Physical weathering, or mechanical weathering, is the process by which rocks are broken down into smaller pieces through mechanical forces. Chemical weathering is not an example of physical weathering.
Explanation:Physical weathering, also known as mechanical weathering, is the process by which rocks are broken down into smaller pieces through mechanical forces. This is typically caused by factors such as temperature changes, freeze-thaw cycles, or the growth of plants and animals. It does not involve any chemical reactions. Examples of physical weathering include frost wedging, salt crystal growth, sheeting, and biological activity.
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What do you think might be the evolutionary benefit of the milk production regulation mechanism in breastfeeding?
Answer: The benefit of evolutionary benefit of milk production regulation mechanism is that the mother will not waste her energy or nutrients required for her growth and development to produce milk at that time rather it will be during her late pregnancy stage where the mammary gland will be activated by oxytocin hormone to produce milk automatically to breast feed infants at birth.
Explanation:
Evolution of lactation or mammary glands in mammals has been very important to feed the infants.
Lactation refers to production of milk from the mammary glands. Lactation normally occur at post pregnancy stage. At this stage the infants are born and milk is been secreted by the mammary glands to breast feed the young ones. Lactation doesn't occur all the time but it occur during late to post pregnancy stage. This is activated by oxytocin which help to implant embryo in the uterus and help in lactation.
The milk production regulation mechanism in breastfeeding is a complex process that is controlled by a variety of hormones, including prolactin, oxytocin, and estrogen. This mechanism ensures that the mother produces enough milk to meet the needs of her baby, but not so much that it becomes a burden.
There are several possible evolutionary benefits of this mechanism.
First, it helps to ensure that the baby is always fed, even if the mother is not constantly producing milk. This is important for the survival of the baby, as it needs to be fed frequently in order to grow and develop.
Second, the mechanism helps to prevent the mother from producing too much milk, which can lead to mastitis, an infection of the breast tissue. Mastitis can be painful and can interfere with breastfeeding.
Third, the mechanism helps to ensure that the milk is of the right composition for the baby. The composition of milk changes over time to meet the changing needs of the baby. For example, the milk produced in the early days after birth is high in colostrum, which is a nutrient-rich liquid that helps to protect the baby from infection.
Hence, the milk production regulation mechanism in breastfeeding is a complex and efficient process that has evolved to ensure the survival and health of the baby.
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The question below refers to the following information: 1. reception 2. transmission 3. transduction 4. perception 5. amplification For the events above, which of the following is the correct sequence as it relates to being able to hear sound
Answer:
Reception, Amplification, transmission, transduction, perception.
Explanation:
In the ear, the external ear collects sound pressure waves (reception) and funnels them toward the tympanic membrane, The vibrations (amplification) from the eardrum/tympanic membrane set the ossicles into motion. The middle ear transmits sound from the outer ear to the inner ear (transmission). The middle ear consists of three bones: the hammer (malleus), the anvil (incus) and the stirrup (stapes), the oval window, the round window and the Eustrachian tube. Sound vibrations are then transduced into electrical energy/impulses by hair cells in the inner ear. These electrical impulses are finally perceived by the brain as sound.
Pigment in mouse fur is only produced when the C allele is present. Individuals of the cc genotype are white. If color is present, it may be determined by the A, a alleles. AA or Aa results in agouti color, while aa results in black coats. What are the phenotypic ratios in the F2 generation of mice if an AACC parent and an aacc parent are bred to yield F1 mice which are then crossed to generate an F2 generation? Enter your answer as a fraction in the blank next to the appropriate coat color.
agouti ______
colorless_______
black________
Answer:
Agouti: colourless:black=9:4:3 so the agouti coat color would be 9/16 , Colourless coat color would be 1/4 , and Black would be 3/16 .
Explanation:
The type of genetic pattern that is followed in this cross is epistasis, in this pattern the phenotypic expression of a particular gene in one locus changes the effect of the gene on other locus.
In this case allele C present will leads to the color but homozygous C denotes colorless and A denotes agouti where aa denotes black. With the first cross all F1 individual will be AaCc . Hence, in the second cross for F2 would be AaCc × AaCc
Thus, the phenotypic ratio of all three type of coat color woyld be agouti: colourless:black=9:4:3 so the agouti coat color would be 9/16 , Colourless coat color would be 1/4 , and Black would be 3/16 .
20. During filtration, which of the following does NOT enter the bowman's capsule from the bloodstream?
A. glucose
B. water
C. ions
D. plasma proteins
E. amino acids
Answer: option D.
Plasma proteins.
Explanation:
Filtration is the transfer of water and solutes like glucose, ions, amino acids from the plasma to the renal tubules in the renal corpsicule plasma . Renal corpsicule help to filter blood. The fluid present in the blood in the glomerulus is pass to the Bowman's capsule to form glomerular filtrate , plasma protein do not enter Bowman's capsule. The filtrate is processed to form urine in the nephron.
Fill in the blanks, your answers should be lower case. The process of makes an RNA copy from a DNA template. The RNA is made by the enzyme which makes the RNA from (3 or 5)' to (3 or 5)'. The process of makes a protein from an mRNA. The protein is made by the which catalyzes the peptide bond between each amino acid. In bacterial cells these two processes occur in the cytoplasm. However, in eukaryotes the process occurs in the nucleus and the process occurs in the cytoplasm
The process of transcription makes an RNA copy from a DNA template.
The RNA is made by the enzyme RNA polymerase which makes the RNA from (3 or 5)' to (3 or 5)'.
The process of Translation makes a protein from an mRNA.
The protein is made by the peptidyl transferase in Ribosome which catalyzes the peptide bond between each amino acid.
In bacterial cells these two processes of transcription and translation occur in the cytoplasm.
However, in eukaryotes the process of transcription occurs in the nucleus and the process of translation occurs in the cytoplasm
Explanation:
The process of translation and transcription comprises the central dogma of molecular biology. By these two process the information stored in the genes flows into the proteins.
The process of creating an RNA copy from a DNA template is known as transcription, conducted by the enzyme RNA polymerase. The translation of an mRNA into a protein is facilitated by the ribosome. While both processes occur in the cytoplasm in bacteria, in eukaryotes, transcription occurs in the nucleus and translation in the cytoplasm.
Explanation:The process of transcription makes an RNA copy from a DNA template. The RNA is made by the enzyme RNA polymerase which makes the RNA from 3' to 5'. The process of translation makes a protein from an mRNA. The protein is made by the ribosome which catalyzes the peptide bond between each amino acid. In bacterial cells, these two processes occur in the cytoplasm. However, in eukaryotes, the transcription process occurs in the nucleus and the translation occurs in the cytoplasm.
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In organisms with large genomes, inversions are more likely to be tolerated if the breakpoints occur in: reciprocal translocations. noncoding DNA. open reading frames. coding DNA. closed reading frames.
Answer:
Non-coding DNA.
Explanation:
Inversion is a type of chromosomal abnormality in which the sequence of a segment of a chromosome is inverted or rotated at an angle of 180 degrees. This type of abnormality can change the reading frame of the gene and can cause mutations.
But if the genome sequence is non-coding that is not involved in the formation of protein synthesis than even if the reading frame is inverted will not affect the phenotype of the cell. Also, the non-coding sequences are removed by the splicing mechanisms.
Thus, Non-coding DNA is correct.
Inosine monophosphate is a branch point for the synthesis of a variety of purine nucleotides. Match the appropriate reaction for the synthesis of either AMP or GMP:
1) AMP A) condensation with Asp
2) GMP B) oxidative hydration
C) transamination with Gln
D) release of fumarate
1. --> 1:A, C; 2: B, D
2. --> 1:C, D; 2: A, B
3. --> 1:A, B; 2: C, D
4. --> 1:A, D; 2: B, C
Answer:
1:A, D; 2: B, CExplanation:
1.Purine Bases Can Be Synthesized by two pathways: i) de Novo and ii) Salvage Pathways.
2.Purine nucleotides synthesis starts with Phosphoribosyl pyrophosphate (PRPP), and it leads to the synthesis of nucleotide, inosine 5'-monophosphate (IMP). Inosine monophosphate is a branch point for the synthesis of many purine nucleotides amd leads to the synthesis of a variety of purine nucleotides.
3.The first reaction of purine synthesis is catalyzed by the enzyme glutamine phosphoribosylpyrophosphate amidotransferase.
g Imagine two different sigma factors with different promoter recognition sequences. What would happen to the overall gene expression profile in the cell if one sigma factor were artificially overexpressed?
Answer:
Transcription is affected via the modulation of the concentrations of the different types of holoenzymes, so saturated promoters are only weakly affected by sigma factor competition. However, in case of overlapping promoters or promoters recognized by two types of sigma factors, we find that even saturated promoters are strongly affected. Active transcription effectively lowers the affinity between the sigma factor driving it and the core RNAP, resulting in complex cross-talk effects. Sigma factor competition is not strongly affected by non-specific binding of core RNAPs, sigma factors and holoenzymes to DNA. Finally, we analyze the role of increased core RNAP availability upon the shut-down of ribosomal RNA transcription during the stringent response. We find that passive up-regulation of alternative sigma-dependent transcription is not only possible, but also displays hypersensitivity based on the sigma factor competition.
"Suppose that two parents are both heterozygous for sickle cell anemia, which is an autosomal recessive disease. They have eight children. Use the binomial theorem to determine the probability that three of the children have sickle cell anemia and five of the children are healthy. Round your answer to the nearest tenth."
If two parents are both heterozygous for sickle cell anemia, the probability that three of the children have sickle cell anemia and five of the children are healthy is approximately 0.1 or 11.0%.
What is the probability of sickle cell anemia?The probability of having 3 children with sickle cell anemia and 5 healthy children out of 8 total children is explained below.
The probability of any individual child having sickle cell anemia with heterozygous parents is 1/4; the probability of any individual child being healthy is 0.75.
Using the binomial theorem,
P(3 children with sickle cell anemia and 5 healthy children)
= [tex]P^8[/tex]₃ ×[tex](1/4)^3[/tex] ×[tex](3/4)^5[/tex]
= P(3 children with sickle cell anemia and 5 healthy children) = 0.110 that is nearly 11%.
Hence, if two parents are both heterozygous for sickle cell anemia, the probability that three of the children have sickle cell anemia and five of the children are healthy is approximately 0.1 or 11.0%.
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Final answer:
The probability that three out of eight children of two heterozygous parents for sickle cell anemia have the disease and five are healthy is approximately 20.7%.
Explanation:
The question posed involves the application of the binomial theorem to a genetic problem involving sickle cell anemia, an autosomal recessive disease. To find the probability that three of the children have sickle cell anemia and five of the children are healthy when both parents are heterozygous for the trait, we can use the binomial probability formula:
P(X=x) = ⁿCₓ × (p)ˣ × (q)ⁿ⁻ˣ
Where 'n' is the total number of children (8), 'x' is the number of children affected with sickle cell anemia (3), 'p' is the probability of having sickle cell anemia (1/4), and 'q' is the probability of being healthy (3/4). The term 'ⁿCₓ' represents the binomial coefficient, which can be calculated using combinations.
P(3 have sickle cell) = ⁸C₃ × (1/4)³ × (3/4)⁵
Calculating further:
P(3 have sickle cell) = 56 × (1/64) × (243/1024)
P(3 have sickle cell) = 56 × (243/65536)
P(3 have sickle cell) = 13608/65536
Converting to a percentage and rounding to the nearest tenth gives us approximately:
P(3 have sickle cell) = 20.7%
in general terms, what two general factors in a medium affect small speed of sound?
Please explain! I will give brainliest!
Answer:
Density of the medium, temperature of the medium, and stiffness of the medium.
Explanation:
Medium
Medium has a huge effect of the speed of sound. When most people discuss the “speed of sound” they are talking about the propagation of sound waves through the medium of “Air”. For anyone who has gone underwater and listen to people talking above it is likely that one would notice the muted an “odd” way that voices sound underwater. This is because the “medium” of water greatly bends, distorts and changes the speed of sound wave.
There is a whole aspect of science that measure and defines the effect of different mediums (gaseous and liquid) on the speed of sound. This is called Fluid Dynamics. Underwater communication is possible if you understand how this wave propagation as well as another important factor (pressure).
Because of elasticity of materials sound will, as a rule of thumb, generally travel faster in solids than in liquids and faster in liquids than in gases.
♦ Temperature
Temperature has a large effect on the speed of sound. Not as much as the “Medium” does, but far more than anything else. Temperature affects the speed of sound because temperature can affect the “elastic” qualities of different mediums. At the very basics lower temperatures will decrease the speed of sound while higher temperatures will increase the speed of sound, all other factors being equal.
♦ Pressure
Pressure is the final factor that has a significant impact on the speed of sound. The effect of pressure on the speed of sound is due to the materials inertial properties. In short, the more pressure that is applied to the material or medium the denser it becomes and the greater the “inertia” becomes. This makes any interactions between particles slower. Therefore the speed of sound throughout the medium is slowed due to the greater pressure.
Answer:
Elasticity and density
Explanation:
The speed of sound depends on elasticity and density of the medium through which it is travelling. The sound travels faster in liquids than gases. The sound travels faster in solids than in liquids. The lesser the elasticity and higher the density , the sound travels slower in a medium.
Speed = elasticity / density
25] Discuss what your analysis above indicates about the applicability of the Hardy-Weinberg criteria to this population. Which assumptions, if any, of the Hardy-Weinberg criteria are violated
Answer: The Hardy–Weinberg principle, also known as the Hardy–Weinberg equilibrium, model, theorem, or law, states that allele and genotype frequencies in a population will remain constant from generation to generation in the absence of other evolutionary influences.
Explanation: the population is in Hardy-Weinberg equilibrium for a gene, it is not evolving, and allele frequencies will stay the same across generations. There are five basic Hardy-Weinberg assumptions: no mutation, random mating, no gene flow, infinite population size, and no selection.
which of the following is not associated with translation
ribosome
dna polymerase
anti codons
mrna
Answer:
The DNA polymerase
Explanation:
In translation, the mRNA produced after transcription, tRNA that bring the brings the amino acids, ribosome; the organize involved in translation and all other factors eg initial factors...are involved in the process of translation. DNA polymerase is involved in the process of replication of the DNA.
Answer:anti codons
Explanation:
Mechanical energy is the energy of _____.
Answer:
Mechanical energy is the energy of potential and kinetic energy i.e sum of potential and kinetic energy.
Explanation:
Mechanical energy is the energy posses by an object which enable it to work due to it's potion or movement.
Mechanical energy can be inform of potential energy( energy stored or due to object position) or kinetic energy( object in motion).
Mechanical energy is the sum of potential energy and kinetic energy. This energy is associated with object's position or motion.
One advantage to the Palo Verde Nuclear Power station to the ASU West solar array is that the nuclear power station can generate electricity at night, but the solar array does not generate electricity when the Sun is down.
a. True
b. False
The statement 'nuclear power station can generate electricity at night, but the solar array does not generate electricity when the Sun is down' is TRUE.
A nuclear power plant is a station in which the energy from nuclear fission is used to create heat in a nuclear reactor, which is then used to generate electricity.Conversely, a solar station is a large-scale photovoltaic system capable of generating electricity by using solar panels.Solar panels cannot produce energy at night because photovoltaic cells need the light energy from the Sun to generate electricity.In conclusion, the statement 'nuclear power station can generate electricity at night, but the solar array does not generate electricity when the Sun is down' is TRUE.
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You are an epidemiologist in charge of county health services for low income, HIV positive individuals in NJ. Which data would be most important for planning the long-term health care needs of this community, incidence or prevalence?
Answer:
The correct answer is - prevalence.
Explanation:
Prevalence is the number of individidual cases that are alive with the disorder at a particular time frame or particular time. It provides the measure more precise.
The given question is comes under Prevalence as the factor of prevalence of HIV cases was based, on average, on number of infected people with HIV in New Jersey by ethnicity, race and gender for the most recent 1 year time frame.
Thus, the correct answer is - prevalence.
Prevalence is the most important data for planning the long-term health care needs of a low-income, HIV positive community.
Explanation:Prevalence and incidence are two important measures in epidemiology. Prevalence refers to the total number of both new and existing cases in a population over time, while incidence measures the number of new cases of a disease during a specific time period. When planning the long-term health care needs of a low-income, HIV positive community, prevalence would be the most important data to consider. This is because prevalence gives an indication of the overall burden of the disease in the community and helps in estimating the resources and services needed to address the health care needs of the population over time.
After you performed the freeze-thaw cycle to lyse the bacteria, you pelleted the debris using a centrifuge. What two reasons did this pellet fluoresce green? (Yes, I know it contained GFP…. but why?)
Answer:
Green Fluorescent Protein Labelled Bacteria
Explanation:
Green Fluorescent Protein labeling is very useful in studying prokaryotes. It is highly likely that the bacteria was also labelled. That's why after lysis and density level differentiation in a centrifuge the pellet start glowing green.
Why does it make biochemical sense that chaperones recognize hydrophobic surface area? What catastrophic event are chaperones meant to prevent in cells?
Answer:
Molecular chaperons in the cells helps in protein folding. These are the group of proteins that have functional similarity and they also assist protein folding.
They have the ability to prevent the non specific binding and aggregation by the binding of the non native proteins.
Molecular chaperons helps in recognizing the hydrophobic surfaces of the unfolded proteins because they themselves are hydrophobic in nature and will combine to the hydrophic binding and bonding.
This helps in guiding the protein to folding.
Chaperones recognize hydrophobic surface areas to prevent protein aggregation during folding, which could lead to cellular dysfunction. They help proteins fold correctly and prevent catastrophic events such as the formation of protein aggregates.
Explanation:Chaperones recognize hydrophobic surface areas because many proteins require assistance in the folding process to prevent them from aggregating during folding. Hydrophobic regions on the protein's surface tend to be exposed and can interact with other hydrophobic regions, leading to protein aggregation. Chaperones bind to these hydrophobic regions and help the protein fold correctly by preventing aggregation. This prevents catastrophic events such as the formation of protein aggregates, which can lead to cellular dysfunction and disease.
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It can be desirable to produce eukaryotic proteins in prokaryotes such as E. coli. To do this several DNA sequences must be joined or cloned together. Place the labels in the appropriate position to allow for the expression of the insulin protein in a prokaryotic cell. cDNA is DNA that is synthesized from the mature mRNA from eukaryotic cells using the enzyme reverse transcriptase.
5' end_____ _____ _____ _____3' end
1. Poly A signal sequence
2. Genomic clone of insulin
3. Rho terminator
4. -95 CAT box
5. -10 TATA
6. -25 TATA box
7. cDNA of insulin
8. -35 Sequence
Final answer:
To express human insulin protein in E. coli, assemble a DNA sequence that contains a -35 Sequence, a -10 TATA box for bacterial promoter activity, followed by the cDNA of insulin, and ends with a Rho terminator for transcription termination.
Explanation:
To generate a recombinant version of the human insulin protein in a prokaryotic cell, such as E. coli, we need to assemble a proper expression cassette. The cassette needs to have a eukaryotic gene sequence converted to cDNA, combined with prokaryotic promoter and terminator sequences to allow expression in the host cell.
The correct order for cloning the insulin cDNA for expression is:
-35 Sequence-10 TATA box (also known as Pribnow box)cDNA of insulinRho terminator or a similar prokaryotic terminatorThe -35 Sequence (-35 Sequence) and the -10 TATA box (-10 TATA) are part of the bacterial promoter necessary for initiating transcription. The cDNA of insulin (cDNA of insulin) is inserted downstream of the promoter, allowing it to be transcribed. Finally, a termination sequence such as the Rho terminator is necessary to stop transcription.
Why these specific elements? The -35 and -10 sequences are binding sites for the RNA polymerase in prokaryotes, which begins the process of transcription. The cDNA of insulin, obtained through reverse transcription of the mature mRNA, excludes any introns that would be present in the genomic clone (thus, we do not use the Genomic clone of insulin or the Poly A signal sequence which are found in eukaryotic expression systems). The Rho terminator is a sequence that facilitates the termination of transcription in prokaryotes.
Many chemotherapy drugs target rapidly dividing cells, such as cancer cells. Why might cancer cells divide, and therefore evolve, more quickly than other cells
Answer:
The faster that cancer cells divide, the more likely it is that chemotherapy will kill the cells, causing the tumor to shrink. They also induce cell self-death or apoptosis. Chemotherapy drugs that kill cancer cells only when they are dividing are called cell-cycle specific.
Explanation:
Given that the mean cuteness of the current herd is 39.7 OMGs, the BAF workers picked the 16 cutest adults and allow them to reproduce with each other. The mean cuteness of these chosen individuals is 41.5 OMGs. What is the predicted mean cuteness of the next generation of alpacas in the BAF environment?
a. 39.9
b. 40.4
c. 41.5
d. 42.5
Answer: 41.5 OMGs is the predicted mean cuteness in the next generation.
Explanation:
Cuteness is a way to measure relative ability of individuals with a certain genotype to reproduce successfully.
Mean cuteness is given as the summation of individual cuteness. Mean cuteness also changes in the next generation.
The mean cuteness of the current herd was given as 39.7 OMGs and after 16 individuals were chosen, the mean was given as 41.5.
This implies that there was a change already as mean cuteness will either increase or decrease in the next generation after selection. Here, 39.7 increased to 41.5 .
The predicted mean cuteness is therefore 41.5 OMGs.
Do we expect a son/daughter to have a trait if the trait is inherited in the following way (for each, please answer Y/N/P/W for yes, no, possibly, or ‘what!? – that doesn’t make sense!!’). Please fill out the grid for each type of inheritance. For example, a X-linked dominant female-limited trait would look like this:
Answer: incomplete
Explanation:
Restriction enzymes, which are extensively used in molecular biology, are products of bacterial "immune" system. Since bacterial genomes span several million base pairs (E. coli > 4 million bps), and the presence of restriction site within a genome is more than likely, how does a bacteria manage to protect itself from innate REs?
Explanation:
In the term of disease by intracellular microscopic organisms, they have the ability and repeated inside phagocytic cells, which causes the circulating antibodies to be distant to intracellular bacteria The natural safe reaction against these microscopic organisms is intervened basically by phagocytes and NK cells Innate immunity additionally arrives in a protein substance structure, called innate humoral immunity. for example, the body's supplement system and substances called interleukin-1 (which causes fever) and interferona) There are abiotic and biotic factors that keep population size of species under control Provide two examples that affect (or have affected) population size of native species in WA (one example for density dependent and one for density independent). You need to pick two real examples, provide specific species names and habitats/ecosystems where those species occur. You need to use different examples from the ones I used in lectures
b) Explain why species that overlap a great deal in their fundamental niches have a high probability of competing. Now explain why species that overlap a great deal in their realized niches and live in the same area, probably do not compete significantly.
c) What are two important pieces of ecological knowledge that you could apply to the decisions you make about your own life? Mainly from an ecological point of view, but it is good that you make a connection with other aspects of your life (personal or professional).
Answer:
Answer A: A key example of biotic and abiotic protagonists is the presence of rocky structures that exist in the mountains of southern Argentina (Patagonia) that thanks to them exist the ideal temperatures for the procreation of animals such as the Patagonian condor and the possibility of form nests in these rocky structures that secure their young.
Answer B:
The explanation for this is that they have populations that increase in number and have their niches realized and constant resources necessary to develop the species in this way, considering the dominant species in the area and the one that predominates, in the case of fundamental niches. , they are necessary and even obligatory niches for many populations, which if populations do not go to them, they die, extinguish, or decrease their number, that is why in the face of this reality made up of a fundamental niche since it is fundamental to persist life of the population.
Answer c:
Two key principles are: one, maintain the equilibrium, that is why one seeks to have an environmentalist position preserving nature, and two considering that we are part of that balance but that does not give us the right to break it with environmental contamination, intervention of natural food systems and many other factors.
Explanation:
In the "arms race" between plants and herbivores, herbivores (insects) can overcome the toxic effects of plant specialized metabolites through natural selection. Which of the following would be an adaptation that would promote insect herbivory?
Plants express a novel gene that encodes for an inhibitor of larval development.
Insects express a gene coding for a new enzyme that degrades the plant toxin.
Plants develop a mutation that promotes the development of hairs and spines.
Insects evolve new taste receptors that trigger an avoidance reaction to the toxin.
The following that be an adaptation that would promote insect herbivory is :
D) Insects evolve new taste receptors that trigger an avoidance reaction to the toxin.
"Adaptation of Insects"The following that be an adaptation that would promote insect herbivory is the Insects evolve new taste receptors that trigger an avoidance reaction to the toxin.
Plants are regularly fed upon herbivorous creepy crawlies. to get security against such assaults plant has created a few instruments like generation of poisons and unstable substance that either slaughter the bother or pull in the normal adversaries of the creepy crawlies respectively.
Thus, the correct answer is D.
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Final answer:
An adaptation promoting insect herbivory in the plant-herbivore arms race is insects evolving an enzyme that degrades plant toxins, allowing them to counteract plant defenses and feed on previously toxic plants.
Explanation:
In the "arms race" between plants and herbivores, an adaptation that would promote insect herbivory is insects expressing a gene coding for a new enzyme that degrades the plant toxin. This represents a direct adaptation by the herbivores to overcome the plant's defenses, enabling them to feed on plants that would otherwise be toxic to them. The coevolution observed in nature often results in such adaptations, where insects evolve mechanisms to detoxify or evade plant defenses, such as specialized enzymes that can neutralize toxic compounds. This form of adaptation allows insects to exploit a wider range of host plants, giving them a selective advantage in habitats where their preferred food sources might be scarce or heavily defended.
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Linda wrote a few steps to describe how carbon circulates between the atmosphere and living organisms. Step 3 is missing.
Step 1: Carbon enters the atmosphere as carbon dioxide from respiration and combustion
Step 2: Plants use carbon dioxide to make glucose
Step 3:
Step 4: Animals exhale carbon dioxide
Which of the following best describes step 3?
a. Dead animals are buried underground and form fossil fuels.
b. Carbon present in glucose is converted into oxygen by animals.
c. Carbon present in glucose is transferred into plant-eating animals.
d. Dead animals decompose and recycle the carbon in the form of carbon dioxide.
Answer:D
Explanation:decomposition of dead plants and animals.
Answer:
c
Explanation:
edit: yup its c i just did the test and got it right
Case Study: Parents of a 3-year-old boy noticed that their son had a waddling gait, fell frequently and had difficulty getting up again, and was not able to run. By age five, there was progressive muscular weakness and muscle wasting, and weakness of the trunk muscles led to abnormal posture. By age 9 he was confined to a wheelchair. Contractures appeared, first in the feet, as the gastrocnemius muscles tightened.1. This hereditary X-linked recessive disease characterized by progressive muscular weakness is ______.2. What does dystrophy mean? Why is this term used to describe this case?
Answer:
Muscular dystrophy
Explanation:
Muscular dystrophy is an inheritable genetic condition that involves mutations in genes that encodes the production of muscle proteins to build and to preserve healthy proteins. The disease is characterised by progressive weakness and loss of muscle mass.
Dystrophy is referred to as a disorder in which the a tissue or an orgasm progressively wastes away. Thé dystrophy term is used because the disease involves the gradual and progressive wasting away of the muscle which is a type of tissue causing muscular weakness allowing the child to experience these symptoms.
Have your partner focus on an object on the far side of the room (e.g., the chalkboard or a chart) for 1 minute. Then have your partner switch focus to an object in your hand (e.g., a pencil). Watch your partner's pupils carefully as the point of focus is changed from far to near.
Answer:
Eye changes or adjusts itself according to the focus of our eye.Explanation:
Pupil changes its shape when the eye focuses on near or far object. When we observe something far away, the pupil gets widen. But when we look at something near to us, pupil get smaller. So, pupil adjust itself according to the focus of our eye. If the partner switches from far to near, the pupil changes from wide to small. Pupil changes its shape according to bright as well as dim light also.
A 35-year-old man presents with anemia, neutropenia, thrombocytopenia, myeloblasts with the presence of Auer rods, and one or two distinct nucleoli and promyelocytes. Cytochemistry examination demonstrates peroxidase and Sudan black B (SBB) positive and TdT terminal eoxynucleotidyl transferase (TdT) negative. This hematologic picture is consistent with: A) Acute lymphoblastic leukemia (ALL) B) Acute myeloblastic leukemia (AML) C) Chronic myelocytic leukemia (CML) D) Chronic lymphocytic leukemia (CLL) E) None of the above
Answer:
B) Acute myeloblastic leukemia (AML
Explanation:
It is a type of cancer that affects the blood as well as the bone marrow. It causes a situation whereby there is excessive immature white blood cells thus affecting the production of matured or normal white and red blood cells, blood platelets through the interference of the myeloid cells.
It is treatable medical condition that requires medical diagnosis. It requires laboratory tests with imaging always needed. It is also known as Acute myelogenous leukemia, acute granulocytic leukemia and acute nonlymphocytic leukemia.
Answer: The hematologic picture is consistent with Acute myeloblastic leukemia (AML). Therefore the correct option is B.
Explanation:
Leukaemia is a type of cancer that affects the blood forming tissues such as the bone marrow, leading to excessive or over production of abnormal blood cells ( usually the white blood cells). They occur in two categories; it can be Acute leukaemia or Chronic leukaemia.
There are different forms of acute leukaemia which is classified according to the lineage ( myeloid or lymphoid). They include:
- Acute Myeloblastic Leukaemia (AML)and
- Acute Lymphoblastic Leukaemia ( ALL).
In AML, the following Laboratory diagnosis differentiates it from other leukaemia, these includes:
- myeloblasts with the presence of Auer rods, and one or two distinct nucleoli and promyelocytes, and
-peroxidase and Sudan black B (SBB) positive and TdT terminal eoxynucleotidyl transferase (TdT) negative. Therefore the correct option is B
Meselson and Stahl designed an experiment that would allow them to discern whether DNA replication occurs in a dispersive, semiconservative, or conservative manner. They started with E. coli that had been growing for many generations in medium containing 15N. They then transferred the bacteria into medium containing only 14N, and allowed the bacteria to undergo two rounds of DNA replication. After each round of replication, the scientists performed density-gradient centrifugation of the DNA. The scientists reasoned that each of the three models would predict different DNA banding patterns after the two rounds of replication. Can you identify the banding patterns predicted by each model after the first round of replication
After the first round of replication in the Meselson-Stahl experiment, conservative replication predicts two bands -- one heavy and one light. Semiconservative replication predicts one intermediate band, as does dispersive replication.
Explanation:The Meselson-Stahl experiment was designed to test the three theories of DNA replication: dispersive, conservative, and semiconservative. After the first round of replication in 14N medium:
Conservative replication would predict two bands -- one at the heavier 15N density and one at the lighter 14N density. This represents the original parent 15N DNA and the new 14N DNA separately. Semiconservative replication would predict one band at an intermediate density. This represents the hybrids of one strand of 15N-parent DNA and one new 14N strand. Dispersive replication would also predict one band at an intermediate density, though the model suggests that each individual DNA strand would be a mixture of old and new DNA, unlike the semiconservative model. Learn more about Meselson-Stahl experiment here:
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Imagine that zika virus has a 1% incidence in the population. A test for the virus has a 3% false positive rate and no false negative rate. If someone takes the test and gets a positive result, what is the chance that they are infected?
Answer:
There is 97% Chance that the person is infected.
Explanation:
According to the question, the test for the virus has 3% false positive and there is no false negative rate.
Therefore if someone takes the test and gets a positive result, the chances of the person to be infected ( True positive ) is
100 - 3 = 97%
Note: False positive result is a result that indicates that a given condition is present when it is not actually present.