Answer:
Explanation:
1. There are six lanes in total out of which three are being showing self-splicing activity and the remaining three are being provided with splicing enzymes. As can be seen in case of RNA A, in the absence of a nuclear extract or when no splicing enzymes are added, there would be no splicing which can be reflected in lane 1 showing formation of only a single band which indicates that splicing does not occur on its own, although when the nuclear extract containing the splicing enzyme, there will be separation of the exons and the introns leading to the formation of two bands. So, the first two lanes specific for RNA A will be having a Spliceosomal Intron.
2. Now the task remains to identify which out of the remaining RNA's will be either Group I or Group II introns. There is no data regarding the addition of Guanosine nucleotide which is usually required in case of Group 1 splicing. Now, as we know that the splicing mechanism in case of Group II and spliceosomal intron is similar being carried out in a way in which an Adenine which is located in the branch site shows binding to the 5'- splice site which shows a similar conformational change for both the groups of introns. So, the Ribosome 2 which will act as a ribozyme and specifies lane 3 and 4 should be Group II intron and thus the last two lanes namely 5 and 6 will specify Group I intron.
An organism in which a parasite or virus lives is called a
Answer:
the answer is a parasite
Answer:
A Host
Explanation:
If a parasite is living, its not going to live on something it can't benefit from, this is called a host.
Think of a cat with fleas, would a flea (the parasite) live on another parasite? NO
THEY WOULD LIVE ON A HOST
Meselson and Stahl were trying to determine if DNA replication was semiconservative, conservative, or dispersive by labeling E. coliDNA with a regimen of heavy nitrogen (H) for one round of replication and then transferring these cells to light nitrogen (L) for two more rounds of replication. Which of the following statements wouldnotbe true within the context of this experiment?a. If DNA replication were conservative, no DNA molecules of intermediate density (H-L) would have been seen.b. If DNA replication were dispersive, only DNA molecules that were of intermediate density (H-L) would have been seen.c. If DNA replication were semiconservative, the DNA molecules that were made would all continue to be heavy density (H-H).d. If DNA replication were semiconservative, the DNA molecules would consist of one parental strand basepaired to one newly replicated strand.e. If DNA replication were semiconserative, a higher proportion of DNA molecules from future divisions would have been low density (L-L)
Answer:
Option C
Explanation:
All of the following options are correct except option C. in the Meselson Stahl experiment, after the first round of replication, the DNA molecules made consisted mainly of one parental strand and one new strand each which is the DNA molecules of intermediate density (H-L). Then future replications will produce more of the low density L-L density. This form of replication is thus termed to be the semi conservative form of replication
Which major environmental factors can affect health?
genetics and obesity
poor water quality and obesity
poor air and water quality
genetics and poor air quality
Answer:
The answer is poor air quality and water quality
Answer:
poor air and water quality
Explanation:
Desquamation occurs in which layer
Answer:
Desquamation occurs in the outermost layer of the skin.
Explanation:
Desquamation is also known as skin peeling and it occurs in the outermost layer or membrane of a tissue such as the skin .
Which of your cells have all of your DNA code?
Which of the following best explains the current state of the majority of the worlds fisheries
Answer:
That they are currently overfished.
Explanation:
OPTIONS
A)they are being sustainably harvested
B)they are currently overfished
C)they are currently underfished
D)they are already depleted
Answer:
B. they are currently overfished
Explanation:
i just got it right on a test
The forward and reverse PCR primers used for the amplification of a VNTR locus on chromosome number 4 are 20 bases and 21 bases long, respectively. The 3’ base of the forward primer is 53 bases away from the first repeat. The 3’ base of the reverse primer is 28 bases away from the first repeat in that direction. Each repeat unit is 60 bases in length. A person has, on one of their number 4 chromosomes, 6 repeats. How long (in bp) should its PCR product be?
The repeats which occur together on a chromosome are referred to as VNTR. The length of each repeat is 60bp.
The first primer is 20bp and is 53 bp away from 1st repeat. Therefore, the total length of sequence amplified by the first primer till 1st repeat = 20+53 = 73bp. Beyond this point 6 repeats on 60 bp are present, hence, the length becomes = 73 + (60X6) = 433bp.
Now the reverse primer which is 21bp in length is located 28bp from the repeat on its side. Reverse primer comes from the opposite direction, so it must be present after the 6 repeats.
The total length of the amplified region = 433 + 28 + 21 = 482 bp.
The PCR product for the given VNTR locus on chromosome number 4 should be 482 base pairs long.
Explanation:The PCR product length can be calculated by adding the lengths of the forward primer, the reverse primer, and the number of repeat units, including the sequences between them. The forward primer is 20 bases long, and its 3' base is 53 bases away from the first repeat. Therefore, the forward primer covers 20 + 53 = 73 bases. Similarly, the reverse primer is 21 bases long and its 3' base is 28 bases away from the first repeat. Hence, the reverse primer covers 21 + 28 = 49 bases.
Each repeat unit is 60 bases long, and the person has 6 repeats. So, the length of the repeat units is 60 * 6 = 360 bases.
To calculate the PCR product length, add the lengths of the forward primer, the reverse primer, and the repeat units: 73 + 49 + 360 = 482 bases. Therefore, the PCR product should be 482 base pairs long.
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Which of the following would increase the effect of erosion in a watershed? *
1 point
floodplains
steep slope
lack of soil
abundant plant life
Which safety measures should you follow during a hurricane? Check all that apply.
Avoid low-lying areas.
Go to low-lying areas.
If your home is on low ground, go to the roof.
If your home is on low ground, go to a shelter.
If you are inside, move away from windows.
If you are inside, open doors and windows.
Answer: Avoid low-lying areas
If your home is on low ground, go to a shelter
If you are inside, move away from windows
Explanation: Low-lying areas are flood zones. The flood water could contain debris that can harm people. With the windows, the wind could slam debris into it and break it. Closed doors can save your life if debris slams into it as well.
Answer:
1, 4, 5. on edge.
Explanation:
Explain two reasons why having an oxygen-rich atmosphere is important for pife on Earth?
Answer:
Humans need oxygen to breathe.
Majority of animals that humans depend on for food need oxygen to breathe.
Answer:
First, oxygen makes up the ozone layer, blocking harmful radiation from the sun, and all living things need oxygen to survive!
Explanation:
Earth's tectonic plates
A.
do not move.
B.
move only up and down.
C.
are constantly in motion.
D.
move only once.
Answer:
The answer is C
Explanation:
Tectonic plates, the massive slabs of Earth's lithosphere that help define our continents and ocean, are constantly on the move.
Which of the following genotypes would be
present if a recessive trait is expressed in the
phenotype?
Answer:
dd, both parents would also be (dd) as well.
Explanation:
Answer:
One of Gregor Mendel's greatest contributions to the study of heredity was the concept of dominance. Mendel observed that a heterozygoteoffspring can show the same phenotype as the parent homozygote, so he concluded that there were some traits that dominated over other inherited traits
What is a cell that would have a high rate of cell division
Explanation:
The highest rate of cell division is a human embryo. Once the sperm finds a viable egg in fallopian tube, it travels to the uterus. Attaches to the uterus where it is called a zygote then begins to multiply rapidly, along with the placenta (from which the developing embryo gets o2, food source, all nutrients a fetus needs to grow,to develop the unborn fetus..
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The promoters of positively controlled operons require activator proteins because The promoters of positively controlled operons require activator proteins because they are needed to bind to the allosteric site of RNA polymerase. RNA polymerase easily recognizes the consensus sequence. the promoters bind RNA polymerase weakly and utilize activator proteins to help RNA polymerase recognize the promoter. they are required to inactivate the repressor proteins.
Answer:
the promoters bind RNA polymerase weakly and utilize activator proteins to help RNA polymerase recognize the promoter.
Explanation:
The promoters of positively controlled operons require activator proteins because the promoters bind RNA polymerase weakly and utilize activator proteins to help RNA polymerase recognize the promoter.
How does carbon become locked inside the earth?
A) tree roots expel carbon into the soil as they grow
B) humans use machines to pump carbon into large underground tanks
C) rocks become richer in carbon as time goes on
D) the carbon-rich bodies of plants and animals decay into the earth after they die
Answer:
a) tree roots expel carbon into the soil as they grow.
Explanation:
Roots offer physical anchoring to the soil of the tree, with water uptake and inorganic nutrition as its principal activity.
The absorption and the transfer of ions and the growth and maintenance of the roots themselves require energy, therefore releasing CO2 into the ground and into the air.
A diverse ground microbiota exists underneath the earth in addition to the plant roots. Their work results in carbon dioxide being released into the soil, which flows into the air. This soil microbiota is responsible for decomposing the litter and, in general, any dead organic matter in the soil. It may breathe more than half of the total soil respiration. Soil temperature and humidity strongly affect root and microbial activity. In warm and moderately moist soils, soil breathing is highest.
The correct answer to how carbon becomes locked inside the Earth is option D: the carbon-rich bodies of plants and animals decay into the earth after they die.
Explanation:This natural process is part of the carbon cycle, where carbon is stored in soil as organic carbon due to the decomposition of living organisms by decomposers or from the weathering of terrestrial rock and minerals. Over long periods, this material might be buried and converted into fossil fuels, which are a type of non-renewable resource because they are regenerated very slowly, if at all. These fuels include coal, oil, and natural gas, which are the anaerobic remains of plants and algae that lived millions of years ago.
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behavioral physiologist is studying the homeostatic control of blood pH. In a trial, a lizard runs on a treadmill for a set amount of time and the blood pH is measured. The blood pH drops as carbon dioxide is released into the bloodstream. The body uses peripheral chemoreceptors to regularly measure blood pH. These chemoreceptors are the ____ in this example.
Answer:
sensor
Explanation:
Homeostasis refers to the ability of an organism to maintain the internal conditions of the body irrespective of the external conditions.
The homeostasis is maintained by the homeostatic mechanism which involves the 3 components which are the receptor or sensor, the integrating centre and the effector or organ.
In the given question, the peripheral chemoreceptors measures and sense the pH of the blood therefore the peripheral chemoreceptor will be considered as the "sensor component" of the homeostatic mechanism.
Thus, the sensor is the correct answer.
Enzyme B helps in the process of breaking molecule A into molecules C and D Which of the following outcomes can be expected?
Answer:
A
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Enzyme B acts as a catalyst to convert molecule A into molecules C and D by reducing the activation energy needed for the reaction. The process could follow a sequential or ping-pong mechanism, where the enzyme binds substrates and facilitates their transformation into products.
Explanation:Enzyme B facilitates the breakdown of molecule A into molecules C and D. This process is an example of an enzymatic reaction where the enzyme lowers the activation energy, thus speeding up the reaction without being consumed in the process. The expected outcome would be the conversion of substrate A into products C and D, with enzyme B acting as a catalyst.
According to the provided information, enzymes work by reducing the activation energy (Option B), not by increasing it or altering the exergonic or endergonic nature of reactions. This reduction allows the enzymatic reaction depicted, A+B=C+D, to proceed more efficiently.
Enzymes may operate via several mechanisms, including sequential reactions where substrates bind in an ordered or random fashion. In the case of Enzyme B, if it follows a ping-pong mechanism, it would alternate between two states, without A and C bound simultaneously, to facilitate the reaction between molecules A and B to form C and D.
A beaker is divided by a membrane that is permeable to water and
impermeable to sodium ions. The sodium-water solution on side A is more concentrated than the sodium-water solution on side B. Which of the following statements will be true when equilibrium is reached across the membrane?
a) The absolute number of water molecules will be different on the two sides of the membrane, whereas the final solute concentration will be equal.
b) Equal numbers of water molecules will exist on each side of the membrane at equilibrium.
c) Since the solute can't cross the membrane, the solution will not reach equilibrium.
d) The solute will maintain the original concentration on both sides of the membrane. e) All answers are correct.
Answer:
a) The absolute number of water molecules will be different on the two sides of the membrane, whereas the final solute concentration will be equal.
Explanation:
According to the concept of osmosis, solvent molecules pass via a semi permeable membrane from area of lower solute concentration to higher solute concentration. Here, solution A and B have different concentration and they are separated by a semi permeable membrane which is only permeable to the water molecules.
Water will move from side B to side A since side B is less concentrated so it has more free water molecules. Eventually, the concentration will become equal on both sides and net movement of water molecules will become zero. However, absolute number of water molecules will be different in both the sides because originally there was more solute in side A so to make the concentration equal, more water molecules will now be present here than side B.
Which is the answer for it? I will appreciate the help
Use the following information to answer the following question below. Moss invades and establishes itself on bare rock, accumulating the beginnings of soil. After several years, enough soil has become established that grasses begin to grow where there was once bare rock. Without the moss building up soil, the grasses would have had no chance. The establishment of the mosses on bare rock and eventual addition of grasses represents intermediate succession. secondary succession. a climax ecosystem. primary succession.
Answer:
Primary succession.
Explanation:
Ecological succession refers to the changes involved in the species structure of a community that is established in a particular habitat over time. The succession is of two types: the primary succession and secondary succession.
When the succession takes place on the barren land then succession is considered primary succession.
In the given question, the moss invades and colonizes the rocks and breaks them into the soil. Here the moss species is known as the pioneer species and since the moss grows on the barren rock, therefore, is considered the primary succession.
Thus, primary succession is correct.
For both clinical and cosmetic reasons, plastic surgeons inject substances into connective tissue underlying the skin epidermis. This plumps up areas deficient in soft tissue, and is used to reduce surgical scars or wrinkles, for example. Which normal connective tissue components are good candidates to be injected as fillers in this type of procedure?Choose one or more:
A.) cadherin
B.) hemidesmosome
C.) collagen
D.) glycosaminoglycan
Answer:
C.) collagen
D.) glycosaminoglycan
Explanation:
Cadherins are the main cell adhesion molecules. These molecules are transmembrane glycoproteins responsible for cell-cell junctions to maintain the integrity of animal tissues. Collagen is a protein molecule or protein that forms fibers, the collagen fibers. They are secreted by connective tissue cells such as fibroblasts, as well as by other cell types.Hemidesmosomes are cell-binding structures that connect epithelial cells to the basement membrane. They are especially important in tissues subjected to mechanical stress. Glycosaminoglycans are a type of functional and structural biomolecules, mainly present in connective, epithelial and bone tissue, as well as in the intercellular environment.LESSON REVIEW
1. Which trophic level provides nourishment for the most organisms in the
biome?
amers
B. secondary consumers
C. producers
D. decomposers
2. A seed is eaten by a chipmunk. The chipmunk is eaten by a snake. The
snake is earen by an acle. Which of the statements about this acnee
Answer:
1.C
2.its a secondary consumer
Explanation:
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Research suggests that eating lots of fruits and vegetables may provide some protection against several types of cancer. Studies have revealed that the risk of getting cancer associated with the lowest intakes of fruits and vegetables is twice as high as that associated with the highest intakes. This association holds for several types of cancer, including cancers of the breast, colon, pancreas and bladder.
Answer:
So is; It has been shown, for example, in breast cancer that women who eat a lot of fruits and vegetables every day may have a lower risk of breast cancer, especially aggressive tumors, than those who consume less vegetables, according to a new study conducted by researchers at the TH Chan School of Public Health in Boston, United States. Cruciferous vegetables such as broccoli and yellow and orange vegetables stand out especially.
Explanation:
The plant-based foods that the researchers studied for cancer prevention are:
Cruciferous vegetables. They include broccoli, cauliflower, cabbage, Brussels sprouts, Chinese cabbage, and green cabbage. Frequent consumption of these foods is associated with a reduced risk of cancer.
Studies show that cruciferous vegetables provide protection against:
- Head and neck cancer.
- Esophagus cancer.
-Stomach cancer.
Several laboratory studies suggest that cruciferous vegetables help regulate enzymes that protect against cancer. Studies also show that cruciferous vegetables can stop the growth of cancer cells in other ways. But these effects can differ between cells, laboratory animals, and people.
Lycopene. This carotenoid is found in products derived from tomato. Other important sources of lycopene are pink grapefruit, watermelon, and apricot.
Studies show that lycopene can provide cancer protection from:
-Lung
-Stomach
-Prostate
-Colon
Given that generating cleaner energy is costly which fact illustrates the likelihood of developing nations will not be able to use the cleanest technologies
Answer:B the use of electricity is growing much faster than the economies are, as measured by GDP
Explanation: I just did the question right now
Which statement below is TRUE?
A)Crossing over shuffles genes between non-homologous chromosomes.
B)During meiosis, cells arrange their alleles so that beneficial alleles are passed on together.
C)Without crossing over, offspring would be genetically identical to parents.
D)New combinations of alleles arise from the random events of crossing over, independent assortment, and mutation.
Answer:
I think it is C.
Explanation:
C)Without crossing over, offspring would be genetically identical to parents.
Without crossing over in meiosis , offspring would be genetically identical to parents.
Meiosis is a type of cell division which occurs twice and forms four unique
daughter cells which have only one copy of chromosome from the parent.
This is as a result of the crossing over of the chromosome so that the parent
and offspring can have unique genetic makeup.
Meiosis is involved in sexual reproduction during the production of sperm and egg cells.
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Kaylee wants to test her hypothesis that she performs better on tests after getting more sleep. In which way will she best be able to do this?
A. by analyzing her test scores when she gets the same amount of sleep
B. by asking her friends if sleep affects their test scores
C. by analyzing her average test scores after nights with different amounts of sleep
D. by reading online about the relationship between sleep and test scores
Human pathogens typically fall into the group called
Human pathogens typically fall into the group called microorganisms which include bacteria, viruses, fungi, and protozoa. Pathogens are responsible for causing various diseases in humans.
Explanation:Human pathogens typically fall into the group called microorganisms. Microorganisms are tiny living organisms, which are usually microscopic and include bacteria, viruses, fungi, and protozoa. These pathogens are responsible for disease in their host.
For example, bacteria, such as Escherichia coli, can cause food poisoning, and viruses, like the influenza virus, can cause the flu. Alternatively, fungi can lead to infections like Athlete's foot and protozoa can result in diseases such as malaria.
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Lipid bilayers formed between two aqueous phases have thisimportant property: they form two-dimensional sheets, the edges ofwhich close upon each other and undergo self-sealing to formliposomes.
a. What are the forces that drive bilayerformation? What are the consequences of this property for thestructure of biological membranes?
b. Explain why phosphoglycerides are capable of spontaneouslyassembling into the bilayer structure found in biological membranesbut triacylglycerols are not.
Answer:
a-The assembly process is driven by interactions between hydrophobic molecules (also called the hydrophobic effect), an increase in interactions between hydrophobic molecules (which causes clustering of hydrophobic regions) allows water molecules to bond more directly between Yes, the entropy of the system increased. This complex process includes non-covalent interactions, tales like Van der Waals, electrostatic bonding, and hydrogen bonding.
b-The fluidity of the membranes depends mainly on the length and the level of unsaturation of the phospholipids. Indeed, the shorter and more unsaturated the membrane lipids, the more fluid the membrane is. This is due to the fact that long chains show a greater association with each other. Specifically, for each -CH2- group added, the free energy of interaction between two chains decreases by 0.5 kcal mol-1, making the interaction stronger. When the chains are shorter, the interaction between the alkyl groups of the lipids is reduced.
Explanation:
The lipid bilayer is a thin polar membrane made up of two layers of lipid molecules, membranes are flat sheets that form a continuous barrier around cells and their structures
An overproducer E. coli strain is an E. coli strain, which carries a recombinant plasmid with an ORF of a protein of interest under the control of a very strong promoter, therefore it produce large quantities of the protein of interest. For example, recombinant human insulin is produced using an overproducer E. coli strain, which carries a plasmid with the ORF of human insulin.
(a) You want to make an overproducer E. coli strain, which will make a protein from bacterium Salmonella, YFPS (your favorite protein from Salmonella). Below is a restriction map of a fragment of Salmonella genomic DNA, which contains ORF of YFPS and a restriction map of a plasmid, which you are using to make the overproducer. Which pair of restriction enzymes would you use?
b.) You want to use the same strategy to produce a human protein in E. coli cells, YFPH (your favorite protein from humans). You have two potential sources to make the insert: a fragment of human genomic DNA, which contains YFPH gene, and a fragment of human cDNA, which contains YFPH ORF. Which one would you use?
c.) What enzyme synthesizes DNA using mRNA as a template during cDNA construction?
Answer:
a) Pair of restriction enzymes: EcoRI and BamHI
b) The fragment of cDNA that contains YFPH ORF
c) The reverse transcriptase
Explanation:
a) Restriction enzymes are defined as those enzymes that cut DNA. These enzymes recognize a target sequence and cut DNA near those sequences. Choosing these two restriction enzymes allows obtaining a full length protein.
b) An intron is defined as a part of the gene that does not code for any amino acid. The regions in a sequence that are expressed in proteins are called exons because they are expressed, and human genomic DNA has introns, which would not be part of the desired protein.
c) Reverse transcriptase is a DNA polymerase type enzyme, its function is to synthesize double-stranded DNA using single-stranded RNA as a template. This enzyme aids in the formation of the double-stranded DNA when RNA has undergone reverse transcription in the cDNA strand. Reverse transcription means the synthesis of DNA from RNA.
a. Bacteriophage MS2 carries its genetic information in RNA. Its chromosome is analogous to a polygenic molecule of mRNA in organisms that store their genetic information in DNA. The MS2 mini-chromosome encodes four polypeptides (i.e., it has four genes). One of these four genes encodes the MS2 coat protein, a polypeptide 129 amino acids long. The entire nucleotide sequence in the RNA of MS2 is known. Codon 112 of the coat protein gene is CUA, which specifies the amino acid leucine. If you were to treat a replicating population of bacteriophage MS2 with the mutagen 5-bromouracil, what amino acid substitutions would you expect to be induced at position 112 of the MS2 coat protein (i.e., Leu —> other amino acid)? (Note: Bacteriophage MS2 RNA replicates using a complementary strand of RNA and base-pairing like DNA.)
b.Would such mutations occur if a non-replicating suspension of MS2 phage was treated with 5-bromouracil?
Final answer:
Amino acid substitutions at codon 112 in MS2 phage could occur due to mispairing caused by the mutagen 5-bromouracil during replication, potentially resulting in a different amino acid being coded. However, mutations would not occur in a non-replicating suspension of MS2 since 5-bromouracil incorporation requires active replication.
Explanation:
If you were to treat a replicating population of bacteriophage MS2 with the mutagen 5-bromouracil, you might expect amino acid substitutions at codon 112 of the MS2 coat protein, which commonly specifies the amino acid leucine (Leu). Given that 5-bromouracil can substitute for thymine in DNA replication, and considering RNA's analogous base uracil, during RNA replication, this mutagen could potentially cause uracil to mispair, leading to a substitution mutation. Specifically for codon 112 (CUA), a possible mutation could turn it into UUA which would likely also code for leucine, or into CUG, which also encodes for leucine due to the degeneracy of the genetic code. A more significant change, such as substituting cytosine with uracil, creating UUA or UUG, could result in a phenylalanine being placed instead of leucine, since these codons specify phenylalanine.
For the second part of the question, mutations from 5-bromouracil would not occur in a non-replicating suspension of MS2 phage since mutations require nucleic acid replication for incorporation into the genetic material, and 5-bromouracil would not be incorporated without active replication.