Answer:
Asexual lineages, being either unicellular organisms or organisms with a small number of germ-line cell divisions, survive due to the very high fidelity of DNA replication, which is enough for the reliable self-reproduction.
Explanation:
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Human cells normally have 46 chromosomes. For each of the following stages, state the number of chromosomes present in the particular cell. a) metaphase of mitosis b) metaphase I of meiosis c) telophase of mitosis d) telophase I of meiosis e) telophase II of meiosis
Final answer:
During metaphase of mitosis and metaphase I of meiosis, cells have 46 chromosomes. Following telophase of mitosis, cells also contain 46 chromosomes. However, after telophase I and telophase II of meiosis, cells contain 23 chromosomes.So ,option b) metaphase I of meiosis is correct.
Explanation:
Human cells normally have 46 chromosomes, comprising 23 pairs, with one set from each parent. This diploid state (2n) changes during the process of cell division.
a) metaphase of mitosis: Each cell has 46 chromosomes. The chromosomes align in the center of the cell, but each one still consists of two sister chromatids at this stage.
b) metaphase I of meiosis: Each cell also has 46 chromosomes. However, these are in the form of 23 pairs of homologous chromosomes that will be separated in the next phase.
c) telophase of mitosis: The cell still contains 46 chromosomes, but they are now being divided into two new daughter nuclei.
d) telophase I of meiosis: The two cells created after this phase each have 23 chromosomes, as the homologous pairs have been separated.
e) telophase II of meiosis: After this final phase of meiosis, there are four cells, and each cell has 23 chromosomes—these are the haploid gametes.
According to existing research, chimpanzees Select one: a. do not use tools. b. have complex grooming and courtship behaviors. c. cannot solve technical problems. d. All of the choices are correct.
Answer:
option b. have complex grooming and courtship behaviors is correct answer.
Explanation:
According to a research paper published in American journal of primatology provided evidences that chimpanzee males, as promiscuous breeders with minimal costs to mating, should show little or no preference when choosing mating partners (e.g. should mate indiscriminately).
Reference: Proctor, D. P., Lambeth, S. P., Schapiro, S. J., & Brosnan, S. F. (2011). Male chimpanzees' grooming rates vary by female age, parity, and fertility status. American journal of primatology, 73(10), 989–996. doi:10.1002/ajp.20964
Which of the following is an example of an abiotic factor?A.the number of individuals of a particular species living in a communityB.the interactions between different species in a communityC.the diversity of prey and predator species in a communityD.the climate of the community in which the species mentioned above inhabit
Answer: Option D) the climate of the community in which the species mentioned above inhabit
Explanation:
Abiotic refers to non-living factors. Thus, the abiotic factors in an habitat where living organisms dwells include the atmospheric conditions such as climate, weather; inorganic nutrients such as phosphorus, carbon compounds, water etc present in the soil of such habitat.
Thus, Option D represent the abiotic factor
Energy:
a) recycles continuously through an ecosystem
b) is used over and over again
c) flows in only one direction through an ecosystem
d) tends to be concentrated by living organisms
Answer: Option C.
Energy flows in one direction in ecosystem.
Explanation:
In the ecosystem, energy flow in one direction. Energy is transfered from one trophic level to another.
Energy is transfered from producers, majorly plants,convert sunlight to chemical energy during photosynthesis and transfer it to herbivores( goat, sheep ) when they eat plants,the herbivores after using the energy transfer energy to the carnivores when the carnivores feed on the herbivores and to decomposer who feed on dead animals.
Energy flows in one direction through an ecosystem, from producers to consumers, and eventually dissipates as heat, while matter is recycled, such as through biogeochemical cycles.
Explanation:Energy in ecosystems behaves differently than matter. It flows in only one direction: from the sun or inorganic chemicals to producers, and then on to consumers. Producers, such as plants, are capable of capturing energy directly through photosynthesis and chemosynthesis. Consumers then acquire energy by feeding on these producers or on other consumers. This flow of energy is depicted using models like food chains and food webs. In contrast, matter is recycled within ecosystems, ensuring that elements such as carbon and nitrogen are continuously available in one form or another.
It is also pertinent to highlight the process of biogeochemical cycles, such as the water cycle, carbon cycle, and nitrogen cycle, whereby nutrients and water are constantly recycled, interacting with both biotic components and abiotic factors of an ecosystem. While matter recycles and maintains a certain balance, energy does not recycle but rather dissipates as heat with each transfer through the trophic levels, eventually exiting the ecosystem.
Tropism can be defined as the Choose one: A. ability to infect a particular type of cell within the host. B. mechanism of entry into a host cell for bacteriophages. C. emergence of a new type of virus. D. ability to infect a broad range of hosts.
Answer:
A. ability to infect a particular type of cell within the host.
Explanation:
In plants, tropism refers to the movement of a plant part in a particular direction in response to any stimulus. In microbiology, tropism refers to the ability of pathogen to infect a particular type of cell in the host.
The tropism in which a particular cell is infected is called cell tropism. The tropism in which a particular tissue is infected then it is cell tissue tropism. Host tropism is also seen in pathogen when they infect specific host. Stimulus is required for any tropism. So the right answer is A.
Answer:
A) ability to infect a particular type of cell within the host.
Explanation:
Viral tropism refers to the ability of a virus to infect a particular cell, tissue and a host species.
The viruses have to interact with a variety of negative and positive factors in the cell so the viral tropism is determined at different levels of the viral-host interaction that is during replication and viral progeny production.
The viral tropism occurs at two-step and based on this is of two types- the receptor-independent tropism which takes place intracellularly and the receptor-dependent tropism which takes place at the cell surface.
Thus, Option-A is the correct answer.
Sort the molecules in the glycolysis pathway based on whether they are intermediates or products in the first half of the pathway that requires energy, or are intermediates in the second half of the pathway that produces energy. Drag the appropriate items to their respective bins.
The given question is incomplete. The complete question is:
Sort the molecules in the glycolysis pathway based on whether they are intermediates or products in the first half of the pathway that requires energy, or are intermediates in the second half of the pathway that produces energy. Drag the appropriate items to their respective bins.
glucose-6-phosphatefructose-6-phosphatefructose-1,6-bisphosphateglyceraldehyde-3-phosphatedihydroxyacetone phosphate1,3-bisphosphoglycerate3-phosphoglycerate2-phosphoglyceratephosphoenolpyruvate
Answer:
The procedure in which glucose gets dissociated into pyruvate and generates energy in the form of ATP is known as glycolysis. The process takes place in the cytosol and can get differentiated into two phases, that is, the preparatory phase that required energy in the form of ATP and the pay-off phase in which the production of ATP takes place.
Post glycolysis, the process of cellular respiration takes place in which the eventual outcomes, that is, two pyruvates are further used. Of the mentioned intermediates, the intermediates that are used in the first half of the pathway are glucose-6-phosphate, fructose-6-phosphate, fructose-1,6-bisphosphate, dihydroxyacetone phosphate and glyceraldehyde-3-phosphate.
The intermediates used in the second half of the pathway are 1,3-bisphosphoglycerate, 3-phosphoglycerate, 2-phosphoglycerate, and phosphoenolpyruvate.
Discuss the concept of fluoroscopically guided positioning and explain why this is an unacceptable practice.
Answer: Fluoroscopy is used as positioning device prior to obtaining film radiographs. It is used for examination where the type of projection and habitus if the patient present difficulties
It is not an acceptable practice because it exposes the skin and underlying tissues to radiation induced injuries and there is a possibility of future occurances of radiation induced cancer in the future
Explanation:
Pellicle ___ Uniform fine turbidity QUESTIONS a. Evenly cloudy throughout b. Growth at top around the edge c. Growth on the bottom d. Membrane at the top e. Suspended chunks or pieces 2 What factors besides physical growth characteristics are important when recording data about an organism
Answer: (d) Membrane at the top
Explanation:
Pellicle: surface membrane produced by organisms floating on top of the medium. Sediment: organisms that sink to the bottom. Uniform fine turbidity: cloudy. Flocculent: Organisms that clump/suspended pieces or chunks.
(2) Other important factors apart from
physical growth too include would be the growth medium, incubation temperature and atmosphere (aerobic or anaerobic). Growth medium,incubation temperatureandatmosphere(aerobicoranaerobic).
A pellicle describing uniform fine turbidity means that the microorganisms are evenly dispersed throughout a medium, giving it a uniformly cloudy appearance. Apart from physical characteristics, environmental conditions, genetic predisposition, presence of competitors or predators, and availability of resources are crucial factors to consider when recording data about an organism.
Explanation:The term 'pellicle' in biology refers to a thin layer supporting the cell membrane in various protozoa. When it describes a uniform fine turbidity, it refers to how the organisms in a liquid medium are dispersed. In this case, 'a. Evenly cloudy throughout' pertains to a uniform fine turbidity, meaning that the microorganisms are evenly dispersed throughout the medium, giving it a uniformly cloudy appearance.
Beyond physical growth characteristics, other factors are important to consider when recording data about an organism. These may include the organism's environment, its genetic predisposition, the presence of predators or competitors, availability of resources such as food and space and potential changes in these conditions over time. All these elements contribute to the overall understanding of the organism and its behavior.
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Inversion of DNA sequences within chromosomes is a common process in evolution. The following gene arrangements in a particular chromosome are found in four different species.
1st sequence STUVWX
2nd sequence UVXTSW
3rd sequence UVWSTX
4th sequence SWVUTX
1. Assuming that the arrangement in 1 is the original arrangement, in what evolutionary order did the four species arise, such that there is only one inversion between each species. (Place 1st sequence first in your ordering.)
Answer:
2, 3, 4, 1
Explanation:
Inversion of DNA sequences refers to the rearrangement of the sequences where a segment is reversed end to end.
For instance in order for the gene of species to have the sequence STUVWX
The order of evolution will be
sequence UVXTSW will be the first sequenceWhen XTSW segment is inverted WSTX
sequence UVWSTX will be the result of the inversion.section UVWS is inverted to SWVU
sequence SWVUTX is formed.when WVUT undergoes inversion to TUVW
It becomes the original arrangement of STUVWX .Which of the following structures does not exist as a pair? Select one: a. The kidney. b. The sphincter muscle. c. The ureter. d. The urethra.
Structures does not exist as a pair d. The urethra Therefore , d. The urethra is correct .
Explanation:
a. The Kidney:
The kidneys are paired organs in the human body, each individual having two kidneys.
They are essential in the urinary system, primarily responsible for filtering waste products and excess substances from the blood to form urine.
b. The Sphincter Muscle:
Sphincter muscles are circular muscles that surround various openings in the body, including the digestive and urinary tracts.
While there are multiple sphincter muscles, they exist as pairs, serving to regulate the flow of substances through the associated openings.
c. The Ureter:
The ureters are a pair of tubes that carry urine from the kidneys to the urinary bladder.
Humans typically have two ureters, one connected to each kidney, allowing the flow of urine from both kidneys to the bladder.
d. The Urethra:
Unlike the kidneys and ureters, the urethra is a single tubular structure. It is the tube that carries urine from the urinary bladder to the exterior of the body.
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d). The urethra is the correct answer because it is a singular structure, unlike kidneys and ureters, which exist in pairs. Sphincter muscles are specialized muscles and not paired organs.
The question asks us to identify which of the following structures does not exist as a pair: The kidney, the sphincter muscle, the ureter, or the urethra.
Kidneys: Humans have two kidneys, which filter blood to produce urine.Sphincter muscles: Multiple sphincter muscles exist in the human body, but they are not paired in the same way as organs like kidneys. Instead, they are specialized muscles to control the passage of substances through various openings and are found singularly based on their function.Ureters: There are two ureters, one for each kidney, which carry urine from the kidneys to the bladder.Urethra: There is only one urethra, which is a tube that carries urine from the bladder out of the body.Thus, the correct answer is d. The urethra, as it is a singular structure.
Suppose that two parents are both heterozygous for sickle cell anemia, which is an autosomal recessive disease. They have seven children. Use the binomial theorem to determine the probability that three of the children have sickle cell anemia and four of the children are healthy. Round your answer to the nearest hundredth.
Huntington\'s disease is an inherited autosomal dominant disorder that can affect both men and women. Imagine a couple has had seven children, and later in life, the husband develops Huntington\'s disease. He is tested and it is discovered he is heterozygous for the disease allele, Hh. The wife is also genetically tested for the Huntington\'s disease allele, and her test results show she is unaffected, hh. What is the percent probability that the first child of this couple will have Huntington\'s disease?also..
What is the probability that any two of the seven children will have Huntington\'s disease?
Answer:
Part A
[tex]0.108[/tex]
Part B
B.1
Fifty Percent
B.2
[tex]16.4[/tex] %
Explanation:
Part A
Given -
Both the parents are heterozygous for the sickle cell anemia
Let the genotype of the parents be
Ss
where S is the allele for presence of sickle cell anemia
and s is the allele absence of sickle cell anemia
If the above two parents are crossed, following offspring are produced
Ss * Ss
SS, Ss, Ss, ss
The probability of children with sickle cell anemia is equal to
[tex]\frac{3}{4}[/tex]
The probability of children with sickle cell anemia is equal to
[tex]\frac{1}{4}[/tex]
Probability that three of the children have sickle cell anemia and four of the children are healthy
[tex]\frac{7!}{3! * 4!} * \frac{3}{4}^4 * \frac{1}{4}^3\\ = 35 * \frac{81}{256 *64} \\= 0.108[/tex]
Part B
B.1
Hh * hh
Hh, Hh, hh, hh
The first child of the parent has a probability of fifty percent to have Huntington\'s disease
B.2
[tex]\frac{7!}{2!*5!} * \frac{1}{2}^2 * \frac{1}{2}^5[/tex]
[tex]= 16.4[/tex]
Which of the following is NOT true regarding the sounds associated with the heart? Select one: a. The "lub-dub" sound originates from the physical act of the heart contracting. b. The "lub-dub" sound originates from blood turbulence through the heart.
The 'lub-dub' sound we hear is the sound of the heart valves closing, not the contraction of the heart muscle. The 'lub' is the sound of the mitral and tricuspid valves closing, while the 'dub' is the sound of the aortic and pulmonic valves closing.
Explanation:The statement 'a. The "lub-dub" sound originates from the physical act of the heart contracting' is NOT true regarding the sounds associated with the heart. The "lub-dub" sound we hear is actually made by the closing of heart valves, not the contractions of the heart muscle itself. Specifically, the "lub" is the sound of the mitral and tricuspid valves closing at the beginning of systole (an active phase when your heart pumps blood), while the "dub" is the sound of the aortic and pulmonic valves closing at the beginning of diastole (a resting phase when your heart fills with blood).
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Final answer:
The incorrect statement is that b. the "lub-dub" sound originates from blood turbulence through the heart. These sounds are actually the result of the closing of the AV valves (for "lub") and the semilunar valves (for "dub"), which can be heard with a stethoscope.
Explanation:
The question is asking which statement is NOT true about the origins of the heart sounds. The answer to this is b, "The "lub-dub" sound originates from blood turbulence through the heart." It is not correct because the "lub-dub" sounds, known medically as S1 and S2, are primarily caused by the closing of the heart valves, not by blood turbulence.
The "lub" or the first heart sound, S1, occurs when the atrioventricular (AV) valves close at the beginning of ventricular systole, while the "dub" or the second heart sound, S2, happens when the semilunar (SL) valves close at the end of ventricular systole. When a medical practitioner uses a stethoscope to listen to the heart, these sounds provide critical information regarding the heart's condition.
In a mono hybrid cross of heterozygous pea plant with round (R) yellow (Y) seeds there will be genotype ratio of 9 _____ 3,____ 3,____ and 1____ PLEASE PLEASE HELP IM KIND OF CONFUSED
The genotype ratio for a monohybrid cross of heterozygous pea plants with round (R) and yellow (Y) seeds is 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green. This ratio comes from the dihybrid cross of RrYy plants, reflecting the principle of independent assortment.
Explanation:In a monohybrid cross of heterozygous pea plants with round (R) and yellow (Y) seeds, the genotype ratio you are asking about is derived from the principle of independent assortment and dominance, leading to a phenotypic ratio of 9:3:3:1 in a dihybrid cross. When focusing on one trait, a monohybrid cross, the ratio simplifies to a 3:1 ratio because each trait segregates independently.
For the dihybrid cross of RrYy (round, yellow) × RrYy, the ratios of expected phenotypes are 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green, which corresponds to the same proportions in genotype.
Therefore, the correctly completed sentence would be a genotype ratio of 9 round yellow (RRYY, RRYy, RrYY, RrYy) : 3 round green (RRyy, Rryy) : 3 wrinkled yellow (rrYY, rrYy) : 1 wrinkled green (rryy).
Final answer:
In a mono hybrid cross of heterozygous pea plant with round (R) yellow (Y) seeds, the genotype ratio is 9:3:3:1 for round, yellow seeds : round, green seeds : wrinkled, yellow seeds : wrinkled, green seed.
Explanation:
The mono hybrid cross of a heterozygous pea plant with round (R) yellow (Y) seeds would result in a genotype ratio of 9 round, yellow seeds : 3 round, green seeds : 3 wrinkled, yellow seeds : 1 wrinkled, green seed. This ratio is obtained by applying the product rule to the independent assortment and dominance of the alleles for seed shape and color.
Why does this chapter on culture include a section that describes similarities and differences between humans and apes, our closest relatives?
Answer: Explanation:
To emphasize culture's evolutionary basis
Please describe the environmental issue in Borneo concerning conservation of the rainforest (a global hot spot) and the orangutans. What grassroots tools were used to improve the conservation and allow the regrowth of the rainforest?
Answer:
Borneo Island was once home to one of the most majestic forests in the world. During the 80s and 90s, Borneo underwent a profound transformation. Their forests were demolished at an unprecedented speed. Its rainforests ended in countries like Japan and the United States in the form of garden furniture, paper pulp and chopsticks. Today, the remaining forests are threatened by the new biofuel market, especially palm oil. As a result, large tracts of land are being transformed to oil palm plantations.
Explanation:
The jungles of Borneo were considered one of the wildest and pristine jungles of the planet, home to nomadic tribes and important populations of orangutans, pygmy elephants and rhinos. Currently, the traditions of these tribes have disappeared, rhinos are almost extinct, and orangutans and elephants are in danger. Otherwise, the rainforests of Borneo have gone from being a net carbon sink, which absorbed greenhouse gases, to being a carbon source, thus contributing to climate change along with deforestation and fires.
Conservation is a priority in Borneo, especially in biologically diverse regions that have escaped intensive logging and fires. The initiative called "Heart of Borneo" is an example of what can be achieved. It is essential that forests be restored. The use of native tree species should be encouraged through financial incentives and education programs, especially with the help of external governments, NGOs and private foundations. In addition, there is a possibility that under future climate agreements, reforestation could pay direct dividends stimulating the local economy and entrepreneurship in the villages.
23. Select all accurate statements
A. All chordates will have a muscular post an*l tail as adults
B. All chordates will have a notochord as adults
C. All chordates have pharyngeal slits or clefts that function in swimming.
D. All chordates are bilaterally symmetrical animals.
E. All chordates have a ventral, hollow nerve cord.
Answer:
E.All chordates have ventral l, hollow nerve cord
D.All chordates are bilaterally symmetrical animals
Below is the DNA strand template for making an mRNA needed to produce a small peptide. a.) Transcribe the sequence into mRNA. (0.5 pt.) b.) Translate the mRNA sequence into protein. (0.5 pt.) c.) Transcribe the mRNA and translate into protein with a mutant sequence that has a transition at position 11. What type of mutation does this cause
Answer:
a) mRNA: AUG ACC GGC AAU CAA CUA UAU UGA
b) Protein: Methionine, Threonine, Glycine, Asparagine, Glutamine, Leucine, Tyrosine, Stop.
Explanation:
The Question was missing the DNA Strand. I have looked up the question and found this to be the DNA strand related to this question. A picture for converting codons to amino acids is also attached. A protein is simply a strand of Amino Acids.
DNA (3 to 5 sequence):
TAC TGG CCG TTA GTT GAT ATA ACT
a). Transcribing to mRNA (5 to 3 sequence):
AUG ACC GGC AAU CAA CUA UAU UGA
b). Related Amino Acids:
1st Codon: Methionine
2nd Codon: Threonine
3rd Codon: Glycine
4th Codon: Asparagine
5th Codon: Glutamine
6th Codon: Leucine
7th Codon: Tyrosine
8th Codon: Stop Codon
c.) DNA with Transition Mutation at Position 11 (5 to 3 sequence):
In genetics, Transition is a point mutation which converts a purine nucleotide to another i.e. A ↔ G, or a pyrimidine nucleotide to another pyrimidine i.e. C ↔ T. Hence, after transition at 11 position, T will be changed to C.
TAC TGG CCG TCA GTT GAT ATA ACT
Transcribing to mRNA (5 to 3 sequence):
AUG ACC GGC AGU CAA CUA UAU UGA
Related Amino Acids:
1st Codon: Methionine
2nd Codon: Threonine
3rd Codon: Glycine
4th Codon: Serine
5th Codon: Glutamine
6th Codon: Leucine
7th Codon: Tyrosine
8th Codon: Stop Codon
Explain what is happening to the whitebark pine in Yellowstone National Park as a result of climate changeResearchers often study ecosystems for a long period of time. Dr. Hadly has studied Yellowstone’s ecology for 30 years and the amphibians for 20 years. What is the value of long- term studies to advancing scientific understanding?
The given question is not complete. The complete question is:
Watch the film “Liz Hadly Tracks the Impact of Climate Change in Yellowstone (Links to an external site.)Links to an external site.” from the Scientists at Work series. Prior to watching the film, read the questions below and think about how you might answer them. You do not need to turn in your answers from before you watch the film.
Questions
Explain what is happening to the whitebark pine in Yellowstone National Park as a result of climate change
Researchers often study ecosystems for a long period of time. Dr. Hadly has studied Yellowstone’s ecology for 30 years and the amphibians for 20 years. What is the value of long- term studies to advancing scientific understanding?
Answer:
1. As a consequence of climate change, the whitebark pine is been targeted by pine beetles, which can thrive in winter conditions. An attack led by various beetles for one to two days has generated holes in the pine leading to its destruction.
2. The objective of long term studies is to find the connections between animals and plants and to witness how they are associating with each other in their micro-environment. Like in the documentary it was the association between the squirrel, pine trees and bear.
The beer needs to consume high content nutritious seeds of the pine tree, which safeguarded them in the winters due to the high amount of fat present in them that offered warmth. However, the beers were not able to reach the pine trees as they were far from their reach and thus, they took help from squirrels, which serve the objective for the bears.
Thus, it can be comprehended that all three creatures, that is, squirrel, tree and bear are associated with each other and developing a food web. All these studies need an ample amount of time and observation. Hence, Dr. Hardley took thirty years examining them and also took twenty years examining the influence of climate on amphibians living in the pond. Their species diversity declined substantially due to global warming that resulted in the drying up of the pond.
A 1475 kg car with a speed of 66.5 km/h brakes to a stop. How many cal of heat are generated by the brakes as a result?
Answer:22.180
Explanation:divide the two together then use the first three number after the decimal.
.
Answer:
Explanation:
Let's start by stating that the work done here would be determined by taking into account the following:
-External forces
-Change in kinetic energy
Work Energy theorem is fully going to guide the solution to this problem.
Let's go now:
Mass of car = 1475kg
Velocity of car= 66.5km/h
Velocity = 66.5km/h * 0.278
V = 18.472 m/s
Kinetic Energy (K.E) = 1/2(mv^2)
K.E = 0.5 * 1475 * 18.472^2
K.E = 737.5 * 341.214
K.E = 251645.90 kgm^2/s^2
1kgm^2/s^2 = 1 Joule
So therefore,
K.E = 251645.90 Joules
K.E = 60.144KCals
A moth learns to visit nectar-filled white flowers, and the good food source allows it to lay more eggs and have more surviving offspring than other moths in the area. Which of the following can you conclude?a. This moth has white colorationb. This moth has high fitnessc. The ability to find nectar-filled flowers is heritabled. This moth has evolved.
Answer: This moth has evolved
Explanation:
Evolutionary patterns show different patterns. One of the patterns is that: one species gradually transform into another species.
The above mentioned pattern is responsible for the moth's ability to lay more eggs and have more surviving offspring than other moths in the area. The moth is gradually transforming, so also its offspring will inherit these new traits, and it would further demonstrate evolution of one species.
Thus, it can be concluded that the unique ability of this moth over other moths is because the moth has evolved.
Beta ( ???? ) sheets are a type of secondary structure in proteins. A segment of a single chain in an antiparallel ???? sheet has a length of 94.5 Å . How many residues are in this segment?
Answer:
The correct answer is 27 residues.
Explanation:
In an antiparallel beta sheet, there is a repeat for every 7 Angstrom units and the number of residues is two for every repeat. In an antiparallel sheet that exhibits a length of 94.5 Angstrom, the number of residues can be determined by using the following formula:
2 residues per every 7 Angstrom,
Length of the given segment is 94.5 Angstrom
Number of residues in this segment are:
= (number of segments) * (residues per segment)
= (94.5/7) * 2
= 13.5 * 2
= 27 residues.
Thus, the number of residues in this segment is 27.
Beta sheets are a type of secondary structure in proteins where segments of the polypeptide chain are aligned side by side. The number of residues in a segment of an antiparallel beta sheet can be calculated using the formula number of residues = length / distance between adjacent residues.
Beta sheets are one of the two major types of secondary structure in proteins, along with alpha helices. In a beta sheet, segments of the polypeptide chain are aligned side by side, forming a sheet-like structure. The length of a segment in an antiparallel beta sheet can be calculated using the formula length = number of residues * distance between adjacent residues. In this case, the length of the segment is given as 94.5 Å, so we can rearrange the formula to find the number of residues: number of residues = length / distance between adjacent residues.
It's important to note that the distance between adjacent residues in a beta sheet can vary, but a common value used for calculations is 3.5 Å. So, using this value, we can calculate the number of residues: number of residues = 94.5 Å / 3.5 Å = 27 residues.
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Which of the following statements is/are correct? (MORE THAN ONE ANSWER)
A. Nematodes are also known as 'segmented' worms.
B. Members of the Phylum Nematoda are all hermaphrodites
C. Given the following on the organisms:
I. An animal
II. Multicellular
III. has tissues
IV. Has a digestive tract
V. exhibits bilateral symmetry as an adult
This organism could be a member of the Phylum Nematoda
D. Nematodes possess a cuticle.
E. Some members of the phylum Nematoda are parasites
Answer:
E
Explanation:
some of the nematodes (non-segmented ) are parasites
Answer:
The statements which are correct about nematodes are options C, D and E:
C: Organisms from the phylum nematoda are animals, multicellular in nature, has tissues, have a complete digestive tract and exhibit bilateral symmetry.
D. Nematodes possess external structure called cuticle.
E. Some members of the phylum nematoda are parasitic, free living organisms that feed on other living materials.
30. Select all accurate statements
A. All chordates have pharyngeal slits or clefts
B. All chordates have a dorsal, hollow nerve cord
C. All chordates are bilaterally symmetrical animals
D. All chordates will have a muscular post an*l tail
E. All chordates will have a notochord as adults.
Answer:
All the options A,B,C,D and E statements are correct.
Chordates are animals that posses a notochord, dorsal, hollow nerve chord, pharyngeal slits and they are bilaterally symmetrical.
The branch like structures on the heads of neurons that receive signals from other neurons are called ____.
Dendrites
hope this helps
Answer: the correct answer is dendrite.
Explanation: Dendrite is actually what receive signals from axon of other neurons, synapse is an area which joins axon and Dendrite. Signals move across the axon in the form of action potential, coming from the cell body.
1. Explain why the communication skills and techniques used within a business unit (department) are not always effective in communicating across business units or up and down the corporate ladder.
Answer:
Communication can be complicated in any organization. This is due to the fact that discussing peer to peer in a corporate office, of the institution, is going to be a lot unlike discussing peer to peer in a low-level office setting.
Also, cultural diversity can take a form in how efficient communication is. It is also very important that varying forms of communication are employed. Businesses that consists of many varying departments won't be dependent wholly on face to face communication. Therefore, it is necessary to have other forms of communication being utilized often. Examples of such Forms: email, phone, radio, etc.
In order to establish an effective route of communication that each sector of the organization feeds from, skilled communication professionals is necessary to be placed at every sector or place of business. This will permit corporate to possess “eyes on the target” per say. In other words, they can possess actual people all through every department or unit that can convey messages back and forth. International seems to do a good job at this. Their communications directors are involved with the relay of messages, permitting corporate’s suggestions which is then tailored to fit their particular department
Duroc Jersey pigs are typically red, but a sandy variation is also seen. When two different varieties of true-breeding sandy pigs were crossed to each other, they produced F1 offspring that were red. When these F1 offspring were crossed to each other, they produced red, sandy, and white pigs in a 9:6:1 ratio. Explain this pattern of inheritance in terms of number of genes, and any allele or gene interactions.
Answer:
The sandy variation could be as a result of a homozygous recessive allele at one of two varying genes in these two different types of sandy pigs. Let's refer to them as genes A and B.
A kind of sandy pig may be aaBB and the variety AAbb. The F1 generation in this cross produces heterozygotes for both genes to give all red. This indicates that the A and B alleles are dominant. In the F2 generation, 6 out of 16 give us homozygous for either the aa or bb alleles and produces sandy.
Out of 16 will we have one that would be doubly homozygous and be white. The other 9 will possess at least one dominant allele for both genes.
What is the difference between biomass and standing crop
Answer:
Biomass: can be defined as the total number of viable organisms, such as plants, animals in a given area.
While,
Standing crop: can be defined as the amount of biomass at a particular time or it is the amount of above-ground plant biomass.
One method for separating polypeptides makes use of their different solubilities. The solubility of large polypeptides in water depends upon the relative polarity of their R groups, particularly on the number of ionized groups: the more ionized groups there are, the more soluble the polypeptide.
1. Which of each pair of the polypeptides that follow is more soluble at the indicated pH?
(a) (Gly)20 or (Lys-Ala)3 at pH 7.0:
O (Gly)20
O (Lys-Ala)3
(b) (Glu)20 or (Phe-Met)3 at pH 7.0
O (Glu)20
O (Phe-Met)3
Answer:
(a) (Glu)zo or(Phe-Met)3 at pH 7.0
O (Glu)zo ✔
O (Phe-Met)s ❌
(b) (Gly) zo or (Lys-Ala)3 at pH 7.0:
O (Gly12) ❌
O (Lys-Ala)✔
(c) (Ala-Asp-Gly)s or (Asn-Ser-His)s at pH 3.0:
O (Asn-Ser-His)s ✔
O (Ala-Asp-Gly)s ❌
(d) (Ala-Ser-Gly)s or (Asn-Ser-His)s at pH 6.0:
O(Ala-Ser-Gly)s ❌
O (Asn-Ser-Hish)s ✔
Explanation:
Polypeptides that has polar or charged side chains are more soluble than polypeptides with nonpolar side chains.
(a) At ph 7.0
(Glu)20 is negatively charged at pH 7 and more soluble
(Phe-Met)3 is observed to be less polar and less soluble
(b)At ph 7.0
(Lys-Ala)3 is positively charged (polar) and more soluble
(Gly)20 is uncharged as only the amino- and carboxyl-terminal groups are charged as its less polar and less soluble too.
(c) At pH 6.0
(Asn-Ser-His)5 has polar Asn side chains and partially protonated His side chains and it's more soluble unlike the (Ala-Asp-Gly)s at that pH.
(d) At pH 3.0
(Asn-Ser-His)s as partially protonated carboxylate groups of Asp residues and it is also neutral but the imidazole groups of His residues are fully protonated and positively charged. Hence it's more soluble than the (Ala-Ser-Gly)s at that particular pH.
Proteins A, B and C bind to each other to form a complex, ABC. Under equilibrium the concentrations of A, B, C and ABC are 10-2 M. The equilibrium constant and the standard free energy of this association reaction at T=300 K are, respectively,
a) 10-6 M2 and -8.3 kcal/mol.
b) 10-4 M2 and -5.5 kcal/mol.
c) 10-3 M2 and -4.1 kcal/mol.
d) 10-2 M2 and -2.8 kcal/mol.
e) 10-1 M2 and -1.4 kcal/mol.
Answer: b) [tex]10^{4}M2[/tex] and [tex]-5.5 kcal/mol[/tex]
Explanation:
Equilibrium concentration of [tex]A[/tex] = [tex]10^{-2}M[/tex]
Equilibrium concentration of [tex]B[/tex] = [tex]10^{-2}M[/tex]
Equilibrium concentration of [tex]C[/tex] = [tex]10^{-2}M[/tex]
Equilibrium concentration of [tex]ABC[/tex] = [tex]10^{-2}M[/tex]
The given balanced equilibrium reaction is,
[tex]A+B+C\rightleftharpoons ABC[/tex]
The expression for equilibrium constant for this reaction will be,
[tex]K_c=\frac{[ABC]}{[A][B][C]}[/tex]
Now put all the given values in this expression, we get :
[tex]K_c=\frac{10^{-2}}{(10^{-2})^3}[/tex]
[tex]K_c=10^4M^2[/tex]
[tex]\Delta G^o=-2.303\times RT\times \log K_c[/tex]
where,
R = universal gas constant = 2 cal/K/mole
T = temperature = 300 K
[tex]K_c[/tex] = equilibrium constant = [tex]10^4[/tex]
[tex]\Delta G^o=-2.303\times 2\times 300\times \log (10^4)[/tex]
[tex]\Delta G^o=-5527.2cal/mol=-5.5kcal/mol[/tex]
Thus the equilibrium constant and the standard free energy of this association reaction at T=300 K are [tex]10^4M^2[/tex] and [tex]-5.5kcal/mol[/tex]
The incidence of Down syndrome, also known as trisomy 21, increases with increasing maternal age. Which of the following errors most likely produces this condition?A. Nondisjunction during meiosis II in either the male and female gameteB. Nondisjunction during either meiosis I or II in the female gameteC. Nondisjunction during either meiosis I or II in the male gameteD. Nondisjunction during meiosis I in either the male and female gamete
Answer:. Nondisjunction during meiosis I in either the male and female gamete
The failure of homologous chromosome to separate at mitosis, or failure of sister chromatids to separate at Anaphase of meiosis I or ii is called non-disjunction.
In this cell division anomaly the unsegregated chromosomes are separated into cells as single chromosomes, therefore extra copies are therefore inherited.This result is one of the cell having an extra copies of chromosomes than others( chromosomes 21.
If this extra chromosome copies is segregated into a sperm or egg, the resulting zygote cells will have extra copy as chromosome 21, and hence Down syndrome.
→It is more common in Meiosis 1 than ii,because of longer duration of the first stage of division than meiosis ii,given rooms to errors.
→Most common in women(with age increase) than men because of exchange of telomeres in them,which worsen with age. Their Meiotic mechanism which weaken with age, and more prone to errors compare to men,
Explanation: