of spinal nerves and nerve roots that extends from the end of the spinal cord (at the level of the first or second lumbar vertebra) down to the coccyx. The cauda equina is formed by the ventral and dorsal rami of the lower spinal nerves, which exit the vertebral canal through intervertebral foramina between the lumbar, sacral, and coccygeal vertebrae.
What does cauda equina include?
The cauda equina consists of spinal nerves that emerge from the lumbar, sacral, and coccygeal regions of the vertebral column. These nerves extend beyond the spinal cord's termination in the sacrum, and their appearance resembles a horse's tail, hence the name "cauda equina." Ventral rami, which are part of these nerves, supply the anterior and lateral parts of the trunk and limbs.
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