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.
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:
In this exercise, you will research the effects of excess greenhouse gases in the atmosphere. You will apply the scienbific method and m ow greenhouse gases warm the planet by covering a jar with plastic wrap. You will then relate the findings of your experiment to global warming Procedure
In Exercise 1, we observed that gases have different properties In this section we will further investigate gases. How are greenhouse gases related to global warming?
Answer:
By heat trap mechanisms...
Explanation:
The main green house gases in our atmosphere are Carbon dioxide, Nitrous oxide, water vapors and ozone. All of these gases have a property of heat trap that trap the heat waves coming from the sun and heat up our atmosphere as a result. This is the reason why Venus is farther from sun but is more hotter than mercury because of it's dense Carbon Dioxide atmosphere.
Which of the following statements about phospholipids is TRUE? A Option A: Each one has two fatty acid chains and the glycerol backbone is bonded to a small polar group. B Option B: Each one has three fatty acid chains. C Option C: The glycerol backbone is bonded to a small nonpolar group. D Option D: Their biological function remains unknown.
Answer:
Each one has two fatty acid chains and the glycerol backbone is bonded to a small polar group.
Explanation:
Phospholipid is a unique form of lipid. The bonding of the glycerol backbone to the polar phosphate group makes phospholipid to have dual solubility unlike general triglycerides.
The polar head is said to be hydrophillic that is water loving, while the two carbon chains that retained lipid features are hydrophobic water hating.
Therefore if a phopholipid is placed in water, in relation to its functions as component of cell membrane, it forms a bi-layer in which the water loving portion hydrophilic head points into the surrounding watery medium, while the hydrophobic layer points inwards far away from the watery medium into the internal cellular layer to form an impermeable barrier to hydrophilic (polar) substances.
This forms the basis of the phospholipd bilayer of the cell membrane. And it controls the permeability of the cell membrane to influx substances into the cells.
Phospholipids consist of a glycerol backbone bonded to two fatty acid chains and a small polar group, making it amphiphilic. They form the plasma membrane of cells, with the polar heads facing the water environment and the non-polar tails facing away from the water.
Explanation:The correct answer is Option A: Each one has two fatty acid chains and the glycerol backbone is bonded to a small polar group. Phospholipids consist of a three-carbon glycerol backbone with two fatty acid molecules attached to carbons 1 and 2, and a phosphate-containing group attached to the third carbon. This structure gives the phospholipid molecule an amphiphilic, or "dual-loving," characteristic with a polar (charged) head and non-polar (uncharged) tail. Phospholipids play a crucial role in forming the plasma membrane of cells, where the hydrophilic (water-loving) head faces towards the water environment and the hydrophobic (water-avoiding) tail faces away from the water. Examples of phospholipids include Phosphatidylcholine and Phosphatidylserine.
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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
The house fly, Musca domestica, has a haploid chromosome number of 6. How many chromatids should be present in a diploid, somatic, metaphase cell?
Answer:
12
12
24
Explanation:
Musca domestica is the scientific name of an animal popularly known as Housefly. From the question, it is known that this organism has a haploid chromosome number of 6.
In a diploid chromosomes, the number of chromatids that will be present will be twice of that in haploid.
Since haploid (n) = 6, diploid (2n) will be; 2 × 6 = 12
Somatic cells on the other hand also exhibit diploid number of chromosomes, again it means we will have (2n) = 2 × 6 = 12
In metaphase cell, cells do make sure they complete the S'phase of the cell cycle before cellular division.Therefore, the DNA present in the chromatids in the S'phase actively engage in chromatids doubling, as a result, 24 chromatids (i.e 12 × 2= 24) exists in the metaphase cell.
The chromatids should be present in a diploid, somatic, metaphase cell-
diploid - 12somatic - 12metaphase cell- 24Chromatids at different stageThe scientific name of the fly Housefly is Musca domestica. It is known that this organism has a haploid chromosome number of 6. In the diploid cells, the number of chromatids that will be present will be twice that in haploid.
Diploid genome (2n)
= [tex]2\times6=12[/tex].
Somatic cells also exhibit diploidy (2n)
= 12
Metaphase stage cells contain 24 chromatids as cells before entering cell division complete S-phase of the cell cycle in which DNA present in chromatids become double due to replication.
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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.
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:
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.
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
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.
Each of two parents has the genotype red / blond, which consists of the pair of alleles that determine hair color, and each parent contributes one of those alleles to a child. Assume that if the child has at least one red allele, that color will dominate and the child's hair color will be red.
a. List the different possible outcomes. Assume that these outcomes are equally likely.
b. What is the probability that a child of these parents will have the blond/blond genotype?
c. What is the probability that the child will have red hair color?
The possible outcomes for the child's genotype are Rb, Rb, bR, and bb. The probability of the child having the Blond/Blond genotype is 25% while the probability of the child having red hair is 75%.
Explanation:In this genetics-based question, we're dealing with a common system of inheritance where the red hair allele is dominant over the blond hair allele. Let's denote the red hair allele as 'R' and the blond hair allele as 'b'. Since both parents have the genotype Red / Blond, their genotypes can be represented as 'Rb'.
a. The different possible outcomes for their children's genotypes (which determine hair color) can be determined via a Punnett square. The possibilities are: Rb (Red/Blond), Rb (Red/Blond), bR (Red/Blond), and bb (Blond/Blond).
b. The probability that a child of these parents will have the genotype Blond/Blond (bb) is 1 out of 4, or 25%.
c. If a child has at least one red allele (R), they will have red hair, owing to the dominance of the red allele. In this scenario, there are 3 out of 4 possibilities where the child has at least one red allele, so the probability that the child will have red hair is 75%.
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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.
Slow release of bacteriophage progeny from a bacterial host cell A. does not kill the host cell B. is a feature particular to filamentous phage C. causes cell lysis D. B and C are correct. E. A and B are correct.
Answer:
C. causes cell lysis
Explanation:
The slow release of the bacteriophage progeny from a bacterial host cell causes bacterial cell lysis and further kills the host cell which is the bacteria, this results to the liberation of progeny viruses which then infect new bacterial cells.
Only a type of bacteriophage called the filamentous bacteriophage reproduces without killing the bacterial host cell. The filamentous phages infect majorly gram negative bacteria.
The correct statements regarding the slow release of bacteriophage progeny are that it is characteristic of filamentous phages and does not kill the host cell. Therefore, the correct answer is E. A and B are correct.
The slow release of bacteriophage progeny from a bacterial host cell is a feature particular to filamentous phages and does not result in cell lysis. Therefore, the correct answer to the question is E. A and B are correct. Filamentous phages can exit the host cell without lysing it through a process called extrusion, where new virions are released from the cell membrane without causing cell death. Unlike lytic phages, which replicate and cause lysis of the host cell, filamentous phages exhibit a different life cycle that allows the host cell to survive the release of new phage particles. Therefore, the statement that slow release of bacteriophage progeny causes cell lysis is incorrect, because this slow release is characteristic of filamentous phages and does not kill the host cell.
The gene frequency for a dominant allele is 0.68 and the recessive allele is 0.39. What percent of the population has a homozygous recessive genotype?
Answer:
15.21 %
Explanation:
If we recall the basic formula of Hardy-Weinberg's equilibrium ; we have the following below:
p + q = 1
p² + 2pq + q² = 1
where;
p = frequency of the dominant allele in the population
q = frequency of the recessive allele in the population
p² = percentage of homozygous dominant individuals
q² = percentage of homozygous recessive individuals
2pq = percentage of heterozygous individuals
Given that p= 0.68 and q = 0.39
the percentage of the homozygous recessive genotype (q² ) will be
(0.39)² = 0.1521
= 0.1521 × 100
= 15.21 %
∴ the percentage of the population that has a homozygous recessive genotype = 15.21 %
Approximately 15.21% of the population has a homozygous recessive genotype.
Explanation:To calculate the percent of the population with a homozygous recessive genotype, we need to square the frequency of the recessive allele. In this case, the frequency of the recessive allele is 0.39, so squaring it gives us 0.39 * 0.39 = 0.1521. To convert this to a percentage, we multiply by 100, giving us 15.21%. Therefore, approximately 15.21% of the population has a homozygous recessive genotype.
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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.
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.
Watch the animation, DNA Replication. After DNA replication, how many double-stranded DNA molecules exist?
Answer:
The DNA is a double helix or made up of two strands. The strands are separated during replication, each serving as a template to produce a complementary strand of each of the separated single strands. Therefore after replication, two double stranded DNA molecules will be present.
Final answer:
Two double-stranded DNA molecules are produced following DNA replication, with each new molecule containing one old and one new strand, reflecting the semiconservative nature of the replication process.
Explanation:
After DNA replication, two double-stranded DNA molecules exist. This is because DNA replication is semiconservative, meaning each new DNA molecule consists of one original and one newly synthesized strand. During DNA replication, the double helix of the parent DNA molecule is unwound by enzymes. Then, DNA polymerase synthesizes a new complementary strand for each of the two single strands from the parent molecule.
The entire replication process ensures that each daughter cell receives an exact copy of the DNA. Mechanisms such as DNA proofreading and the repair of mistakes, which involve enzymatic correction during replication, help maintain the accuracy of the genetic information passed on to each new cell.
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.
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.
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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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.
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.
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 .Samples of rejuvenated mitochondria are mutated (defective) in 3% of cases. Suppose 18 samples are studied, and they can be considered to be independent for mutation. Determine the following probabilities. (a) No samples are mutated. (b) At most one sample is mutated. (c) More than half the samples are mutated.
Answer:
a) [tex]0.58[/tex]
b) [tex]0.598[/tex]
c) [tex]0[/tex]
Explanation:
Given -
Total sample i.e n [tex]= 18[/tex]
Probability (p) [tex]= 3[/tex] % [tex]= 0.03[/tex]
We will use binomial distribution theory for determining the probability of mutated sample
Let X be the number of mutated sample
a) No samples are mutated i.e [tex]X = 0[/tex]
[tex]P(X=0) = 0.03^0 * 0.97^{18}\\= 0.5779 = 0.58\\[/tex]
[tex]0.58[/tex]
b) At most one sample is mutated
[tex]P(X=0) = 0.58 + 0.03^1 * 0.97^{17}\\= 0.598[/tex]
c) More than half the samples are mutated.
[tex]P(X = 10) + ........+ P(X = 18) = 0[/tex]
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
Which characteristics of DNA polymerase I raised doubts that its in vivo function is the synthesis of DNA leading to complete replication?
The question is incomplete. The complete question is as follows:
Which characteristics of DNA polymerase I raised doubts that its in vivo function is the synthesis of DNA leading to complete replication?
its composition of a single polypeptide chain .
deficiency of enzyme in some organisms that are still capable of DNA synthesis .
requirement of Mg2+ presence in order for the enzyme to work .
low stability under normal physiological conditions.
Answer:
Deficiency of enzyme in some organisms that are still capable of DNA synthesis
Explanation:
The DNA polymerase I may be defined as the important enzyme that play an important role in the DNA replication of prokaryotes. DNA pol I is the replicating enzyme, DNA repair enzyme and can also acts as the exonuclease.
DNA pol I has been studied invitro and Arthur Korenberg explain the discovery of the DNA pol I. This DNA pol I plays an important role in DNA repair rather than the replication process. This explained invivo by the fact that some in some organisms the deficiency of this enzyme do not halt the process of replication. If the DNA pol I acts as the main replaicating enzyme, the DNA synthesis must be stopped in the organisms that lack DNA synthesis.
Thus, the correct answer is option (2).
is rectus femoris NOT a muscle of the the medial forelimb of the fetal pig that you will be dissecting?
Answer:
Yes it in not a muscle of the medial forelimb of the fetal pig
Explanation:
Rectus femoris is not the part of forelimb of the fetal pig.
However, rectus femoris in fetal pig originates from the hip and that to through the ilium and extends towards the knee and insert into its tendon. These are responsible for flexing the hip and extension of lower knee and legs. This muscle is a part of muscles of thigh and lies at its top. In quadriceps these muscles are the most anterior muscles.
Answer: Rectus femoris is not a muscle of medial forelimb of the fetal pig.
Explanation:
Pigs are mammals like humans. Pigs have similar structures with humans but there are slight differences in these structures because pigs are quipedal and humans are bipedal.
Rectus femoris is not found in the medial forelimb of fetal pigs instead it is found in the medial hindlimbs of fetal pigs and is one of the quadriceps muscles. It starts from the hip through the ilium and extend to the knee.Rectus femoris function in knee extension, its extend the lower leg and it flexes the hip.
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.
pH has a major influence on protein structure by altering electrostatic interactions. In order to illustrate this point, let us think about polylysine (a polypeptide chain consisting of only L-lysine residues). At pH 10 and above, polylysine forms an ahelix. At a pH of 7 and below, however, this same polypeptide chain assumes an unfolded conformation (often referred to as ‘random coil’ or ‘random conformation’).
1) Can you explain why this transition occurs at pHs below the pKa of Lys?
2) What other residue(s) might you expect to show a similar pattern of pH-dependent folding and unfolding?
3) The residue(s) you might expect based on charge to be capable of forming α-helices do not do so in water when they alone make up a polypeptide chain. Can you come up with a reasonable explanation for why this might be?
4) Speculate on what you think might happen to the pH-dependence of α-helix formation if you had a polypeptide chain consisting of both glutamate and lysine residues.
Answer:
Explanation:
(1) Remember that this occurred in a polylysine which consists only of L-Lys, the pKa of Lys residues might be different in polylysine as compared with the free amino acid in the solution.
The transition can also occur if more than 50% of the lysine residues need to be charged in order to ‘break’ the helix. Note that there is a 50:50 ratio of the coordinated bond of protons and dissociated forms.
(2) Other residues that are positively charged at neutral pH. They are arginine and histidine.
(3) Both arginine and histidine are more voluminous than Lys. Steric interference can prohibit the formation of an α-helices.
. Given that 3 of the 64 possible codons are stop codons, what is the chance of having a stop codon at any given position, assuming that the sequence is random?
Answer:
[tex]1 - (\frac{61}{64})^n[/tex]
Explanation:
Given
There are three stop codon
TAA, TAG, TGA.
Out of 64 available codons, 3 are stop codon thus the remaining are non-stop codon.
So the probability of choosing a non stop codon is
[tex]\frac{61}{64}[/tex]
Let us suppose the number of trials are "x" then the probability of choosing a non stop codon is
[tex](\frac{61}{64})^x\\[/tex]
Probability of choosing a stop codon is equal to
[tex]1 -[/tex] probability of choosing a non stop codon
[tex]1 - (\frac{61}{64})^n[/tex]
The chance of having a stop codon at any given position in a random sequence of codons can be calculated by dividing the number of stop codons by the total number of possible codons.
Explanation:In a given sequence of codons, the chance of having a stop codon at any given position can be calculated by dividing the number of stop codons by the total number of possible codons. The number of stop codons is given as 3, and the total number of possible codons is 64. Therefore, the chance of having a stop codon at any given position is 3/64, or about 0.046875, assuming that the sequence is random.
Learn more about Stop codons here:https://brainly.com/question/34227988
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