Answer:
Explanation:for a mendelian cross between a Talk(D) and a dwarf individual (d)
DD×dd= Dd Dd Dd Dd
The offspring will be heterozygote all loci i.e all the offspring will be tall this is because one of the parents carries a dominant gene for tallness and the other parent carries a recessive gene for tallness.
This may not follow for the progeny's offspring as segregation will occur at the 2nd filial generation.
The same principle applies for violet and white, progeny will be heterozygous at all loci i.e violet color
When Mendel crossed a plant homozygous for round seeds to another plant homozygous for wrinkled seeds, he found that all the progeny had round seeds. How is this explained?
A) The parent that was homozygous for round seeds underwent self‑pollination.
B) The allele for round seeds is recessive to the allele for wrinkled seeds.
C) The progeny were homozygous for the allele for round seeds.
D) Segregation of alleles in the two parents produced gametes with both alleles.
E) The allele for round seeds is dominant to the allele for wrinkled seeds.
Answer:
E
Explanation:
Gregor Mendel discovered the principles that governs heredity. In one of his experiments, he discovered that an organism receives two forms of a gene called ALLELE from each parent. He realized that one allele is capable of masking the expression of its variant pair in a gene. He called the allele that masks or is expressed, DOMINANT allele while the allele that is masked, RECESSIVE allele. He termed this principle the LAW OF DOMINANCE.
The above explained law of dominance is what applies in the question here. When the homozygous round allele and wrinkled allele were crossed, the allele for round seeds are dominant over the allele for wrinkled seeds (recessive) i.e. in a heterozygous state (combination of the different alleles), the round allele will mask the phenotypic expression of the wrinkled allele, expressing itself over it.
Measles is a disease that causes fever and rash. The graph gives the number of measles cases in the United States between 1954 and 2008. The measles vaccine was introduced in the 1960s. What can you conclude using the information in the graph? a graph giving the number of measles cases in the United States between the years 1954 and 2008 shows a high around 775,000 cases in 1958 and a low of 0 cases in 2008
A. Measles is the leading cause of fevers for children under age 5 in the United States.
B. Visitors who are not vaccinated continue to spread measles in the United States.
C. The measles vaccine had little effect on the population of the United States. D. The measles vaccine eliminated measles in the United States.
Answer:D
Explanation:
Answer:
D. The measles vaccine eliminated measles in the United States.
Explanation:
Just took the test and got it right. :)
A researcher is comparing the sequences of genes encoding cell wall proteins in archaeons with those coding cell wall proteins in bacteria. How similar do you expect these sequences will be?
Answer:
These sequence with not be similar because, the archaeons cell wall lack peptidoglycan.
Explanation:
The peptidoglycan is made up of peptide chain and polysaccharide. They are mostly found on bacterial cell. It gives structure to the organism and provide strength to the outer structure of the organism as well.
The cell wall of most archae are composed of S-layer protein. The cell wall also composed of other sugar proteins like pseudomurein, heteropolysaccharide and methanochondriotin.
In this case, the sequences will be different , simply because of the difference in the composition of their cell walls.
Final answer:
Archaeons and bacteria have some similarities in their cell structures, but there are distinct differences, such as the composition of their cell walls and plasma membranes.
Explanation:
Archaeons and bacteria have some similarities in their cell structures, such as containing a cell wall, cell membrane, nucleoid region, and often a capsule, flagellum, and pili. However, there are distinct differences between the two. One major difference is that bacteria have cell walls made up of peptidoglycan, while most archaeons do not have peptidoglycan in their cell walls.
Another difference is that the plasma membranes of archaeons are made up of lipids that are different from those in bacteria. Additionally, the ribosomal proteins of archaeons are more similar to those in eukaryotic cells, not those in bacteria.
Why is it helpful for scientists to compate substrate samples in the arson case discussed in the video
Answer:
because they need to check if the burnt substrate had a gasoline or other flammable material.
Explanation:
Scientists need to compile substrate samples in the case of arson, because by analyzing the burnt substrate and analyzing the deepest substrate (which may not be burned), scientists can see if the fire started naturally or if it had human influence . This is because, if in the burned substrate they find substances of flammable material, such as gasoline, they will be able to prove that the fire was initiated by the action of human beings on purpose.
Sandy has a huge crush on Casey. When he is nearby, Sandy doesn't pay attention to anything or anyone else. Psychologist Donald Broadbent explained that we selectively attend to the most important information in _________ model he developed.
Answer:
The most important information in this model he developed is the "filter theory."
Explanation:
Filter theory is a description of desirability planned by Kerchoff and Davies (1962). This concept proposes that individuals advance associations by smearing a sequence of filters, such as resemblance of communal demographic issues and arrogance and complementary of wants to slender down the pool of obtainable candidates. Filter theory is a sociological theory concerning dating and mate selection. It proposes that social structure limits the number of eligible candidates for a mate. Most often, this takes place due to homogamy, as people seek to date and marry only those similar to them.
There are two different alleles for flower color, P and p. The image shows a white sweet pea that is labeled with its two alleles for petal color.
A white sweet pea is labeled p p.
What can you surmise from the labeled image?
A.The genotype is heterozygous.
B.White petal color is a recessive trait.
C.White petal color is a dominant trait.
D.The phenotype is heterozygous.
B. White petal colour is a recessive trait.
Explanation:
In Mendelian genetics, the alleles depicted in small letters are recessive traits.
Since both alleles are same and in lower case they are homozygous recessive alleles for a white coloured petal gene.
The genotype of the trait will be tt because if it is crossed either with heterozygous or homozygous P alleles it will appear as tt to have phenotypic trait of white colour.
So the genotype will be tt.
An 8 hour exposure to a sound intensity level of 95.0 dB may cause hearing damage. What energy in joules falls on a 0.850 cm diameter eardrum so exposed?
Answer:
[tex]E=5.0662\times 10^{-3}\ J[/tex]
Explanation:
Given:
time of exposure of eardrum to the specific sound, [tex]t=8\ hr=28800\ s[/tex]intensity of the sound, [tex]\beta=95\ dB[/tex]diameter of the eardrum, [tex]d=0.85\ cm=0.0085\ cm[/tex]We have the relation between the flux density of the sound energy as:
[tex]\beta=10\log_{10}(\frac{I}{I_0} )[/tex] ................(1)
where:
[tex]I_0=[/tex] the minimum flux density of sound energy just audible to human ears [tex]=10^{-12}[/tex] [tex]W.m^{-2}[/tex]
[tex]I=[/tex] the flux density of the sound energy due to the given intensity of sound
[tex]\beta=[/tex] given intensity of sound in decibels
from eq. (1) we've:
[tex]95=10\times \log_{10} (\frac{I}{10^{-12}} )[/tex]
[tex]I=0.0031\ W.m^{-2}[/tex]
The above value is Power per unit area.
We now find the area of eardrum:
[tex]A=\frac{\pi.d^2}{4}[/tex]
[tex]A=\frac{\pi\times 0.0085^2}{4}[/tex]
[tex]A=5.67\times 10^{-5}\ m^2[/tex]
Now the energy reaching the eardrum per second is:
[tex]P=I\times A[/tex]
[tex]P=0.0031\times 5.67\times 10^{-5}[/tex]
[tex]P=1.7591\times 10^{-7}\ W[/tex]
Now the total energy reaching the eardrum in the given time:
[tex]E=P.t[/tex]
[tex]E=1.7591\times 10^{-7}\times 28800[/tex]
[tex]E=5.0662\times 10^{-3}\ J[/tex]
What are the functions of each cell part:
1. Cytoplasm
2. Cell Membrane
3. Nucleus
4. Endoplasmic Reticulum
5. Ribosome
6. Mitochondria
7. Lysosome
8. Golgi bodies
9. Vacuole
10 Chloroplast Cell wall
Answer:
Each part perform specific function for cell.
Explanation:
1. Cytoplasm ; Site for various chemical reactions.
2. Cell Membrane ; Protective covering; Material transport regulation.
3. Nucleus ; Control center of cell; Constitute genetic material
4. Endoplasmic Reticulum ; Material processing; movement inside the cell
5. Ribosome ; Protein synthesis
6. Mitochondria ; power house of cell
7. Lysosome ; Digestion, break down of substances.
8. Golgi bodies ; transport of material and packaging.
9. Vacuole ; waste & water storage.
10 Chloroplast Cell wall; protection, transportation, Photosynthesis.
The functions of each cell part include: Cytoplasm is a gel-like substance that fills the cell and acts as a medium for cellular processes. The Cell Membrane controls the movement of substances in and out of the cell. The Nucleus contains DNA and controls the activities of the cell.
Explanation:Cytoplasm: It is a gel-like substance that fills the cell and acts as a medium for cellular processes.Cell Membrane: It controls the movement of substances in and out of the cell, providing protection and support.Nucleus: It contains DNA and controls the activities of the cell.Endoplasmic Reticulum: It is responsible for the synthesis and transport of proteins and lipids.Ribosome: It is the site of protein synthesis.Mitochondria: They produce energy in the form of ATP through cellular respiration.Lysosome: They contain enzymes that break down waste materials and foreign substances in the cell.Golgi bodies: They modify, sort, and package proteins for transport within and outside the cell.Vacuole: It stores water, nutrients, and waste materials in the cell.Chloroplast: It is found only in plant cells and is responsible for photosynthesis.Cell wall: It provides strength, support, and protection to plant cells.Learn more about Cell Parts Functions here:https://brainly.com/question/12553639
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true or false? Stratified squamous epithelium forms the surface of the skin, the lining of the mouth, and the lining of the esophagus.
Answer: True
Explanation:
The type of epithelium well suits the areas where there is a constant abrasion, and due to this fact the toughest and the thickest layer can be sequentially sloughed off and will be replaced before the layers in the basement will be exposed.
It is the outermost layer of the skin, lining of the mouth, vagina and esophagus.
The given statement is true, that the area like esophagus, vagina and mouth are covered with stratified squamous epithelium.
The ____ monitors all the information about eye, head, and body positions and passes it on to brain areas that control movement. Select one: a. occipital lobe b. central sulcus c. precentral gyrus d. parietal lobe
Answer: Parietal lobe
Explanation:
The parietal lobes are one of the four main lobes of the cerebral cortex. It is positioned above the temporal lobes and behind the frontal lobes.
These lobes serves many important functions such as functioning and processing of the sensory information, body's awareness and spatial orientation.
This lobe processes the information about eye, head and body position.
Hence, the correct answer is parietal lobe.
The parietal lobe is responsible for monitoring the positions of eyes, head, and body, integrating sensory information, and contributing to motion perception and control.
The structure that monitors all the information about eye, head, and body positions and passes it on to brain areas that control movement is the parietal lobe. The parietal lobe interacts with the somatosensory cortex and is involved with motion perception, responding to both visual and vestibular motion cues. Cortical regions like the posterior parietal cortex and other areas receive vestibular signals which are essential for spatial orientation and navigation, coordinating movement with the visual perception of space.
Two populations of birds with somewhat different coloration live on opposite sides of a peninsula. The habitat between the populations is not suitable for these birds. When birds from the two populations are brought together, they produce young whose appearance is intermediate between the two parents. These offspring will breed with each other or with birds from either parent population, and all offspring of these pairings appear intermediate to various degrees.What keeps the two populations separate?A. temporal reproductive isolation B. lack of hybrid viability C. behavioral reproductive isolation D. spatial reproductive isolation
Answer: D. spatial reproductive isolation
Explanation:
The spatial reproductive isolation can be defined as the reproductive isolation in which populations are separated by great distances, this can also occur when the members of the same population inhabit to different parts of the same area.
According to the given situation, the birds of the two populations have been separated by the peninsula. This is creating a spatial reproductive barrier between the two populations. When the members of the spatially separated populations brought together they will not be able to produce offspring. Due to the fact that the members of the population might have developed traits which get expressed in the progeny. Thus the offspring appears to be the intermediate of the two.
Meiosis guarantees that in a sexual life cycle, offspring will inherit one complete set of chromosomes (and their associated genes and traits) from each parent. The transmission of traits from parents to offspring is called heredity. Another important aspect of meiosis and the sexual life cycle is the role these processes play in contributing to genetic variation. Although offspring always resemble their parents, they are genetically different from both of their parents and from one another. The degree of variation may be tremendous. The following processes are associated with meiosis and the sexual life cycle: DNA replication before meiosis crossing over chromosome alignment in metaphase I and separation in anaphase I chromosome alignment in metaphase II and separation in anaphase II fertilization Sort each process into the appropriate bin according to whether it contributes to heredity only, genetic variation only, or both. (Note that a bin may be left empty.)
Answer:
HEREDITY ONLY:
none
GENETIC VARIATION ONLY:
none
BOTH:
all
Explanation:
In all the organism that reproduce sexually, produced offspring are genetically varied. As this genetic variation continues, it also leads to hereditary evolution of organism through gene flow from one generation to other. Therefore, the processes of DNA replication, crossing over, chromosome separation and alignment etc. ensure flow of genes from one generation to the next. This does not happen in organisms reproducing asexually.
Final answer:
Meiosis and fertilization are essential for heredity and genetic variation in sexual reproduction. Processes like DNA replication contribute solely to heredity, while crossing over and random alignment during meiosis promote genetic variation. Fertilization brings together both aspects by mixing parental genetic material.
Explanation:
In sexual reproduction, meiosis and fertilization play crucial roles in ensuring heredity and generating genetic variation among offspring. DNA replication before meiosis ensures that each chromosome is copied to prepare for division, contributing to heredity by preserving genetic information. Crossing over during prophase I and the random alignment of chromosomes during metaphase I are processes that introduce genetic variation.
The alignment and separation of chromosomes during metaphase II and anaphase II also contribute to heredity as they ensure each gamete receives a complete haploid set of chromosomes. Finally, fertilization restores the diploid chromosome number and combines genetic material from both parents, contributing to both heredity and genetic variation.
DNA replication: Heredity only Crossing over: Genetic variation onlyChromosome alignment in metaphase I and separation in anaphase I: Genetic variation onlyChromosome alignment in metaphase II and separation in anaphase II: Heredity onlyFertilization: Both heredity and genetic variationYou observe two groups of frogs that are closely related. Males from each group "sing" with a species-specific song to attract a female frog. Females only respond to songs sung by males of their species. This is an example of _____ separation and _____ reproductive isolation between two species.
Answer:behavioral pre
Explanation:because their used to thier kind
7.
What are the 3 checkpoints normal cells go through to prevent unregulated cell division?
Answer:
The three cell cycle checkpoints are:
G1 CheckpointG2 CheckpointSpindle CheckpointExplanation:
Cell Cycle Checkpoints:
Cell cycle checkpoints are regulatory points at which the current state of the cell is assessed to determine whether the cell cycle can continue or not.
The G1 checkpoint confirms the cells entry into the mitotic phase; it commits the cells for division. If a cell passes G1 checkpoint, it will continue in the cell cycle until it divides into two daughter cells. This checkpoint checks DNA damage, the presence of adequate nutrients and suitable cell size. The G2 checkpoint operates at the G2/M transition. It checks DNA damage and assesses whether DNA has completely and successfully replicated during the S-phase. If DNA damage is detected, DNA repair mechanisms are activated.The spindle checkpoint assesses whether all chromatids have successfully attached to the spindle fibers. This is a crucial checkpoint as the integrity of cell division depends on the correct number of chromosomes received by each daughter cell.Indicate the oxygen requirements of the microorganisms growing in each thioglycolate tube. The surface of the tube is exposed to oxygen and is aerobic. Oxygen concentration decreases with depth; the bottom of the tube is anaerobic. Drag the appropriate labels to their respective targets.
In a thioglycolate tube, aerobic microorganisms grow at the surface where it is oxygen-rich, anaerobic microorganisms grow at the bottom where it is oxygen-poor, facultative anaerobes can grow throughout the tube and microaerophiles grow below the surface in low oxygen conditions.
Explanation:The oxygen requirements for microorganisms can vary. In a thioglycolate tube, the surface of the tube is aerobic because it is exposed to oxygen, meaning aerobic microorganisms grow here. As the depth increases in the tube, the oxygen concentration decreases, therefore anaerobic microorganisms grow at the bottom part of the tube since they do not require oxygen for growth.
There are also facultative anaerobes which can grow throughout the tube as they can adjust to both oxygen-rich and oxygen-poor environments. Finally, microaerophiles that require low levels of oxygen can be found growing a little below the surface where the oxygen concentration is lower than at the surface but higher than at the bottom.
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DDT was once considered a "silver bullet" that would permanently eradicate insect pests. Today, instead, DDT is largely useless against many insects. What would need to be true for pest eradication efforts to have been successful in the long run?
Answer:
4. None of the individual insects should have possessed genomes that make them resistant to DDT
Explanation:
DDT became ineffective in the eradication of insect pests because insects evolved by natural selection.
Some insects had genes that made them to be resistant to the effects of DDT. Hence, these few usually survive whenever DDT is applied to their environments. The few surviving ones thus reproduce in the environment, pass the DDT-resistant gene on to their offspring and give rise to a population of insect that are resistant to DDT.
Hence, if none of the earlier insects possessed the DDT-resistant gene, it would not have been possible for the insect population to evolve and pest eradication efforts using DDT would have been successful in the long run.
The correct option is option 4.
DDT was once considered a "silver bullet" that would permanently eradicate insect pests. Today, instead, DDT is largely useless against many insects, the option that needs to be true for pest eradication efforts to have been successful in the long run - All individual insects should have possessed genomes that made them susceptible to DDT.
DDT is a pesticide used to eradicate pests but due to its overuse pest developed genes that are resistant to DDT and it became ineffective as insects evolved by natural selection.
The genes that help them to make them be resistant to the effects of DDT would always help them survive whenever DDT is applied to their environments.If all the insects in this environment have only DDT susceptible genes in their genome then DDT would be able to eradicate them as they lack the resistant gene.If there will be no resistant gene then DDT would be effective in the long run.Thus, DDT was once considered a "silver bullet" that would permanently eradicate insect pests. Today, instead, DDT is largely useless against many insects, the option that needs to be true for pest eradication efforts to have been successful in the long run - All individual insects should have possessed genomes that made them susceptible to DDT.
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Penicillin’s mode of action is to inhibit Choose one: A. cell wall cross-link formation. B. DNA replication by allosterically binding to DNA polymerase. C. translation by binding to the 50S ribosomal subunit. D. translation by binding to the 30S ribosomal subunit.
Answer:
A. cell wall cross-link formation.
Explanation:
The bacterial cell wall has peptidoglycan which is a heteropolymer of two different monomers. These are N -acetylmuramic acid and N -acetylglucosamine. Peptide cross-links formed by the enzyme transpeptidase join the chains of these monomers together to make multiple layers of the cell wall. Penicillin is an antibiotic that does not allow the enzyme " transpeptidase" to form the cross-linked. The bacteria treated with penicillin lyse in a hypotonic solution.
What happens to contraction of a muscle cell if some of the Ca2+ that was released during a contraction is still in the cytoplasm (sarcoplasm) when the next stimulus arrives? Group of answer choices The muscle contracts with the same tension generated as during the first contraction, because muscles contract in an all-or-none fashion. The muscle contracts with the same tension generated as during the first contraction, because the number of cross-bridges formed is always the same during a contraction. The muscle contracts with greater tension generated because there will be more Ca2+ in the sarcoplasm after the second stimulation, which will bind to more troponin molecules. The muscle contracts with greater tension generated because more troponin molecules bound means greater myosin-binding sites (active sites) are revealed on the actin, leading to a larger number of cross-bridges formed.
Explanation:
Calcium stays in the sarcoplasmic reticulum until discharged by an improvement. Calcium at that point ties to troponin, causing the troponin to change shape and expel the tropomyosin from the coupling destinations. Cross-connect stick proceeds until the calcium particles and ATP are never again accessible. ATP is basic to get ready myosin for official and to "revive" the myosin. When the actin-restricting destinations are revealed, the high-vitality myosin head overcomes any issues, framing a cross-connect. When myosin ties to the actin, the Pi is discharged, and the myosin experiences a conformational change to a lower vitality state. As myosin consumes the vitality, it travels through the "power stroke," pulling the actin fiber toward the M-line.If toxic fumes are generated while performing an experiment, await instructions from your lab instructor; follow them completely and as quickly as possible. true false
Answer:True
Explanation:
The toxic fumes in the laboratory is very common when performing any experiment inside the laboratory.
There are various types of chemical reaction which can fortunately or unfortunately result in evolution of the toxic fumes.
In this case the person should not panic and just listen to the instructor carefully.
Follow it as soon as possible.Carefully and completely empty the place and go to some place which has fresh air.
The difference between the images in the two eyes, or the ______, is the binocular cue the brain uses to see in three dimensions.
Answer:
Disparity
Explanation:
The disparity in vision refers to the imperfect match between the two eyes caused by the different views perceived at a slightly different angle by two eyes.
This difference between the image formed at the retina at a different angle gets fused and the brain perceives the fused image as a single perception by the eyes. This disparity thus allows the brain to analyse the 3D view of the sample and the binocular disparity between image helps determine the depth of the image.
Thus, Disparity is the correct answer.
Which of the following statements is true? a. The right brain helps people analyze and verbalize. b. Right and left brain concepts are meaningless. c. The left brain helps people understand analogies and imagine things. d. The right brain helps people understand analogies and imagine things.
Answer: The answer is D. The right brain helps people understand analogies and imagine things.
Explanation: The brain when divided symmetrically from the middle can be categorized as right and left cerebral hemispheres. The two hemispheres have different functions.
The right cerebral hemisphere is responsible for the coordination of the left body. It is also responsible for the artistic, creative and imaginative functions. Ability to understand analogies and imagine things is a result of a well developed right cerebral hemisphere.
The term secretion refers to Group of answer choices A.the process by which a cell releases a substance into the extracellular space. B.synthesis of a protein for export from the cell. C. the manufacture and assembly of a material. storage of a material, until it is time for it to leave the cell. D.None of these describe secretion.
Answer:
The term secretion refers to the cell that releases a substance into the extracellular space.
Explanation:
When a cell or organ or gland produce and secretes substance in to extracellular space,this biochemical process is called as secretion.
The secretory products may be hormone, enzymes etc.
This process facilitates the movement of material from one place to another place.
For example, Pituitary gland produce and release prolactin hormone that induces breast cells to produce lactation during sucking process done by baby.
Pituitary gland also secrete different hormones that acts on different glands like adrenal, testes, ovary, thyroid gland etc that in turn produce different hormones like adrenaline,testosterone,estrogen,progesterone, TSH,T3, hormone etc.
A hypothetical endangered species of wildflower has been reduced to a single small population in a mountain meadow. A rare early spring blizzard kills all but 3 of the remaining plants, one of which has a rare mutation. This is an example of:
A. genetic drift (bottleneck).
B. disruptive selection.
C. natural selection.
D. stabilizing selection.
E. gene flow.
Answer:
A. genetic drift (bottleneck).
Explanation:
Genetic drift can cause big losses of genetic variation for small populations. In our case, we already had a small population so killing further reduced the genetic variation in the population.
Disruptive selection is not correct because the plants died randomly (particular phenotype were not favored). Natural selection and stabilizing selection are not correct for same reason.
Gene flow is incorrect because we did not have any individuals coming in into the population!
True/False: Western blotting allows the visualization of proteins with the use of antibodies that specifically bind to the proteins of interest?
Answer:
True
Explanation:
Name the possible ways that diseases are spread. Describe how this happens in communities.
Subject: Anatomy and Physiology
In 1944 Avery, MacLeod, and McCarty performed transformation experiments using live, harmless bacteria and extracts from virulent bacteria treated with various enzymes. Which of the following enzymes were used and why? (A) Proteases and RNases to rule out protein and RNA as the transforming factors (B) Lipase (an enzyme that facilitates the breakdown of lipids) to rule out lipoproteins as the transforming factor (C) Kinase (an enzyme that facilitates transfer of a phosphate group from ATP to a substrate molecule) to show that transformation is phosphorylation dependent (D) ATPase to show that transformation is not dependent on ATP
Answer:
Option-A
Explanation:
Avery, MacLeod, and McCarty performed the experiment on the strains of S. pneumoniae used by the Griffith in the reformation experiment.
To know the chemical nature of the transforming principle which was expected to be a protein at that time, the solution was treated with a variety of biomolecule degrading enzyme-like RNases, DNases and proteases.
They found that the treatments with proteases and RNases has no effect on the process of transformation but the treatment with DNAses has stopped the process of transformation as it degraded the transforming principle and confirmed that DNA is the transforming material and is the heredity material.
Thus, Option-A is the correct answer.
Final answer:
Avery, MacLeod, and McCarty used enzymes including (option A) proteases and RNases to degrade proteins and RNA, showing that neither were the transforming principle. When DNA was degraded with DNAase, the transformation did not occur, identifying DNA as the carrier of genetic information.
Explanation:
In 1944 Avery, MacLeod, and McCarty performed transformation experiments to determine the nature of the "transforming principle" that allowed non-virulent bacteria to become virulent. They used specific enzymes to degrade different molecular components of the virulent bacteria. Proteases and RNases were used to rule out protein and RNA as the transforming factors, as treatments with these enzymes did not prevent the transformation. However, when the extracts were treated with DNAase, which degrades DNA, the transformation did not occur. This indicated that DNA was the necessary component for transformation and thus the carrier of genetic information.
It has been suggested that scientists should attempt to genetically engineer a plant to have black leaves so that it would perform photosynthesis more efficiently. What is the likely reason behind this idea?
Answer:
Black color absorb more light that other color.
Explanation:
Photosynthesis is the process in which leaves absorb solar energy and use it to convert CO2 and H2O into chemical energy called glucose(food).
As photosynthesis is dependent on the sunlight energy therefore it has been suggested to scientists that they should genetically engineer a plant to have black leaves as black color absorbs more sunlight than any other color. Black color absorb all the wavelength of sunlight and it does not reflect any.
So more light absorbance capability of black color should be the reason behind this idea.
Scientific thinking: what clues about biology might road kill provide?
Answer:biology will provide what animals eat road kill
Explanation:because road kill gets eaten
Roadkill can provide several clues about biology and the ecosystem. It helps to determine the species that live in a specific location by looking at roadkill.
Scientists can ascertain the presence and distribution of several species in a specific area by determining the characteristics of the animal, such as its physical qualities, size, and markings.
Population Density and Distribution: Information on the population density and geographic spread of specific species may be gleaned from the frequency of roadkill incidences.
High levels of roadkill incidences may be a sign of increased animal activity or possible habitat fragmentation in certain locations.
Migration Patterns: Roadkill might provide information about an animal's migratory patterns. Identifying the seasonal migrations of particular species, including their routes and timing, may be done by tracking roadkill incidences over time and across different locales.
Impact of Human Activity: Roadkill can bring attention to the effects of human activities, such as building roads or urban development, on wildlife populations.
Roadkill occurrences that occur often may be a sign of growing human-wildlife contact or regions where animal habitats and roadways cross.
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Two alleles of a given gene exist in a population: CTGT and TGTC. (Note: The codes given are only for the transcribed strand of the DNA in the two alleles.) Three possible single crossover events are possible between these two alleles. As a result of the three different crossover events, how many total new alleles (new sequences of DNA) could be produced?
A. 3.
B. 4.
C. 5.
D. 6.
E. 12.
Explanation:
D. 6
There are two forms of the gene, CTGT and TGTC, called alleles. These have undergone 3 crossover events, which may create two separate copies.
Thus 2 × 3 (# of events)= 6 new alleles
The interchange of chromosome segments, including homologous non-sister or identical chromatids. Crossing over happens at chiasmata, where non-sister chromosomes are fused together.
Further Explanation:
DNA variants on chromosomes,may have different forms called alleles. DNA, which is a genotype, is transcribed into mRNA and then converted into amino acids that are linked together by rRNA to form proteins that make up the morphology of the individual.
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Final answer:
The total number of new alleles that could be produced from three possible single crossover events between CTGT and TGTC alleles is 6.
Explanation:
The total number of new alleles that could be produced from three possible single crossover events between CTGT and TGTC alleles is 6.
To calculate this, we need to consider all possible combinations resulting from each crossover event:
New allele 1: CTGT - TGTC
New allele 2: CCGT - TTTC
New allele 3: TCGT - CGTC
New allele 4: CTTC - TGCC
New allele 5: TTGT - CGTC
New allele 6: CCGT - TGTC
Therefore, a total of 6 new alleles can be produced from these crossover events.
Which of the following ancillary or supportive criteria is addressed in this statement: Infants who are put to bed prone less than half the time have an increased incidence of SIDS compared to those who are never put to bed prone. Those who are always put to bed prone have an increased incidence of SIDS compared to those who are put to bed prone less than half the time.
A. Consistency.
B. Biological plausibility.
C. Strength of the association.
D. Dose-response relationship.
Answer:
The answer is D- Dose-response relationship
Explanation:
A dose-response relationship is an association between dose and incidence in which an increasing level of exposure will either lead to an increase or decrease of an incident or outcome. For instance, in the scenario given the dose or exposure will be the time the infants are put to bed prone while the response is the level of incidence of SIDS (Sudden Infant Death Syndrome).
Final answer:
The statement refers to the dose-response relationship in the context of SIDS incidence related to infants being put to bed prone.
Explanation:
The statement you’ve provided addresses the dose-response relationship, which is option (D). This criterion refers to the association between the frequency or intensity of exposure to a certain factor and the likelihood of a particular outcome. In this case, the factor is infants being put to bed prone (on their stomachs), and the outcome is the incidence of Sudden Infant Death Syndrome (SIDS). The statement suggests that there is a dose-response effect, where the more frequently infants are put to bed prone, the greater their risk of SIDS. Those put to bed prone less than half the time have an increased incidence compared to those who are never put to bed prone, and those who are always put to bed prone have an even higher incidence of SIDS.