Answer:
190.
Explanation:
The linking number may be defined as the sum of the twist and writhes present in the DNA molecule.
Linking number = twist + writhes.
Writhe = Total size / base pairs in circular DNA
Writhes = 4200 / 20 = 210.
Writhes must be -210 because the DNA is negatively supercoiled.
Twist number = Total size / 10.4
Twist number = 4200/ 10.5 = 400.
Then, linking number = -210 + 400
Linking number = 190.
Thus, the answer is 190.
The linking number in vivo for a 4200 bp plasmid with 1 negative supercoil every 20 base pairs is 190. This is calculated by first determining the relaxed linking number (400 turns) and then subtracting the number of negative supercoils (210), which are a result of the supercoiling present within the plasmid.
Explanation:To calculate the linking number for a plasmid that is 4200 bp and normally has one negative supercoil every 20 base pairs in vivo, we'll use the given size of the plasmid and the information that there are 10.5 base pairs per turn of the double helix.
Firstly, the relaxed (i.e., without any supercoils) linking number of the plasmid can be determined by dividing the number of base pairs by the number of base pairs per turn. For a 4200 bp plasmid with 10.5 bp per turn, this would be calculated as follows:
Linking number = Number of base pairs / (Base pairs per turn)
= 4200 bp / (10.5 bp/turn)
= 400 turns
This represents the number of turns in the plasmid under relaxed conditions. Considering there is normally one negative supercoil for every 20 base pairs, we can calculate the number of negative supercoils introduced:
Number of supercoils = Total base pairs / Base pairs per supercoil
= 4200 bp / 20 bp/supercoil
= 210 supercoils
As each supercoil changes the apparent linkage number by -1, to find the actual linking number in vivo under the presence of supercoiling, we need to subtract the number of supercoils from the relaxed linking number:
Linking number in vivo = Relaxed Linking number - Number of supercoils
= 400 turns - 210 supercoils
= 190 turns
The linking number in vivo for the 4200 bp plasmid with the given level of supercoiling is therefore 190.
The causative agent of botulism, a deadly form of food poisoning, is an endospore-forming bacterium called Clostridium botulinim. Why might it be difficult to kill this bacterium in contaminated food?
Answer: It is difficult to kill clostridium botulinum in food because the organism grow under low oxygen condition and produce spores and toxin. The bacterium can survive under harsh condition and produce spores that are resistant to heat, drying and chemicals. These spores germinate into bacteria in food when conditions is favourable.
Explanation:
Botulisim is a serious and fatal illness caused by the neurotoxin secreted by clostridium botulinum. This toxin block nerve function and can lead to respiratory and muscular paralysis. The toxin is found in improper processed can food.
Use the following words in multiple sentences to show your understanding of how they connect together.
Answer:
The correct answer is explained in the below mentioned sequence :
Eukaryotic Cell.Nucleus.DNA.Bases.Chromosomes.Genes.Transcription.mRNA.Translation.Codon.Ribosome.Amino acids.Protein.Explanation:
Eukaryotic Cell contains a Nucleus which has a double membrane.The nucleus contains DNA or deoxyribonucleic acid which is the genetic material.The DNA is made up of nitrogenous bases along with phosphate and deoxyribose sugar backbone.The DNA supercoils itself to form the Chromosome.Each chromosome has multiple Genes on then that are represented by specific sequences.The genes are transcribed by DNA-dependent RNA polymerase to generate an mRNA transcript.The mRNA or messenger RNA moved from the nucleus to the cytoplasm where they undergo Translation.Translation is a process in which the message in the mRNA sequence is read by the ribosome to produce a polypeptide chain.The message on the mRNA remains as a three letter code called the Codon. Each codon represents an amino acid.The Ribosome reads each codon and with the help of amino-acyl tRNA synthase uses amino acids to construct a polypeptide chain.The polypeptide chain folds in the proper orientation to generate a functional protein.Use the terms in the answers below to fill in the blanks in the following sentences. "An investigator is studying mutants in methionine synthesis. The _________ mutants are unable to ATP sulfurylase, also known as ______. This protein is the product of the _________ gene. "
Question is incomplete i have added full question in ask for detail section.
Answer:
Option a. met3, Met3p, MET3 is correct answer
"An investigator is studying mutants in methionine synthesis. The _met3_ mutants are unable to ATP sulfurylase, also known as _Met3p_. This protein is the product of the _MET3_gene. "
Explanation:
MET3 encodes ATP sulfurylase, which is a catalyst of first step of the sulfur assimilation pathway. This pathway results in the formation of hydrogen sulfide which is a precursor in the biosynthesis of cysteine, homocysteine, and methionine.
Source: National Center for Biotechnology Information, U.S. National Library of Medicine
Final answer:
The met mutants cannot synthesize methionine due to a non-functional enzyme ATP sulfurylase, product of the MET3 gene. Scientific methods like the use of differential media aid in the study of these mutants and their genetic makeup, which informs on methionine biosynthetic pathways.
Explanation:
An investigator is studying mutants in methionine synthesis. The met mutants are unable to synthesize methionine due to a defective ATP sulfurylase, also known as Met3p. This protein is the product of the MET3 gene. Through research on yeast strains that are unable to synthesize essential sulfur-containing amino acids due to inactivations within the biosynthetic pathway, scientists are able to use a genetic screen method to distinguish which MET genes are missing and thus understand the metabolic pathways for Met and Cys in more detail.
By using selective media containing various sulfur sources and differential media, such as BiGGY agar, these mutants could be characterized based on their growth properties. This method is essential to predict how mutations in genes involved in Met and Cys synthesis will affect the concentrations of metabolites in the pathway.
Lipids provide a significant energy reserve. form essential structural components of cells. help to maintain body temperature. cushion organs against shocks. All of the answers are correct.
Answer: All of the answers are correct
Explanation:
Lipids refers to a group of small biomolecules that do not dissolve in water, but dissolve readily in nonpolar solvents and contain fatty acid, sterols.
They have several:
- phospholipids forms the structural components of cells and biological membranes.
- Triacylglycerides serves as energy reserve (release energy during starvation) in the adipose tissues of animals.
- lipids in the adipose tissues of animals also provide insulation against cold, thereby maintaining body temperature
- lipids such as phospholipids are found in the biological membranes of many organs like the heart, brain etc, where they help to absorb shock or damage during severe hit to such organs.
So, all the answers are correct
Lipids are crucially important as they provide a significant energy reserve, form essential structural components of cells, help maintain body temperature and cushion organs against shocks. All the options are correct.
Explanation:Lipids are a class of organic molecules. These lipids perform several essential functions in the body. Firstly, they serve as a significant energy reserve, meaning that during times when other energy sources like carbohydrates are not available, the body can break down lipids to generate energy.
Second, lipids form essential structural components of cells. They are integral parts of the cell membrane and provide barriers in cellular transportation. Third, they help to maintain body temperature by providing insulation. Lastly, they protect our internal organs by cushioning them against shocks, acting almost like shock absorbers in the body.
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Which is the correct order of egg layer development? Group of answer choices chitin, vitelline, lipid vitelline, lipid, chitin lipid, vitelline, chitin vitelline, chitin, lipid
Answer:
1. Vitelline (The outer most layer)
2. Chitin (The inner layer)
Final answer:
For egg layer development, the correct order is vitelline, chitin, and lipid.
Explanation:
The correct order of prenatal development is zygote, embryo, fetus. This sequence starts with the zygote, which is the initial cell formed when a sperm cell fertilizes an egg cell. Following fertilization, the zygote begins a series of divisions and transforms into an embryo, where basic structures and organ systems begin to form. Eventually, the embryo develops into a fetus, during which time the tissues and organs that formed earlier begin to mature. This stage continues until birth.
In the context of embryonic development, there are three germ layers: the ectoderm, mesoderm, and endoderm. These layers form during a process called gastrulation and will eventually differentiate into all the organs and tissues of the body. The ectoderm gives rise to structures such as the skin and nervous system, the mesoderm forms components like the muscles and circulatory system, and the endoderm develops into internal structures such as the gut and lungs.
Regarding egg layer development, the correct composition of an egg cell layers is vitelline, chitin, lipid. The vitelline layer is immediately outside the egg plasma membrane, followed by a protective chitin layer, and finally, the lipid-rich yolk constitutes a major portion of the egg's content, providing nourishment for the developing embryo.
The Km of your favorite enzyme that operates by normal Michaelis-Menten kinetics is 10-6 M. You start off with a substrate concentration of 10-3 M and an enzyme concentration of 10-9 M. kcat is 10sec-1 . After a short time you have reached steady state.
a)__________ is the concentration of free enzyme.
b)__________ is the concentration of the ES complex.
Please provide an explanation. I am a bit confused.
Answer key says answers are:
a)0
b) 10^-9M
Answer: (a) 0 no free enzyme left
(b) 10^-9M
Explanation:
ANSWER:
Given that
Kcat = 10 sec-1
Km = 10^-6 M
[S] = 10^-3 M
[Enzyme] = 10^-9 M
The reaction follows the following path-
Enzyme (E) + Substrate (S) <=> ES complex -> E + Product (P)
According to the improved model of Michaelis-Menten kinetics, upon addition of substrate and the enzyme, instead of dynamic equilibrium, a steady state is reached. The time taken is very less, almost instantaneously (since Kcat is much higher than the concentrations we are dealing with (10 per second! Whereas we are dealing with concentrations as low as 10-9).
In this steady state, the Enzyme and substrate instead of existing individually, exist as an Enzyme-Substrate complex, or ES complex.
Physically, Km is a measure of how well substrate complexes with an enzyme, i.e. It's binding affinity.
You can imagine this as if 1 unit of the substrate can bind to "Km" units of Enzyme. For the give conditions, 1 M of the substrate requires 10^-6 M enzyme for complete binding. So, 10^-3 M of the substrate will require 10^-3 x 10^-6 = 10^-9 M of the enzyme, which is the exact amount of enzyme added to the reaction mixture.
So it is safe to assume that when the steady state is reached, all of the enzyme is bound to the available substrate producing the ES complex with the concentration equal to the limiting reactant, i.e. the enzyme = 10^-9 M
Hence, there will be no free enzyme left after the short duration of the reaction. And the concentration of the ES complex will be 10^-9 M
Three alleles control the ABO blood types. IA and IB are codominant genes, so the combination of IAIB produces the AB blood type. The third allele Iois recessive to the other two alleles. Indicate which of these parents could produce the given child.O A X AB produce B child.O A X O produce A child.O A X B produce O child.O A X AB produce O child.O A X AB produce B child.O B X B produce O child.O AB X AB produce A child.
Answer:
The ABO blood group represents the phenomena of Co dominance and the multiple alleles.
1. The cross between O A X AB results in the formation of progeny with the genotype OA ( A blood group), OB ( blood group), AA ( A blood group) and AB ( AB blood group). Their cross results in the B progeny.
2. The cross between O A X OO results in the progeny with A blood group ( AO, AO ) and ( OO, OO) and O blood group. Their cross results in the A progeny.
3. The cross between O A X B produce the children with genotype A, B, AB and O if the parent is OB.
4. The cross between O A X AB produces the child with the genotype OA, AA ( A blood group) OB ( B blood group) and AB (AB blood group). No child with O blood type is produced by the parents.
5. The cross between O B X B produce the children with genotype OO, OB,OB and BB if the parent is OB.
6. The cross between AB X AB results in the progeny with genotype AA ( A blood group) , AB, AB (AB blood group) and BB (B blood group). Their cross results in the progeny with A blood group.
When using oil immersion you must start focusing and centering the specimen as usual with the 4X objective, then sequentially with the 10X, and 40X, and finally the 100X objectives.
At what point is the oil introduced?
Answer:
100X
Explanation:
Majorly when using lower magnification microscope objective lenses (4x, 10x, 40x) the light refraction is not usually observed. But, at the 100x objective lens, the light refraction when employing a dry lens is observed. At a reducd light refraction, more light goes in the microscope slide and channelled at the very narrow diameter of a greater power objective lens. In microscopy, the greater the light = clear and vivid images. Immersion oil that has a refractive index same as the glass slide in the region filled with air, channels more light through the objective and a clear and crispy image is produced.
NOTE: OIL WILL DAMAGE THE OTHER LENSES.
Answer:The point at which oil is introduced is when focusing with 100X objectives.
Explanation:
Oil immersion in microscopy involves the use of a highly magnification( such as 100X objectives) and a transparent oil which aims at increasing the amount of light that is passing through the mounted specimen at a short focal length for a better view of the specimen features. The purpose of using oil immersion technique is to reduce the amount of scattered light as magnification is increased; thus, at lower magnification( such as X10,X40) this is not noticed.
Centrioles, cilia, flagella, and basal bodies have remarkably similar structural elements and arrangements. Which of the following hypotheses is most plausible in light of such structural similarities?
a. Loss of basal bodies should lead to loss of all cilia, flagella, and centrioles.
b. Motor proteins such as dynein must have evolved before any of these four kinds of structures.
c. Natural selection for cell motility repeatedly selected for microtubular arrays in circular patterns in the evolution of each of these structures.
d. Cilia and flagella arise from the centrioles.
e. Cilia and flagella coevolved in the same ancestral eukaryotic organism.
Answer:
Cilia and flagella arise from the centrioles
Explanation:
Some cells have projections made up of microtubules and covered by extensions of the plasma membrane.
These projections can be cilia or flagella.
Centrioles are also made of microtubule from which the cilia and flagella arise. It also takes part during cell division
Basal bodies are protein structures found at the base of cilia and flagella. It functions as a site starting point for the growth of microtubule for the cilia and flagella
A scientist at the University of Iowa uses a microscope to observe cells in the brain known as microglia. He makes observations about their structure, location, and activity. The scientist ntually observes the cells undergo a sudden and radical shift in their structure/shape and their motility (ability to move). He asks himself questions about what is causing this shift in behaviors and begins to design an experiment to determine the answer. Briefly describe how the scientist practiced both the exploration and testing aspects of scientific inquiry.
Answer:
Scientific inquiry can be illustrated as a procedure by which one can comprehend and extend the present information. It incorporates deducing of a hypothesis by knowing the current scenario, arranging materials and procedures, performing the experiment, assessing the data attained and providing unbiased outcomes. It is very essential that one goes through each o the steps diligently as leaving even one can modify the anticipated outcomes.
In the given case, the scientist was witnessing microglia. He observed their activity, structure, and location. This is termed as exploration as he is utilizing the accessible resources to extend his knowledge regarding a specific topic. After that, he witnesses that they are shifting briskly, which makes him question the phenomenon.
This can be a previously familiarized information, which he is just witnessing or something which was all new to him. Thus, to know more regarding this or to answer the specific question, he decides to develop an experiment. This is considered as the testing aspect of the scientific inquiry as he is arranging the materials and will determine the reason for this movement methodically.
Given that the horse has a diploid chromosome number of 64 and a zebra has a diploid chromosome number of 46, what would be the expected chromosome number in the somatic tissues if a viable hybrid were possible between these two animals
Answer: 55
The expected chromosome number in the hybrid would be 55
Explanation:
To obtain the hybrid organism:
- the diploid (2n) cell of horse will undergo meiosis to produce a gamete with halved chromosome. This gamete is with haploid number (n)
Thus, 64 / 2 = 32
- the diploid (2n) cell of Zebra will undergo the same pattern as the horse
Thus, 46 / 2 = 23
Then, the gametes of the two animals will be fused to form a zygote of the hybrid.
32 + 23 = 55
Hence, the zygote will form a diploid hybrid organism with an expected chromosome number of 55
Think about the many factors that are used to classify viruses. In order for Lauren to determine the type of virus she may have, she needs to understand some basic concepts about how viruses are characterized. Please sort the following statements as being true or false regarding characteristics of viruses.
Answer:
True statements:
- Viruses are not capable of metabolic activity on their own; they must have a host cell to reproduce.
- The genetic material of a virus can be DNA or RNA.
- Viruses can have different shapes.
- Capsids function to surround and protect the nucleic acid core.
- The genomes of viruses are usually smaller than the genomes of bacteria.
Incorrect Statements
- All viruses have both a capsid and an envelope.
- Capsids are adquired from the host cell during replication or release.
- The intracellular state of a virus is called a virion.
Explanation:
All of this are very important aspects to know in order to identify a virus from a bacteria. And also there are a lot of factors by which a virologist can classify viruses: the type of nuleid acid, the presence or absecence of an envelope, the capsyd, and how the virus reproduce.
The significance of checkpoints can be demonstrated by considering what happens when they are impaired. What would occur if there was a gain-of-function mutation in the promoter for the cyclin E gene such that cyclin E protein was always made at high levels even under conditions in which cyclin E would not normally be made?
The significance of checkpoints can be demonstrated by considering what happens when they are impaired. What would occur if there was a gain-of-function mutation in the promoter for the cyclin E gene such that cyclin E protein was always made at high levels even under conditions in which cyclin E would not normally be made?
a. Cells will pass through the G2/M checkpoint with damaged DNA.
b. Cells will skip the S phase and go directly to G2 phase and not complete DNA replication.
c. Cells will pass through the G1/S checkpoint even if conditions are not ideal for cell division.
d. Cells will pass the M checkpoint with chromosomes unattached to spindles.
Answer:
c. Cells will pass through the G1/S checkpoint even if conditions are not ideal for cell division.
Explanation:
Cyclins are the regulatory proteins that are formed during different stages of the cell cycle. A specific cyclin during each stage bind to corresponding cyclin-dependent kinases (CDKS). The activated CDKs phosphorylate the proteins required for the progression of a cell through a specific stage of the cell cycle.
For example, Cyclin E is synthesized at the peak near the G1 -S phase transition or G1/S checkpoint. They activate the CDK-2 and allow the cells to progress from G1 to the S phase. If the cell is not prepared for DNA replication, cyclin E is not formed or is inhibited by specific protein kinases which in turn does not allow the cell to enter the S phase.
Any gain of function mutation resulting in the constitutive synthesis of cyclin E would allow the cell to pass through the G1/S checkpoint irrespective of the conditions.
Claim: An individual virus docks on the surface of a cell, infects it, hijacks the cellular machinery inside, and replicates itself, sometimes thousands of times. Justification: Based on what you learned about size, scale, and the component parts of a virus, justify with scientific reasoning how a virus is able to accomplish this.
Answer:
Virus is a ultramicroscopic pathogenic particle that consist of nucleic acid (DNA or RNA) enclosed in a protein coat and reproduces by binary fission.
Explanation:
Virus is a ultramicroscopic pathogenic particle capable of passing through bacteriological filters;
- consist of nucleic acid (DNA or RNA) enclosed in a protein coat
- capable of replicating within a living host only and spreading from cell to cell
- infects cells of bacteria, plants and animals
- hijack the metabolism of the infected cell, though they cannot carry out metabolism themselves.
- they reproduce by binary fission i.e without the formation of gametes
The claim is justified based on the understanding of virus structure and the cellular processes it exploits. Virus is able to accomplish by an extremely small infectious agent.
Viruses infect cells by a process known as viral entry. This begins with the docking of the virus to specific receptors on the host cell's surface. The virus's capsid proteins are precisely shaped to bind to these receptors, much like a key fits into a lock. Once bound, the virus can enter the cell through various mechanisms, such as endocytosis or membrane fusion.
After entering the cell, the virus releases its genetic material, which can be either DNA or RNA. This genetic material then exploits the host cell's machinery to replicate and produce viral components. The virus hijacks the host's cellular processes by using its own genetic instructions to direct the synthesis of viral proteins and the replication of its nucleic acid.
The replication process can result in the production of many new virus particles, or virions. The number of virions produced can vary widely, from a few to thousands, depending on the type of virus and the conditions within the host cell. Eventually, the newly formed virions are released from the host cell, often causing the cell to lyse (burst) and die, and the cycle repeats as these new viruses go on to infect other cells.
A virus is an extremely small infectious agent, typically consisting of a nucleic acid molecule in a protein coat called a capsid, and in some cases, an outer lipid envelope. The size of a virus is on the order of tens to hundreds of nanometers, which is several orders of magnitude smaller than a typical eukaryotic cell, which is about 10 to 100 micrometers in diameter.
Proteins are constantly being synthesized in a living cell. Why doesn't the number of protein molecules become too great for the cell to contain, leading to cell destruction?
Answer:
Both eukaryotes and prokaryotes have gene regulation strategies. By gene regulation, the amount of protein is maintained in the living cell.
In prokaryotes, operons are present which are controlled by a regulatory gene that regulates the synthesis of protein in prokaryotes. In eukaryotes, repressor and transcription activators regulate the expression of gene and protein formation.
So by gene regulation cell maintains the number of protein molecules to become too great for a cell. Without gene regulation the survival of any organism is not possible.
After stamping your replica plates, you return to examine the results and see that there are 100 colonies on the strep nal plate from master plate strain B and no growth on the strep nal plate from master plate strain A. Which strain (A or B) is the streptomycin-resistant master plate strain
Answer/Explanation:
An important way of selecting for bacteria carrying specific recombinant DNA is to add a resistance gene (such as antibiotic resistance) to your gene of interest. That means that any bacteria that carry your DNA will be resistant to a specific antibiotic (in this case streptomycin). Therefore, you can grow your bacteria on plates with streptomycin, and in theory, the bacteria will be unable to grow if they don't have the resistance gene and your gene of interest.
Here, strain A does not grow on streptomycin plates, but strain B now has 100 colonies. This suggests strain B can grow on streptomycin, meaning it is resistant.
Strain B is therefore the streptomycin resistant master plate strain.
A ligand is a A. substance that binds to receptor molecules. B. type of electrical stimulus. C. type of drug. D. cholinergic synapse.
Answer: (A) Substance that binds to receptor molecules.
Explanation: Is any molecule which attaches reversibly to a protein. These are typically used in cellular signaling and cellular regulation, but have many other uses.
a. A ligand is a substance that binds to receptor molecules.
What is a Ligand?
A ligand is an ion or molecule, which donates a pair of electrons to the central metal atom or ion to form a coordination complex. It is a substance that forms a complex with a biomolecule to serve a biological purpose.In protein-ligand binding it is usually a molecule which produces a signal by binding to a site on a target protein.A ligand is defined as any molecule or atom that irreversibly binds to a receiving protein molecule, otherwise known as a receptor.For example: In protein-ligand binding-insulin is used to signal various things to the metabolism of each cell.Learn more:
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When the energy flow through an ecosystem is measured at each trophic level, approximately __________ of the available energy passes from one level to the next.
Answer:
5-20%
Explanation:
Only approximately 5-20% of energy is passed from the one trophic level to the next trophic level and 10% law says only 10% energy is transferred. The rest of the energy is lost as heat and some energy is lost as undigested food.
So due to this at each level, most of the energy is lost and very less energy is available for the higher trophic level. This is the reason the number of organisms at higher trophic levels is very less and the trophic level is limited to 4 trophic levels.
Mitosis and meiosis always differ in regard to the presence of a. chromatids. b. homologs. c. bivalents. d. centromeres. e. spindles.
Answer:
c. bivalents.
Explanation:
In Meiosis a bivalents are formed during the first stage of meiosis which is the prophase. The bivalent consist of a paired chromosome and four chromatids i.e two chromosomes in a tetrad. one chromosome comes from each parent.
In mitosis, a bivalent is not formed. There is the presence of chromatids, homologs, centromeres and spindles during mitosis and meiosis.
Mitosis and meiosis always differ in regard to the presence of bivalents (Option c).
A chromosome is a specific linear chain of genetic material (DNA), which is transmitted as a unit during cell division.A bivalent refer to two homo-logous chromosomes that exchange genetic material during recombination.A bivalent occurs only in meiosis where the interchange of genetic material between homo-logous chromosomes occurs.In conclusion, mitosis and meiosis always differ in regard to the presence of bivalents (Option c).
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"In order to determine if the amygdala is needed for rats to learn to avoid an electric shock, Kevin plans to inject a neurotoxin into the amygdala. Which technique is Kevin using?
Answer:
Lesion technique
Explanation:
A lesion is a damage on a region of a tissue or organ.
The lesion method is a technique where a lesion or some damage is deliberately inflicted on the brain causing damage to the neurons.
The main intent being to know the specific function of a certain part of the brain.
These can be done surgically or chemically.
In our case, Kevin wants to know if the amygdala is needed for rats to learn to avoid an electric shock.
He injects a neurotoxin into the amygdala, this cause a lesion on the amygdala.
From these he can get to know the function of the amygdala.
Kevin is using the D. lesion technique to inject a neurotoxin into the amygdala to study its role in fear learning in rats.
The method he is utilizing is known as an injury. Lesioning includes purposefully harming explicit pieces of the mind to concentrate on their capability by noticing the impacts of the harm on conduct or physiology. For this situation, Kevin is disturbing the capability of the amygdala to notice its part in the learning and dread reaction to an electric shock. Studies, similar to those including patient SM or examination with rhesus monkeys, have shown that sores in the amygdala bring about diminished dread reaction, demonstrating its basic job in handling dread and aversive learning.Complete question:
"In order to determine if the amygdala is needed for rats to learn to avoid an electric shock, Kevin plans to inject a neurotoxin into the amygdala. Which technique is Kevin using?
a. MRI
b. fMRI
c. PET
d. lesion
The energy content and biomass of ________ is lowest in a terrestrial food web. Group of answer choices a. detritivores and decomposers b. top carnivores producers c. small carnivores such as spiders and lizards
Answer:
Option b. top carnivores is correct.
Explanation:
organism that found at the top of the PYRAMID OF NUMBERS that feed (preys) on other organisms but which is not itself preyed on are called top carnivore. They actually regulate terrestrial ecosystems. For example, tiger. (See attached Image)
In a terrestrial food web, producers have the lowest energy content and biomass.
Explanation:In a terrestrial food web, the energy content and biomass is lowest in producers. Producers, such as plants, convert solar energy into chemical energy through photosynthesis. They are the primary source of energy in a food web, and as energy flows through the system, some energy is lost at each trophic level, resulting in a decrease in energy content and biomass.
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In a population of rabbits, there are 496 black rabbits and 27 white rabbits. Fur color is determined by a pair of alleles where (B) is the dominant allele which produces black fur and (b) the recessive allele which produces white fur. The frequency of the dominant allele for black fur (B) is 0.8. What is the frequency of the black fur phenotype in the population (the population is in Hardy-Weinberg Equilibrium)? Round your answer to the nearest hundredth. (Hint: remember the numeric format of frequency).
Answer:
[tex]0.96[/tex]
Explanation:
It is given that
B is the dominant allele which represents the black color
and b is the recessive allele which represents the white fur.
B being dominant will result into black color fur for genotype "Bb"
Given -
Frequency of black fur allele (p) is [tex]0.8[/tex]
As per Hardy Weinberg's first law of equilibrium
[tex]p + q = 1\\[/tex]
Substituting the value of p in above equation, we get -
[tex]q = 1-p\\q = 1-0.8\\q= 0.2[/tex]
q represents the frequency for white fur allele
Frequency of white fur phenotype is
[tex]q^2\\= 0.2^2\\= 0.04[/tex]
Frequency of homozygous black fur phenotype (BB) is
[tex]p^2\\= 0.8^2\\= 0.64[/tex]
As per Hardy Weinberg's second law of equilibrium -
[tex]p^2 + q^2 + 2pq = 1\\0.64 + 0.04 + 2pq = 1\\2pq = 1 - 0.68\\2pq = 0.32\\[/tex]
Combined frequency of homozygous and heterozygous black fur phenotype is
[tex]0.64 + 0.32\\= 0.96[/tex]
Multicellular organisms use cell division, mitosis, for growth and the maintenance and repair of cells and tissues. There are few cells in the body that do not undergo mitosis: most somatic cells divide regularly, some more than others. Single-celled organisms may use cell division as their method of reproduction. Regardless of the reason for mitosis, the process ensures genetic continuity. Consider the model of the cell cycle. Which detail(s) from the model best support the argument that cell division promotes genetic continuity?
The model of the cell cycle provides details about how cell division promotes genetic continuity, including DNA duplication, accurate segregation of chromosomes, and cytokinesis.
Explanation:The model of the cell cycle provides several details that support the argument that cell division promotes genetic continuity:
During interphase, the cell grows and the nuclear DNA is duplicated. This ensures that each daughter cell receives an exact copy of the genetic material, maintaining genetic continuity. In the mitotic phase, the duplicated chromosomes are segregated and distributed into daughter nuclei. This process ensures that the genetic information is accurately passed on to the next generation of cells. Following mitosis, the cytoplasm is divided through cytokinesis, resulting in two genetically identical daughter cells. This further ensures that each cell receives a complete set of genetic information. Learn more about Genetic continuity here:https://brainly.com/question/33255930
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The method of scientific inquiry that describes natural structures and processes as accurately as possible through careful observation and the analysis of data is known as A) hypothesis-based science. B) discovery science. C) experimental science. D) quantitative science. E) qualitative science.
Answer: B) Discovery science
Explanation:
Discovery science is the science which is a fact finding inquiry. It describes the natural structures and processes and tries to find the phenomena responsible for these natural structures and processes by keen observation, data collection, analysis of the data and drawing the results and relating it with the natural phenomena.
Discovery science: A method of scientific inquiry that describes natural structures and processes through careful observation and analysis of data.
Explanation:The method of scientific inquiry that describes natural structures and processes as accurately as possible through careful observation and the analysis of data is known as discovery science.Discovery science focuses on gathering information through observation and description of natural phenomena. It aims to uncover new knowledge and develop new theories rather than testing specific hypotheses.
For example, in biology, discovery science can involve observing and studying new species, analyzing ecological patterns, or investigating the behavior and characteristics of organisms in their natural habitats.
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Several factors influence the rate of diffusion and among these factors are temperature, ____________ , electrical currents, and molecular size. For example, as temperature ____________ , the rate of diffusion increases.
Answer: States of matter; increases
Several factors influence the rate of diffusion and among these factors are temperature, STATES OF MATTER, electrical currents, and molecular size. For example, as temperature INCREASES, the rate of diffusion increases.
Explanation:
Factors influencing the rate of diffusion include:
- Temperature: High temperature increases the speed at which molecules move. Thus, as temperature increases, the rate of diffusion increases
- States of matter: Diffusion varies within the three states of matter. The diffusion of gases is much faster than that of liquids and solids, because the gas molecules are freer and therefore faster than the rest.
- Molecular size: The smaller the molecules, the faster the rate of diffusion while the larger the molecules, the slower the rate of diffusion.
Other factors include electrical currents.
When 5110 first appeared, the Grants did not know how to classify him.
Where did he come from? Was he a new species of finch, or was he an infertile hybrid of known species? If birds prefer to mate with other birds from their own species, what would 5110's fate be on Daphne where there were no other birds like him?
Answer:
When a new infertile hybrid appears in an environment, it feels like an anomaly in the environment and there are no possible measures to classify it as a new specie or a member of a specie just by a physical examination. When 5110 subject appears in that environment it is possible to classify it as a piece of a terminated branch of only one member as an anomaly of the closely related species.
His fate is solely relied on better care of himself for a good life span cause there are no other measures to save him. Futhermore added he is also not a proper member of a specie so no mating oppurtunities at all.
Answer:
Genetic testing was carried out on 5110 and the grants concluded that he was a hybrid. He has G. fortis and G. scandens genes.
See below further explanation
Explanation:
When 5110 first appeared, the Grants did not know how to classify him.
Where did he come from? Was he a new species of finch, or was he an infertile hybrid of known species? If birds prefer to mate with other birds from their own species, what would 5110's fate be on Daphne where there were no other birds
Genetic testing was carried out on 5110 and the grants concluded that he was a hybrid. He has G. fortis and G. scandens genes.
He was probably from a G fortis parent and hybrid G scandens .
crossbreeding is when The same species of different features/genetic make up are made to mate with the same of different features/genetic makeup and the product can have an enhanced quality
5110 will be a big bird, if there is an interbreeding between the hybrid G Scandens and G. fortis
Consider the data on cliff swallow mortality. Why is this an example of directional selection?Overall genetic variation decreased.Selection did not occur, because no reproduction occurred (just survival).Individuals with intermediate phenotypes survived best.The average trait value changed in one direction (in this case, larger size).
Answer:
The average trait value changed in one direction.(In this case, larger size)
Explanation:
In evolution a natural selection can be disruptive, directional or stabilizing
In stabilizing no extreme trait is favored hence provides intermediate values .
Disruptive selection both extreme traits are favored over the intermediate trait.
Directional, the enviroment will favor the survival of one trait hence a change in direction either towards the left or the right.
In the case of swallow cliff mortality, selection favored the larger size.
The first cells may have gained energy through a. redox reactions using diatomic oxygen (O2) as the final electron acceptor. b. ingesting algae present in the early oceans. c. light-driven ion pumps. d. the controlled oxidation of petroleum.
Answer:
c. Through light-driven ion pumps
Explanation:
a. Through redox reactions using diatomic oxygen (O2) as the final electron acceptor❌
Oxygen was observed to be lacking on early Earth.
b. Through the controlled oxidation of petroleum❌
Petroleum is categorically a fossil fuel and couldn't be marked as being available or found on early Earth.
c. Through light-driven ion pumps✔✔✔
Light-energy that was derived from the sun was a ready and accessible source of energy that could have directed proton pumps in early or first cells.
d. Through ingesting algae present in the early oceans❌
Algae are known to be eukaryotes and was seen to have been established billions of years after the first cells developed.
Answer:
c. light-driven ion pumps.
Explanation:
According to the records we know that the first cells took energy from the surrounding environment, and lived in the absence of O2, with the shortage of nutrients they began to obtain energy from sunlight and CO2 present in the atmosphere.
What is one method or line of reasoning that scientists could NOT use when testing the hypothesis that dolphins and Ichthyosaurs form a monophyletic group?
Answer:
They cannot use the reasoning, "Dolphins and Ichthyosaurs form a paraphyletic group."
Explanation:
This is because they always form a monophyletic goup. Both of the species have specific morphological similarities that are enough to consider them in monophyletic group instead of a divergent paraphyletic.
Belinda's food choices are lacto-vegetarian. Since vegetarians require 1.8 times more iron to make up for the low bioavailability of their diet, what would her daily iron requirement be?
Belinda's daily iron requirement would be 1.8 times the standard Recommended Dietary Allowance due to the lower bioavailability of iron in her lacto-vegetarian diet. This adjusted amount ensures she receives sufficient iron despite the dietary restrictions.
To determine Belinda's daily iron requirement, we need to consider that the bioavailability of iron in a vegetarian diet is lower compared to diets that include animal products. According to the information provided, the Recommended Dietary Allowance (RDA) for iron is 1.8 times higher for those following a lacto-vegetarian diet due to the low bioavailability of iron from plant sources.
The RDA for iron varies by age and gender. For instance, adolescent boys and adult men typically require less iron than adolescent girls and women of childbearing age, who may need more iron to compensate for loss during menstruation or to support a pregnancy.
Given that Belinda follows a lacto-vegetarian diet, if the standard RDA for her demographic is X mg of iron per day, her requirement would be 1.8X mg/day to account for the lower bioavailability of plant-based iron sources. For example, if the standard RDA for a woman in her age group is 18 mg/day, Belinda would need approximately 32.4 mg/day (18 x 1.8 = 32.4 mg/day).