Answer:
Because of their property of preciseness and rhythmic repetitions like a normal clock sequence...
Explanation:
Evolutionary biologist use these neutral polymorphisms as molecular clock because of their gradual accumulation of mutations that remain on changing gradually in a uniform sequence overlarge number of generations. These changing sequences are highly precise over and over, that's why these are also termed as neutral mutations.
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 nucleus of "Lead-208", 208 82 Pb, has 82 protons within a sphere of radius 6.34×10-15 m. Each electric charge has a value of 1.60218 × 10^-19 C. The Coulomb constant is 8.98755 × 109 N · m^2/C^2. Calculate the electric field at the surface of the nucleus. Answer in units of N/C.
Answer:
2.94 × 10²⁰ N/C
Explanation:
Given that:
The nucleus of "Lead-208 has 82 protons,
with a radius (r) 6.34×10-15 m, &
each electric charge has a value of 1.60218 × 10^-19 C
∴ The formula for calculating an electrical field at the surface of the nucleus is:
[tex]E=\frac{k*q}{r^2}[/tex]
Substituting our values into the equation above, we have;
E = [tex]\frac{8.98755*10^8*82(1.60218*10^{-19C)}}{(6.34*10^{-15}_m)^2}[/tex]
E = 2.93870499×10²⁰ N/C
E ≅ 2.94 × 10²⁰ N/C
To calculate the electric field at the surface of the nucleus, use the formula E = k * |Q| / r^2, where k is the Coulomb constant, |Q| is the absolute value of the charge, and r is the distance from the charge. Plugging in the values for Lead-208, we find that the electric field at the surface of the nucleus is 2.31 × 10^18 N/C.
Explanation:To calculate the electric field at the surface of the nucleus, we can use the formula for electric field created by a point charge. The electric field (E) is given by the equation E = k * |Q| / r^2, where k is the Coulomb constant, |Q| is the absolute value of the charge, and r is the distance from the charge. In this case, the charge of the nucleus is equal to the number of protons (82), so |Q| = 82 * (1.60218 × 10^-19 C). The radius (r) is given as 6.34×10^-15 m. Plugging these values into the equation, we can calculate the electric field at the surface of the nucleus:
E = (8.98755 × 10^9 N · m^2/C^2) * (82 * (1.60218 × 10^-19 C)) / (6.34×10^-15 m)^2
E = 2.31 × 10^18 N/C
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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.
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:
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.
If the pressure angle increases, the minimum number of teeth on a pinion to avoid interference increases true false
Answer:
False
Explanation:
The relation between the pressure angle and the number of teeth on a pinion is given by
[tex]T_n = \frac{2 A_w}{G[\sqrt{1+\frac{1}{G}(\frac{1}{G} +2 }) sin^2\alpha -1]}[/tex]
Here
[tex]T_n =[/tex] minimum number of teeth on a pinion to avoid interference
[tex]A_w =[/tex] Multiplication factor for standard addendum for the wheel
G represents the gear ratio. It is also known as the velocity ratio
[tex]sin \alpha =[/tex] Pressure angle or angle of obliquity
As we can see from this relation, that the pressure angle is inversely proportional to the number of teeth on a pinion. Thus, if the pressure angle is increased the number of teeth must decrease according to the relation represented above.
Hence, the given statement is false.
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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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-cellularA. are buried in the interior of the helix B. provide the hydrogen bonds that form the helix C. are typically polar D. extend outward from the helix spiral E. are located in an alternating arrangement between the inside and the outside of the helix
Answer: D
Explanation: The side chain of amino acids is projected outward from the outer helical surface
How is energy utilized in photosynthesis?
Answer: During photosynthesis, green plants uses light energy from sunlight to convert carbon dioxide and water to glucose and water. Light energy is converted to chemical energy during photosynthesis.
Explanation:
Photosynthesis is the process by which green plants uses light energy from sunlight to produce carbohydrates by converting carbon dioxide and water to carbohydrates and oxygen.
Blue flower color is produced in a species of morning glories when dominant alleles are present at two gene loci, A and B. Purple flowers result when a dominant allele is present at only one of the two gene loci, A or B. Flowers are red when the plant is homozygous recessive for each gene. What flower color ratio is expected from the cross of aaBb to AABb?
Answer:
The flower colour ratio obtained is: Blue : Purple = 3 : 1.
Explanation:
According to the question, the genotype, AABB or AaBb or AABb or AaBB gives blue phenotype.the genotype, AAbb or aaBB or Aabb or aaBb gives purple phenotype.Red phenotype is for the aabb genotype.As only the presence of dominant allele is mentioned but whether it is homozygous or heterozygous is not mentioned, in 1. we consider that presence of a single dominant allele A or B in either locus is capable of showing a blue phenotype, irrespective of the nature of the other allele.In 2. we consider that the presence of a single dominant allele, A or B in one locus with recessive pairs of alleles in the other locus is sufficient to develop the purple phenotype.The aaBb individual will produce the gametes: aB, ab.The AABb individual will produce the gametes: AB, Ab.Crossing them,aB ab
AB AaBB AaBb
(Blue) (Blue)
Ab AaBb Aabb
(Blue) (Purple)
Of the offspring obtained, 3 have blue phenotype and 1 has purple phenotype.Hence, they have the following phenotypic ratio:Blue : Purple = 3 : 1.
From the cross of aaBb and AABb morning glory plants, we can expect that three out of every four offspring will have blue flowers and one out of four will have purple flowers. There will be no red flowers since all offspring will have at least one dominant A allele.
There are two genes to consider, A and B, with the following dominant (capital letter) and recessive (small letter) alleles: blue flower color if both loci have at least one dominant allele, purple if only one locus has a dominant allele, and red if both loci are homozygous recessive.
For the A gene locus: since one parent is aa and the other is AA, all offspring will have the genotype Aa, resulting in the dominant trait.For the B gene locus, the cross is Bb x Bb, leading to a expected ratio of 3:1 where 75% exhibit the dominant phenotype and 25% are homozygous recessive.As all offspring have at least one dominant A allele they won't be red, but we need to see who has at least one dominant B. Three out of four will have B (either BB or Bb) and one will be bb. So, for every four offspring, three will be blue (A-B-), and one will be purple (A-bb). There won't be any red offspring because all have at least one dominant A allele.
Below is a hypothesis about why male white-crowned sparrows sing dialects.
Is this hypothesis a genetic-developmental explanation, a physiological-psychological explanation, an adaptive value explanation, or one relating to evolutionary history?
The ability to sing the local dialect enables a bird to form bonds with others in the area so that they can adjust their total reproductive output, reducing the risk of local overpopulation.
Answer:
The answer is an adaptive value explanation.
Explanation:
Adaptive value is the effect that the behavior of an organism has on the reproductive fitness of the organism.
This can help offspring to cope with their new surrounding or condition.
It is a quantity that can be measured by contribution of an organism to the gene pool of their offspring. This measurement can be releasing of chemicals to avoid predators, sexual mimicry by some animals or trying to imitate so as mix with others.
Why should deepwater shrimp on different sides of the isthmus have diverged from each other earlier than shallow-water shrimp?
Answer:
Reproductive isolation occurs faster in deep-water shrimp than shallow-water shrimp.
Explanation:
Though in the same territory, the blockage caused by the isthmus would quickly and permanently isolate shrimps living in the deeper parts of the water, thus making them unable to breed. This situation would then caused lack of gene flow within the deep-water shrimps , and the emergence of new species that are genetically different (diverge) from one another
The shallow-water shrimp, on the other hand, experience minimal isolation due to the shallowness of water, and could still breed with one another. Thus, they experience a relatively lower reproductive isolation
Answer:
It is because the deep water shrimp and the shallow water shrimp being in different geographical location for a very long time have gotten use to their location and will definitely exhibit divergence.
Integrating (combining) many synaptic inputs over space (surface area of the plasma membrane) is called _____
Answer:
The correct answer is - spatial summation.
Explanation:
Spatial summation is the is the effect or impact of triggering an action potential in a neuron from many synaptic inputs or one or more presynaptic neurons. It takes place due to the multiple post synaptic potential starts simultaneously and a different part of neuron.
Thus, the correct answer is - spatial summation.
Answer:
Spatial summation.
Explanation:
Summation may be defined as the process of the process of adding of the different stimulus and has the ability to generate the action potential. Two main types of summation are temporal summation and spatial summation.
The spatial summation determines the effect that are responsible for the generation of the action potential from the presynaptic neurons. This summation combines the different stimuli on the different neurons over a particular space. The excitatory postsynaptic potential (EPSP) are involved in the spatial summation.
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.
How did you develop from a single-celled zygote to an organism with trillions of cells? How many mitotic cell divisions would it take for one zygote to grow into an organism with 100 trillion cells?
Answer:
The correct answer is "A single-celled zygote develops into an organism with trillions of cells by going trough multiple mitotic processes; a mitotic cell will undergo 47 mitotic divisions to develop into an organism with 100 trillion cells".
Explanation:
It is fascinating how a single-celled zygote develops into an organism with trillions of cells. The key of an organism's development is called mitosis: a cell division process at which a parent cells produces two daughter cells. Apparently producing two cells from one is not much, but it should be taken into account that it is an exponential process. For instance, a mitotic cell will undergo 47 mitotic divisions to develop into an organism with 100 trillion cells (2^47=140 trillion).
We developed from a single-celled zygote to an organism with trillions of
cells through series of cell divisions known as Mitosis.
It will take about 47 mitotic cell divisions for one zygote to grow into an
organism with 100 trillion cells.
Mitosis is a type of cell division that is associated with growth and
replacement of tissues. It helps in the division of cells to enable them
become more specialized in functions and structure.
It will take about 47 mitotic cell divisions for one zygote to grow into an
organism with 100 trillion cells as the cells divides at a higher degree.
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At the end of the paper, the authors state that the Nitrogen of a DNA molecule is divided equally between two subunits and each daughter molecule receives one of these. a. Which component of a DNA nucleotide contains nitrogen?b.How could you label another part of the DNA and repeat these tests with another component?
The nitrogen component in a DNA nucleotide is found in the nitrogenous base, which are adenine, guanine, cytosine, or thymine. Another component of the DNA that could be labeled for tests is the phosphate group.
Explanation:The component of a DNA nucleotide that contains nitrogen is the nitrogenous base. These bases are adenine (A), guanine (G), cytosine (C), and thymine (T). The nitrogenous base is part of the nucleotide, which also includes a deoxyribose (5-carbon sugar) and a phosphate group.
As for labeling another part of the DNA for further tests, you could label the phosphate group. The phosphate group is another key component of the nucleotide and forms part of the DNA's double-helix structure. By labeling it, we could track its distribution during DNA replication, much like nitrogen.
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Final answer:
The nitrogenous base is the component of a DNA nucleotide that contains nitrogen. To repeat tests with another component, isotopic labeling can be used on the phosphate group or the 5-carbon sugar.
Explanation:
The component of a DNA nucleotide that contains nitrogen is the nitrogenous base. DNA nucleotides are made up of three parts: a deoxyribose (5-carbon sugar), a phosphate group, and a nitrogenous base. There are four types of nitrogenous bases in DNA: the purines adenine (A) and guanine (G), and the pyrimidines cytosine (C) and thymine (T).
To label another part of the DNA and repeat the tests, one could use isotopic labeling on either the phosphate group or the 5-carbon sugar component of the nucleotide. This would allow scientists to track these components through the replication process in a similar way to how the nitrogenous bases were tracked.
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.
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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In the mouse, gene A allows pigmentation to be deposited in the individual coat hairs while its allele a prevents such deposition of pigment, resulting in an albino. Gene B gives agouti (wild-type fur) while its allele b gives black fur.The cross between a doubly heterozygous agouti mouse mated with a doubly homozygous recessive white mouse is:AaBb X aabb.
1. What would be the expected phenotypic ratio in the progeny?
Answer:
9 : 3 : 4 = agouti : black : albino
Explanation:
Given,
aa would not allow pigmentation resulting into albino mouse. So,
AaBb = agouti fur
aabb = albino fur
AaBb X aabb :
A_B_ = 9 = agouti
aaB_ = 3 = albino
A_bb = 3 = black
aabb = 1 = albino
Hence, expected phenotypic ratio in the progeny would be =
9 : 3 : 4 = agouti : black : albino
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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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 probability distribution of X, the number of imperfections per 10 meters of a synthetic fabric in continuous rolls of uniform width, is given ascol1 x 0 1 2 3 4 col2 f(x) 0.41 0.37 0.16 0.05 0.01 Find the average number of imperfections per 10 meters of this fabric.
Answer:
E(X) = 0.88
Explanation:
The average number of imperfections to be determined is the expected value of the random variable X.
E(X) = Σxf(x)
X
Of which f is the probability distribution of the random variable X.
Thus, the needed value is:
E(X) = Σxf(x)
X
= 4
Σxf(x)
X_0
= 0 . f (0) + 1 . f (1) + 2 . f (2) +
3 . f (3) + 4 . f (4)
= 0 . 0.41 + 1 . 0.37 + 2 . 0.16 +
3 . 0.05 + 4 . 0.01
= 0.88
E(X) = 0.88
Final answer:
The average number of imperfections per 10 meters of fabric is calculated by multiplying each possible number of imperfections by its probability and summing these products, yielding an average of 0.88.
Explanation:
To find the average number of imperfections per 10 meters of fabric, we calculate the expected value of the probability distribution given. The expected value (or mean) of a discrete random variable X is computed as E(X) = μ = Σ[x * f(x)], where x represents the value and f(x) the probability of that value. In our case, we will multiply each number of imperfections by its corresponding probability and sum up these products.
So, the calculation will be as follows:
(0 * 0.41) + (1 * 0.37) + (2 * 0.16) + (3 * 0.05) + (4 * 0.01)
= 0 + 0.37 + 0.32 + 0.15 + 0.04
= 0.88
Therefore, the average number of imperfections per 10 meters of this fabric is 0.88.
Imagine that each allele at the BXP007 locus is found at exactly the same frequency in a population. Since there are 8 possible alleles at the BXP007 locus, what is the frequency of any one allele from this locus in this population?
Answer:
[tex]\frac{1}{8}[/tex]
Explanation:
The variant forms of a gene are produced when they are located at the same position or gene locus on any chromosome.
So frequency of an allele is calculated by dividing the number of times a particular allele is observed in a given population by the total number of copies of all the alleles that can occur at that genetic location for that population.
Based on above explanation, the frequency of any one allele from this locus in this population is equal to
[tex]\frac{1}{8}[/tex]
In a population with eight equally frequent alleles at the BXP007 locus, the frequency of any one allele is 1/8 or 0.125, indicating a 12.5% presence in the population.
Explanation:If we assume that each allele at the BXP007 locus is found at the same frequency in a population and that there are eight possible alleles at this locus, we would determine the frequency of any one allele by dividing the total number of alleles by the number of different possible alleles.
Since there are eight alleles, the frequency of any one allele would be 1/8 or 0.125. This means that each allele would have a frequency of 12.5% in the population.
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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.
True or False The difference between spirochetes and spirillum bacteria is that the spirillum bacteria contain endoflagella or axial filaments
Answer:
False
Explanation:
The difference between the spirochetes bacteria from the spirillum is that spirochetes have axial filaments while spirillum posses an external flagella.
The statement is true. The primary difference between Spirochetes and Spirillum bacteria is that Spirochetes utilize endoflagella or, axial filaments, while Spirillum uses flagella located at the pole ends for movement.
Explanation:The statement you are asking about is True. Spirochetes and Spirillum bacteria are indeed different in that respect. The Spirillum is a type of bacteria that has flagella located at the pole ends while Spirochetes possess endoflagella or axial filaments. An endoflagellum, also called an axial filament, wraps around the cell, between the cytoplasmic membrane and outer membrane in spirochetes, and produces cell movement. In contrast, the flagella of most other bacteria, including Spirillum, protrude from the cell body and make whip-like movements to generate locomotion.
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"Henri, a 5-star chef in a French Restaurant, has been diagnosed with leukemia. He is about to undergo chemotherapy, which will kill rapidly dividing cells. He needs to continue working between bouts of chemotherapy. What consequences of chemo might affect his job as a chef?"
Answer:to die
Explanation: because yes
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).
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 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.