Answer:
D preventing a lake from freezing over
Answer: Option D) preventing a lake from freezing solid
Explanation:
This is based on the difference in density of water in solid form as against it's liquid form
Ice, which is water in solid form is less dense to water as liquid water. Hence, ice floats on liquid water, thereby preventing the total freezing of the lake which would lead to death of aquatic organisms living in polar regions
Using your knowledge of DNA recombination events to complete the following: Propose two ways in which antibiotic resistance may develop in a bacterium Describe how bacterial cells acquire the ability to produce toxins (Use the following terminology in your answer: recombination, DNA, horizontal gene transfer, conjugation, transformation, transduction, pilus, F factor, transposable elements, transposons, pathogenicity islands)
Answer:
A) Propose two ways in which antibiotic resistance may develop in a bacterium?
Antibiotic resistance may develop in a bacterium through A GENETIC MUTATION and BY REQUIRING RESISTANCE FROM ANOTHER BACTERIUM
B.) Describe how bacterial cells acquire the ability to produce toxins?
The virulent strains of bacteria as in Corynebacterium diphtheria, and Streptococcus pyogenes all manufacture toxins with weighty physiological impacts, unlike the nonvirulent strains which do not manufacture toxins. The toxins are generated by a bacteriophage gene that has been received by transduction.
Using the knowledge of DNA recombination events to complete the -
two ways in which antibiotic resistance may developbacterial cells acquire the ability to produce toxins1. Antibiotic resistance means some bacteria can grow and survive in the presence of one or more antibiotics like tetracycline, ampicillin, or others. It can be developed by 1) horizontal gene transfer 2) Mutation
Horizontal gene transfer is a process that helps bacteria to transfer genes with each other.It is called horizontal gene transfer due to genetic exchange/ F factor, this process is also known as recombination.There are three methods of recombination: Conjugation, Transduction & Transformation.A mutation is a change in the genome of the organisms due to various reasons.2. Pathogenic bacteria may produce toxins. Toxins are of two types; exotoxins & endotoxins.
Exotoxin is released by lipopolysaccharides and protein which are associated with the bacterial cell wall.Endotoxins are associated with the structural mechanism of the bacterial cell. Pathogenicity islands are acquired by microorganisms by horizontal gene transfer.Thus, the explanation is given above about the way resistance develops and the production of toxins.
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Was the decrease in the frequency of bb individuals between successive generations always the same? Why or why not?
Answer:
The frequency of bb individuals between successive generations wasn't always the same because of natural selection. The capacity of reproduction decreased by a small portion.
Explanation:
Natural selection is the reason why the reproduction of bb generations wasn't always the same. The randomly selection of mates to reproduce was also variable.
Final answer:
The decrease in the frequency of bb individuals between successive generations is not consistent due to genetic drift, natural selection, and population size variations. Factors such as the survival advantage/disadvantage of the bb genotype and the population size significantly influence these changes, making the Hardy-Weinberg Equilibrium principle's assumption of constant allele frequency not always applicable.
Explanation:
The decrease in the frequency of bb individuals between successive generations is not always the same due to factors such as genetic drift, natural selection, and population size. Genetic drift, particularly in small populations, can lead to significant fluctuations in allele frequencies over time. Natural selection can either increase or decrease the frequency of bb individuals depending on whether the bb genotype confers a survival advantage or disadvantage. Additionally, the size of the population plays a crucial role; smaller populations are more susceptible to changes in allele frequency due to random events.
Therefore, the Hardy-Weinberg Equilibrium principle's assumption about constant allele frequency does not always hold in natural populations. The frequency of the bb genotype can vary across generations due to selection, genetic drift, and changes in population size. Each generation's genetic makeup is determined by the alleles passed down from the parents, with the frequency of bb potentially rising or falling based on these evolutionary forces.
The observed numbers for dominant and recessive types in an F2 generation are 154 and 46. What would the expected number of individuals with a heterozygous genotype?
Answer: Expected heterozygous genotype = 2×46 = 92
Explanation:
According to the hypothesis of segregation of paired genes in heterozygous F1 generation crosses produces 1:2:1 genotype ratio. This means that all these three possible genotypes should be produced in the F2 generation meaning that we have 1(Dominant homozygous) :2 (Dominant heterozygous): 1 (Recessive Homozygous).
Phenotypic ratio shared (dominant: recessive) = 154 and 46
If recessive homozygous phenotype = recessive homozygous genotype = 46
The expected dominant heterozygous genotype= 2 × (recessive homozygous genotype)
= 2 × 46 = 92
In an experiment, a certain colony of bacteria initially has a population of 50,000. A reading is taken every 2 hours, and at the end of every 2-hour interval, there are 3 times as many bacteria as before. a. Write a recursive definition for A(n), the number of bacteria present at the beginning of the nth time period. b. At the beginning of which interval are there 1,350,000 bacteria present?
Answer:
1) Recursive definition: [tex]p_n = (50,000)3^n[/tex]
2) At the beginning of the 4th interval
Explanation:
1)
The initial population of the bacteria at time zero is
[tex]p_0 = 50,000[/tex]
Here we are told that the reading is taken every two hours; we call this time interval "n", so
[tex]n=2 h[/tex]
And also, after every time interval n, the number of bacteria has tripled.
This means that when n = 1,
[tex]p_1 = 3 p_0[/tex]
And when n=2,
[tex]p_2 = 3 p_1 = 3(3p_0)=9 p_0[/tex]
Applied recursively, we get
[tex]p_n = 3^n p_0[/tex]
And substituting p0,
[tex]p_n = (50,000)3^n[/tex] (1)
2)
Here we want to find at the beginning of which interval there are
[tex]p=1,350,000[/tex]
bacteria.
This means that we can rewrite eq.(1) as
[tex]1,350,000=(50,000)3^n[/tex]
By simplifying,
[tex]27=3^n[/tex]
Which means that
[tex]n=3[/tex]
However, this means that the number of bacteria is 1,350,000 after 3 time intervals; therefore, at the beginning of the 4th interval.
Imagine that you are Gregor Mendel and you want to assure yourself that the F1 generation plants you generated from crossing true-breeding plantsactuallyare heterozygotes. You perform a testcross by mating all of your F1 smooth pea plants with plants that are homozygous recessive (wrinkled pea plants).
You are examining the phylogenic relationship of a newly discovered plant species (Species 2). You amplify the RUBISCO barcode and sequence the DNA. After entering your sequence into BOLD the following comparison comes up.Species 1. ATGCAAATTTGGGCATCCGAATGGTTGCAASpecies 2. ATGCAAATTTTTTGGGCATCCGAATGGCAAWhat DNA modifications have occurred in Species 2 that makes it different from Species 1? Check all that apply.a. Inversionb. Duplicationc. Deletion
Answer:
a. Inversion
b. Duplication
Explanation:
Inversion has the name suggest, has to do with a segment of DNA being reversed from end to end.
In this case here,
Inversion is taking place here.
species 1 ATGCAAATTTGGGCCCATGAATGGTTGCAA
species 2 ATGCAAAAATTTTGGTACGCCGAATGGTTGCAA
Therefore, the sequences in bold in species 1 are observed to be reversed end to end in species 2.
Deletion ❌❌
I am sure it's not feasible because deletion entails removal of a few sequences.
It can be seen that species 2 is longer than species 1, which gives another reason why deletion is not feasible too, as no sequences are seen to be deleted.
I believe duplication is feasible since AATT sequences are repeated once.
Our final answer,
inversion and duplication occur here.
A single newt is part of all the newts in a pond. The pond contains many more organisms as well as abiotic factors. The pond is part of a bigger picture, an ecosystem.
Which term BEST describes all the newts in a pond?
A) community
B) individual
C) population
D) species
Answer: Population
All the newts in the pond best describes population
Explanation:
Population is the total number of organisms of the same species living together in a given area at a particular time. In an ecosystem, the community is made up of many populations of different species of organisms relating with their environment.
Therefore, of all the different species of organisms present in the pond, the newts alone best describes a population
Answer: Option C.
It describes population.
Explanation:
Population is the sum total of living organisms of the same species that live in a particular geographical area, interacting with each other and are capable of interbreeding. The newts in the pond is described as population. Because there are many number of newts that are living and interbreeding in the ponds.
Changing the pH of the binding reaction mixtures can have a dramatic effect on ligand-protein binding. Altering the pH of the reactions between warfarin (which binds to site IIA of HSA via hydrophobic interaction) or ibuprofen (which binds to site IIIa of HSA via an ionic interaction) and HSA is likely to:A. Affect both reactions equallyB. Affect warfarin-HSA binding but not ibuprofen-HSA bindingC. Affect ibuprofen-HSA binding but not warfarin-HSA bindingD. Not affect either interaction
Answer:
B
Explanation:
Changing the pH of the binding reaction mixtures can have a dramatic effect on ligand-protein binding. Altering the pH of the reactions between warfarin (which binds to site IIA of HSA via hydrophobic interaction) or ibuprofen (which binds to site IIIa of HSA via an ionic interaction) and HSA is likely to Affect warfarin-HSA binding but not ibuprofen-HSA binding.
Answer:
B
Explanation:
The Earth is about 4.6 billion years old. However, the oldest sea floor is only about 180 million years old. What do you think is the reason for this? (Hint: Remember that new seafloor is constantly being created, but the Earth is not getting bigger with time.)
Answer:
The seafloor is only 180 million years old due to the process of subduction. The floor of the sea's tends to get colder and denser with the passage of time. At a certain time, the seafloor becomes so dense that it sinks in the upper mantle. The Earth's crust cannot undergo this process and hence has oldest rocks. We can say that the seafloor is less than 180 million years old because it is typically recycled back into the mantle of the Earth.
The oldest sea floor is only 180 million years old because new seafloor is constantly being created through seafloor spreading at mid-ocean ridges.
Explanation:The reason that the oldest sea floor is only about 180 million years old, while the Earth is 4.6 billion years old, is because new seafloor is constantly being created through a process called seafloor spreading. This process occurs at mid-ocean ridges, where tectonic plates are moving apart.
As the plates separate, magma from the Earth's mantle rises and solidifies, forming new oceanic crust. Over time, this new crust moves away from the ridge and gets older, while new crust forms at the ridge. Therefore, the oldest sea floor is relatively young compared to the age of the Earth.
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The Calvin cycle produces a versatile chemical compound called ____________ , which can be converted to many carbohydrates, as well as fatty acids and amino acids. Compared to animal cells, both algal and plant cells have enormous ____________ capabilities.
The Calvin cycle produces Glyceraldehyde-3-phosphate (G3P) that can be converted to other compounds such as carbohydrates, fatty acids, and amino acids. Algal and plant cells, compared to animal cells, have greater carbon fixation abilities due to their capacity to execute photosynthesis.
Explanation:The Calvin cycle produces a versatile chemical compound called Glyceraldehyde-3-phosphate (G3P), which can be converted to many carbohydrates, as well as fatty acids and amino acids. The Calvin cycle harnesses energy in the form of ATP and NADPH to produce G3P. During this light-independent reaction of photosynthesis, carbon dioxide from the atmosphere is converted into carbohydrates using an enzyme called RuBisCO.
After three cycles, some G3P molecules leave the cycle to become part of a carbohydrate molecule, while the remaining G3P molecules stay in the cycle to be regenerated into RuBP, ready to react with more CO₂. Both algal and plant cells have enormous carbon fixation capabilities compared to animal cells due to their ability to perform photosynthesis, which forms an energy cycle with the process of cellular respiration. This allows plants to function in both light and dark conditions and to interconvert essential metabolites.
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Consider a diploid cell that contains three pairs of chromosomes designated AA , BB, and CC. Each pair contains a maternal and a paternal member (e.g., Amand Ap, etc.)
In mitosis, what chromatid combination(s) will be present during metaphase?
1. AmAm
2. CpCp
3. BpBp
4. AmAp
5. BmBp
6. CmCp
Answer:
The chromatid combinations present will be AmAm, CpCp and BpBp
Explanation:
In mitosis, there is no exchange of materials between homologous chromosomes, thus the paternal and the maternal member of chromosomes DO NOT exchange materials at all.
This simply yields chromatids combinations that are singly maternal as AmAm OR singly paternal as CpCp and BpBp
During metaphase in mitosis, the diploid cell will have the chromatid combinations AmAm, CpCp, and BpBp.
Explanation:In mitosis, each pair of chromosomes duplicates itself resulting in two sister chromatids. These sister chromatids are identical copies of each other and are joined together at a point called the centromere. During metaphase, the chromosomes line up along the equator of the cell, and each sister chromatid is attached to a spindle fiber. In the given scenario, the diploid cell contains the following pairs of chromosomes: AA, BB, and CC. Therefore, during metaphase, the chromatid combinations present are:
AmAm: This represents the two sister chromatids of the AA chromosome pair.CpCp: This represents the two sister chromatids of the CC chromosome pair.BpBp: This represents the two sister chromatids of the BB chromosome pair.Learn more about mitosis here:https://brainly.com/question/31626745
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We are interested in determining whether plumage for the Guinea hens follow a common epistatic relationship. The observed phenotypes are dull, bright and brilliant. Which mode of inheritance would most likely explain the data below? Phenotype Dull 136 Bright 94 Brilliant 13 Total 243Select One:a. dominant/recessive epistasis b. duplicate dominant epistasis c. duplicate recessive epistasis d. duplicate genes with cumulative effect e. single recessive epistasis f. single dominant epistasis
Answer:
single dominant epistatits
Explanation:
When a dominant allele at one locus can mask the expression of both alleles (dominant and recessive) at another locus, it is known as dominant epistasis. In other words, the expression of one dominant or recessive allele is masked by another dominant gene. This is also referred to as simple epistasis
Final answer:
Epistasis is the genetic phenomenon influencing the expression of one gene by another gene. In this case, the most likely mode of inheritance for the plumage phenotypes is duplicate recessive epistasis (c).
Explanation:
Epistasis is a genetic phenomenon in which the expression of one gene is influenced by another gene. In this case, since the plumage phenotypes dull, bright, and brilliant do not follow a simple dominant or recessive inheritance pattern, the most likely mode of inheritance would be duplicate recessive epistasis (c). This means that two recessive alleles at different loci are required to produce a specific phenotype.
Calculate the average metabolic rate of a 65-kg person who sleeps 8.0h, sits at a desk 6.0h, engages in light activity 6.0h, watches TV 2.0h, plays tennis 1.5 h, and runs 0.50h daily.
Final answer:
To calculate the average metabolic rate of a 65-kg person with specified activities, one must understand the metabolic rate, including the basal metabolic rate and the additional energy expended during physical activities.
Explanation:
The question involves calculating the average metabolic rate of a 65-kg person based on their daily activities, including sleeping, sitting at a desk, engaging in light activity, watching TV, playing tennis, and running. This calculation requires understanding the concept of the metabolic rate, which describes the rate at which the body uses energy for basic functions at rest (known as basal metabolic rate, or BMR) and activities. The BMR varies with age, gender, body weight, and muscle mass. Athletes, for example, have a higher BMR. Additionally, the energy expended in physical activities contributes to the total daily energy expenditure, which can be significantly higher than the BMR alone. The daily energy needs thus depend on both the BMR and the energy spent on various activities throughout the day.
A horse has 64 chromosomes and a donkey has 62 chromosomes. A cross between a female horse and a male donkey produces a mule, which is usually sterile. How many chromosomes does a mule have? Can you think of any reasons for the fact that most mules are sterile?
Answer:
The horse has 64 number of chromosomes and the donkey has 62 number of chromosomes.
At the time of recombination the number of chromosomes which will be formed will be odd in number.
32 from horse and 31 from donkey which combines to form 63 pair. This is a odd number so there can be no equal number of segregation of chromosomes.
This is the fact the offspring produced by crossing a donkey and horse is sterile.
SMC (structural maintenance of chromosomes) proteins are associated with chromatin in many types of organisms. Two of the major complexes that eukaryotes possess are cohesins and condensins.
Sort the following phrases as describing cohesins, condensins, or both. Note: If you answer any part of this question incorrectly, a single red X will appear indicating that one or more of the phrases are sorted incorrectly.
1) generally dimeric, forming a V-shaped complex
2) hold sister chromatids together after replication, until chromatid separation
3) each terminus contains part of a site for ATP hydrolysis
4) contain two coiled-coil motifs connected by a hinge domain
5) associated with kleisin
6) essential for DNA replication and cell division
7) contributes to the compaction of chromosomes during mitosis and enables proper chromatid separation during anaphase
These have been sorted here:
The SMC (structural maintenance of chromosomes) proteins encompass two major complexes in eukaryotes: cohesins and condensins.
Cohesins maintain sister chromatid cohesion after replication until separation, associated with kleisin, and crucial for DNA replication and cell division.
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Final answer:
Cohesins and condensins are SMC protein complexes essential for chromosome maintenance in cell division. Cohesins primarily hold sister chromatids together, while condensins contribute to chromosome compaction during mitosis. Both share structural features and are crucial for DNA replication and cell division.
Explanation:
The SMC (structural maintenance of chromosomes) proteins are crucial for the proper regulation of chromosome structure and segregation during cell division. Specifically, in eukaryotes, cohesins and condensins are two major complexes involved in chromosome maintenance. The function and structure of these complexes can be described by the following phrases:
generally dimeric, forming a V-shaped complex - Both cohesins and condensinshold sister chromatids together after replication, until chromatid separation - Cohesinseach terminus contains part of a site for ATP hydrolysis - Both cohesins and condensinscontain two coiled-coil motifs connected by a hi-nge domain - Both cohesins and condensinsassociated with Klein - Both cohesins and condensinsessential for DNA replication and cell division - Both cohesins and condensinscontributes to the compaction of chromosomes during mitosis and enables proper chromatid separation during anaphase - CondensinsIf you want to use PCR technique to amplify markers located on six unlinked locations on the chromosomes, how many total unique primer sequences do you need? A. 1 B. 3 C. 6 D.8 E. 12
Answer:
The correct answer is option E, that is, 12.
Explanation:
It is mentioned that the markers are situated in six unlinked sites, the unlinked signifies that they are located on six distinct chromosomes. If one requires to augment the six markers, which are devoid of any definite end sequence, then only six forward primers are adequate. However, if one needs to augment a particular region on the chromosome, then both the reverse and forward primers are needed for each marker. Thus, a sum of 12 primers is required in such a case.
In case, if all the markers are situated in a single chromosome inside a particular region, then only one forward or both reverse and forward primers, that is, two primers would suffice. Generally, the technique of PCR is used to intensify particular fragments and both end and start sequences are illustrated. In such a case, the markers are specified.
Therefore, both reverse and forward primers are needed to augment every marker. Hence, it can be concluded that 12 primers are needed to augment all six markers situated on six unlinked sites.
The transfer of genetic information flows from DNA to protein synthesis in a complex process that involves many different types of biomolecules. Which of the following statements about this process are true? 1) Transcription is the process of transferring information from RNA to protein synthesis. 2) DNA utilizes tRNA to control the amino acid sequence of proteins. 3) mRNA catalyzes the formation of peptide bonds, linking amino acids together to form proteins. 4) Replication is the process of preparing mRNA from template DNA.
Answer:
. 2) DNA utilizes tRNA to control the amino acid sequence of proteins.
3) mRNA catalyzes the formation of peptide bonds, linking amino acids together to form proteins.
Explanation:
The formation of peptide bonds of polypeptide chains to form protein is by enzyme peptidyl transferase. This enzyme is located in the large ribosome sub-units;(ribosomes is made up of large and small sub units),and it made up of RNA.
Thus option 3 that mRNA catalyze the peptide bond for linking amino acids together is correct. The process of mRNA catalysis takes place in the cytoplasm.it begins with the alignment of the mRNA bases(condons) on the ribosomes sub-units, and corresponding alignment of the tRNA, in such a way that the anticodon at its head bonded by hydrogen bonds with the codon on the mRNA. The codon-anticodon bonding continues based on the type of protein attached to the head of the tRNA and and the corresponding condon and anticodon bonding .
The amino acids are joined together by peptide bonds. This extends with each additional amino acids to form polypeptide chains of amino acids The reactions is catalyzed by the (RNA contained peptidyl transferase above).
The main objective of DNA in gene expression is to produce required proteins needed for cellular reactions. Therefore for correct delivery of the DNA coded messages transcribed in the mRNA it must utilize tRNA so that correct protein sequence are positioned by tRNA as anti codons to bond with codons on the mRNA. This regulation is control at the transcription level coordinated by the DNA, under the influence of an enzyme.Therefore the option 2 that DNA utilizes tRNA to control the amino acid sequence of proteins is correct.
NOTE
Option 1,is wrong because transcription is the process of transfering coded information from DNA to mRNA, not as quoted in the question
Option 4. is wrong because the process of producing new DNA replica from original DNA molecule is called Replication, not as quoted,
Answer: The following statements are true:
2) DNA utilizes tRNA to control the amino acid sequence of proteins.
3) mRNA catalyzes the formation of peptide bonds, linking amino acids together to form proteins.
Explanation: mRna carries the information from DNA in the form of codon, each three nucleotide form a codon for one specific amino acids. When mRna binds to ribosomal subunit, translation (protein synthesis) starts.
tRNA carries the anticodon that are complemented to mRna codons and they carry these anticodons to ribosome where translation process occurs, thus, tRNA will help in the protein synthesis by binding that specific amino acid to growing polypeptide.
Snapdragons with red, normally shaped flowers are mated with plants with white, abnormally shaped flowers. In the F1, all the flowers are pink and have normal shape. The F1 intercross yields the following F2:3/16 red, normal6/16 pink, normal3/16 white, normal2/16 pink, abnormal1/16 red, abnormal1/16 white, abnormala) What are the parental genotypes?b) What are the F2 genotypes and phenotypes?c) What conclusions can be made about the allelic and gene interactions?
The parental genotypes for the snapdragons are homozygous red with normal shape and homozygous white with abnormal shape. F2 genotypes show incomplete dominance for color and Mendelian inheritance for shape. Incomplete dominance results in a 1:2:1 phenotypic ratio, while shape follows a dominant-recessive pattern.
Explanation:The student is dealing with incomplete dominance in snapdragons, where red flower color (CRCR) and white flower color (CWCW) blend to produce pink flowers (CRCW). Given the F2 generation's phenotypic ratios, we can infer the parental genotypes that produced the F1 generation with pink, normally shaped flowers.
a) Parental genotypes: One parent must have been red, homozygous normal (CRCR) and the other white, homozygous abnormal (CWCW). This explains the F1 generation of all pink, normal shaped flowers (CRCW).
b) F2 genotypes and phenotypes: The F2 generation shows phenotypes and genotypes based on independent segregation and incomplete dominance of flower color and normality of shape. Using a Punnett square for dihybrid cross, we have a genotypic ratio for color and shape. For color: CRCR (red), CRCW (pink), CWCW (white), and for shape, normal (N) is dominant over abnormal (n). The F2 would be: 3/16 CRCR-NN (red, normal), 6/16 CRCW-NN or -Nn (pink, normal), 3/16 CWCW-NN or -Nn (white, normal), 2/16 CRCW-nn (pink, abnormal), 1/16 CRCR-nn (red, abnormal), 1/16 CWCW-nn (white, abnormal).
c) Conclusions about allelic and gene interactions: The flower color demonstrates incomplete dominance, as neither red nor white is completely dominant. The F1 and F2 ratios suggest that flower shape acts as a simple Mendelian trait, with normal being dominant to abnormal. Therefore, the interaction is that color exhibits incomplete dominance whereas shape follows typical Mendelian inheritance.
The total value of an ecosystem:
a. is composed of the direct economic value and the potential pharmaceutical value of an ecosystem.
b. includes all of the values embodied by the ecosystem, including future uses and non-use values (such as cultural, symbolic, and aesthetic values) of an ecosystem.
c. is the monetary value of all of the beneficial aspects an ecosystem provides.
d. is composed of the utilitarian uses and products an ecosystem provides, such as water storage and filtration, even if they are not directly paid for.
e. is composed of the direct values an ecosystem provides upon which a price can be placed, such as crops and medicinal plants.
Answer:
Includes all of the values embodied by the ecosystem, including future uses and non-use values (such as cultural, symbolic, and aesthetic values) of an ecosystem.
Explanation:
Ecosystem may be defined as the constituent of the living and non living things present in the ecosystem. The living component includes the plants, animal and microorganisms. The non living component includes the water, air and soil.
The ecosystem provides oxygen and different gases important for sustain life. The ecosystem provides the aesthetic and cultural value that are used by future as well. Ecosystem provide medicine, food, furniture, fibers, habitat for the large number of organisms.
Thus, the correct answer is option (b).
he _____ is the division of the autonomic nervous system associated with rest, repair, and energy storage. a. parasympathetic nervous system b. sympathetic nervous system c. somatic nervous system d. endocrine system
Answer: Option A) parasympathetic nervous system
Explanation:
The autonomic nervous system consists of two parts namely
- the sympathetic nervous system SNS, and
- the parasympathetic nervous system PNS
The PNS stimulates the same organs as SNS, but its action is opposite to SNS.
And it acts to return the body to a normal state, thus it is associated with actions such as:
- slowing heart beat
- dilates arteries,
- lowering blood pressure
- stimulating saliva secretion etc
All these actions of the PNS bring about rest, repair, and energy storage.
The xylem cells are square, whereas the phloem cells are round. From which part of the stem tissue is wood made? Select one: a. spongy mesophyll b. the cuticle c. old layers of xylem and phloem d. central parenchyma Flat leaves lose water to the environment more readily than pine needles do.
Answer:
The correct answer is option c. "old layers of xylem and phloem".
Explanation:
Wood is one important good used for construction obtained from the main substance of the outer layer of the trunk or branches of a tree. Biologically, the outer layer of the trunk of a tree is comprised of the old layers of xylem and phloem, which are dead cells that were part of the heartwood or the centre of the tree. These old layers of xylem and phloem form the outer bark structure of the tree.
What are two disadvantages of cephalization?
Answer:
1) Since all the important tissues like the sensory and nervous tissues have been concentrated on the head, a damage to the head will lead to a damage to important organs.
2) The animal would not be able to see activities taking place behind the head.
The mitochondria within eukaryotic cells have their own genomes. Imagine that a mutation arises on the mitochondrial genome and, at the time of cytokinesis of the host cell, 10% of the mitochondria in that cell have that mutation. In the two daughter cells, what percentage of the mitochondria will possess that mutation? A. Although the cytoplasm containing the mitochondria will be equally divided between the two cells, there is no precise mechanism for ensuring that the organelles are equally divided.B. One cannot accurately predict what the percentages will be in each cell.C. 50% eachD. A and B onlyE. None of the above
C) A and B only is the right option.
Explanation:
During the process of cell division as mitosis or meiosis, the random segregation of mitochondria takes place in the resulting daughter cells
It is found that when cell divides, mitochondria present on the opposite side of the cell plate will have daughter cells that are different from the progenitor cell with respect to mitochondria. Due to this reason, it is difficult to predict the percentage of mitochondrial mutation passed on.
Also, it is proved that only maternal cells are capable of passing the mitochondrial DNA.
Final answer:
Mutated mitochondria are distributed randomly during cell division, leading to varying percentages in daughter cells.
Explanation:
When a mutation arises on the mitochondrial genome and 10% of the mitochondria in a cell have that mutation during cytokinesis, the distribution of the mutated mitochondria will be random in the daughter cells. Since mitochondria divide independently and are distributed randomly during cell division, there is no precise mechanism to ensure equal distribution of mutated mitochondria. Therefore, the percentage of mitochondria with the mutation in the two daughter cells will vary and could differ from the initial 10%.
Classifying Life on Earth - Kingdoms
Listed in the Item Bank are key terms and expressions, each of which is associated with one of the columns. Some terms may display additional information when you click on them. Drag and drop each item into the correct column. Order does not matter.
ITEM BANK: Move to Bottom
AnimaliaArchaebacteriaEubacteriaFungiPlantaeProtista
Prokaryotic Unicellular
Eukaryotic Multicellular Autotroph
Eukaryotic Multicellular Heterotroph
Eukaryotic Unicellular/Multicellular Auto/Heterotroph
Animalia: Eukaryotic multicellular heterotroph
Arachaebacteria: Prokaryotic unicellular
Eubacteria: Prokaryotic unicellular
Fungi: Eukaryotic multicellular heterotroph
Plantae: Eukaryotic multicellular autotroph
Protista: Eukaryotic unicellular/multicellular auto/heterotroph
Explanation:
Living organisms of different kingdoms are classified according to their number of cells, type of nutrition and presence or absence of nucleus.
Unicellular: single celled organism; multicellular: organisms with multiple number of cells
Prokaryotic: absence of nucleus or membrane-bound cellular organelles. Eukaryotic: nucleus is present
Autotroph: Prepares their own food. Heterotroph: Depends on other organisms for food
Species belonging to Kingdom Animalia are eukaryotic, multicellular, heterotrophic, motile, reproduce sexually or asexually.
Species belonging to Kingdom Plantae are eukaryotic, multicellular, autotrophic, nonmotile, reproduce sexually or asexually.
Species belonging to Kingdom Protista are eukaryotic, unicellular or multicellular, and can be autotrophic or heterotrophic, reproduce sexually or asexually.
Species belonging to Kingdom Fungi are eukaryotic, multicellular (few are unicellular), heterotrophs – saprophytes or parasites
Species belonging to Kingdom Monera including arachaebacteria and eubacteria are prokaryotic unicellular organisms, reproduce asexually
The classification of living organisms is arranged in the order and they are listed below Archaebacteria, Eubacteria, Fungi, Protista, Plants, Animals.
Explanation:
1. Animals - Eukaryotic Multicellular Heterotroph
Cell type- EukaryoticMode of nutrition- HeterotrophNumber of cells- Multicellular2. Plants - Eukaryotic Multicellular Autotroph
Cell type- EukaryoticMode of nutrition- AutotrophNumber of cells- Multi-cellular3. Fungi - Eukaryotic Unicellular/Multicellular Auto/Heterotroph
Cell type- EukaryoticMode of nutrition- HeterotrophNumber of cells- Multi-cellular /unicellular4. Protista - Eukaryotic Unicellular/Multicellular Auto/Heterotroph
Cell type- EukaryoticMode of nutrition- Heterotroph/AutotrophNumber of cells- Multi-cellular /uni-cellular5. Bacteria - Prokaryotic Unicellular
Cell type- ProkaryoticMode of nutrition- Heterotroph/AutotrophNumber of cells-Uni-cellular6. Archae - bacteria-Prokaryotic Unicellular
Cell type - ProkaryoticMode of nutrition - Heterotroph/AutotrophNumber of cells - Uni-cellularGibson Assembly – Several enzymes are present in the Gibson assembly reaction. For each of the following enzymes, define the role of the enzyme at room temperature (when the reaction is set up) and at 50 °C (the temperature for the 30 minute incubation after set up).
a. Phusion DNA polymerase ,
b. T5 exonuclease ,
c. Taq DNA ligase
Answer with explanation:
Gibson Assembly is a method of molecular cloning which join multiple DNA fragments in a single reaction. It was created by Daniel G. Gibson, Chief Technology Officer and co-founder of SGI-DNA. (See attached picture)
a. Role of Phusion DNA Polymerase
It bring the enzymes closer to DNA fragment and help enzymes to make PCR product with speed and more accurately. Moreover it has the ability of amplifying long templates. (See attached picture)
a. Role of T5 exonuclease
It is an exonuclease enzyme which means it remove the nucleotide from DNA stand in 5´ to 3´ direction. It create nicks in the double stranded DNA for the incorporation of other fragments. Furthermore, it also work fine in single stranded DNA. (See attached picture)
a. Role of Taq DNA ligase
It is a thermostable enzyme which catalyzes the phosphodiester bond formation between 5´-phosphate and the 3´-hydroxyl of two adjacent DNA strands. (See attached picture)
Suppose you want to radioactively label DNA but not RNA in dividing and growing bacterial cells. What radiolabeled molecule would you add to the culture medium? Why would it selectively label DNA but not RNA?
Answer:
Tritiated thymine or tritiated thymidine
Explanation:
It would selectively label DNA but not RNA because of the presence of the uniformly labeled backbone phosphorus atoms in the DNA.
The cell responsible for secreting the matrix of bone is the__________a. osteoclast. b. chondroblast. c. chondrocyte. d. osteoblast.
Answer
D.Osteoblasts
Explanation:
Appositional growth is the increase in the diameter of bones by the addition of bony tissue at the surface of bones. Osteoblasts at the bone surface secrete bone matrix, and osteoclasts on the inner surface break down bone. The osteoblasts differentiate into osteocytes.
The cell responsible for secreting the matrix of bone is the osteoblast, which supports the growth, maintenance, and repair of bones.
Explanation:The cell responsible for secreting the matrix of bone is the osteoblast. Osteoblasts are a type of bone cell that form new bone, or 'osteo', tissue. They do this by secreting a matrix that later mineralizes to become hardened bone tissue. This process is critical for the growth, maintenance, and repair of bones in the body.
In contrast, osteoclasts are involved primarily in the breakdown and resorption of bone tissue, while chondroblasts and chondrocytes are associated with cartilage formation and maintenance, not with bone tissue formation.
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Which of the following statements concerning transcription is true?
A. All types of RNA in the cell are synthesized by transcription, which uses a portion of DNA as a template for copying.
B. Promoter sequences signal the end of a gene and mark the place where transcription stops.
C. During transcription, entire chromosomes are copied because the starting position of genes is unknown.
D. Transcription is the process whereby identical copies of DNA are made in preparation for cell division.
Answer: Option A) All types of RNA in the cell are synthesized by transcription, which uses a portion of DNA as a template for copying.
Explanation:
All types of RNA:
- viral RNA,
- ribosomal RNA,
- transfer RNA,
- messenger RNA and
- double-stranded RNA, have a sequence that is the same as 'antisense' strand of the DNA.
Thus, it is true that all types of RNA in the cell are synthesized by transcription, which uses a portion of DNA as a template for copying.
Option A is the correct statement regarding transcription: All types of RNA in the cell are synthesized by transcription using a portion of DNA as a template.
Explanation:The correct statement concerning transcription is option A: All types of RNA in the cell are synthesized by transcription, which uses a portion of DNA as a template for copying. Transcription is the process by which an RNA molecule is synthesized using a DNA template. Specific sequences called promoters signal the start of a gene, and terminator sequences signal the end of gene transcription, not the other way around as mentioned in option B. The entire chromosome is not copied during transcription; only the specific gene being transcribed is copied.
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Could the way you perform the procedure affect the outcome? If the outcome changes, does it mean the net rate of photosynthesis has changed?
The procedure and conditions under which photosynthesis is conducted can affect the outcome and changes in the outcome may indicate alterations in the net rate of photosynthesis. Several factors related to the procedure, such as light intensity, temperature, carbon dioxide concentration, water availability, chlorophyll content, and duration of the experiment, can impact the outcome. Researchers often conduct controlled experiments to understand the specific factors influencing photosynthesis.
The procedure and conditions under which photosynthesis is conducted can indeed affect the outcome, and changes in the outcome may indicate alterations in the net rate of photosynthesis. Photosynthesis is a complex process that involves several factors, and variations in the experimental setup can influence its efficiency.
Here are a few factors related to the procedure that can impact the outcome:
Light Intensity: The rate of photosynthesis is often directly proportional to the intensity of light. If the light source or its intensity changes, it can affect the rate of photosynthesis.Temperature: Photosynthesis is also temperature-sensitive. A change in the temperature of the environment can influence the activity of enzymes involved in photosynthesis and, consequently, the overall rate of the process.Carbon Dioxide Concentration: Alterations in the concentration of carbon dioxide (CO2) can impact photosynthesis. If the experimental conditions lead to changes in CO2 availability, it can affect the outcome.Water Availability: Water is a crucial component of photosynthesis. If there are variations in water availability or if the plant experiences water stress, it can affect the rate of photosynthesis.Chlorophyll Content: The health and amount of chlorophyll in plant cells play a vital role in photosynthesis. Any changes in the procedure that affect chlorophyll content can influence the outcome.Duration of the Experiment: The duration for which the experiment is conducted can also impact the results. Photosynthesis rates might change over time due to various factors.If the outcome changes, it may indicate a difference in the net rate of photosynthesis. An increase in the outcome might suggest an enhancement in photosynthetic activity, while a decrease could indicate a reduction. Researchers often conduct controlled experiments, adjusting one variable at a time while keeping others constant, to understand the specific factors influencing photosynthesis.
What group of organisms are the most important primary producers in the marine aquatic food web? How deep down in the water column can they be found?
Final answer:
Phytoplankton are the most important primary producers in the marine aquatic food web, performing most of the ocean's photosynthesis and supporting the food web as the main food source for zooplankton, the primary consumers.
Explanation:
The most important primary producers in the marine aquatic food web are phytoplankton. Phytoplankton can be found floating on or near the surface of the water where sunlight can penetrate, and they perform the bulk of the ocean's photosynthesis, contributing to 95% of the ocean's primary productivity. These organisms can typically be found up to the depth where light can still reach, which is known as the euphotic or photic zone.
This zone can extend to depths of about 200 meters but varies depending on water clarity. Phytoplankton serve as the foundation of the marine food web, feeding zooplankton which represent the primary consumer level, followed by secondary consumers such as small fish.
Phytoplankton play a crucial role in both aquatic and terrestrial environments as they are significant consumers of carbon dioxide (CO2) and producers of oxygen through the process of photosynthesis. Their abundance and health are thus critical for marine ecosystems and global carbon cycles.