Answer: cDNA library will be created .
and mRNA from human cells to serve as source of crystallins will be used
Explanation:
To clone a gene, you would need to use a technique called gene cloning or recombinant DNA technology. The basic steps of gene cloning involve isolating the gene sequence of interest, inserting it into a vector, and then transferring the vector into a host cell.
Explanation:To clone a gene, you would need to use a technique called gene cloning or recombinant DNA technology. The basic steps of gene cloning involve isolating the gene sequence of interest, inserting it into a vector, and then transferring the vector into a host cell. In this case, since you want to clone a human crystallin gene, you would need to isolate the specific gene sequence for the human crystallin protein. Once you have the gene sequence, you can use the technique of gene cloning to obtain multiple copies of the gene, which can then be expressed and studied.
The calcium carbonate (CaCO3) stones located on the maculae are called:
a. ampullae
b. stereocilia
c. ossicles
d. ottoliths
Answer:
d. ottoliths
Explanation:
An otolith is a calcium carbonate structure in the saccule or utricle of the inner ear, majorly in the vestibular system of vertebrates.
The otolith organs is chiefly made up of the saccule and utricle
A) Prof. Robo is studying skin color of Banana Slugs. He has isolated 8 recessive mutants that are white instead of yellow. He performs the following complementation analysis. (+) indicates yellow color (-) indicates white color. Based on complementation analysis, how many genes affect skin color? Place the various mutants in the different complementation groups.
Answer:378
Explanation: 378
have been cloned and
The other 207 loci have been mapped out, the true gene identities have yet to be determined.
Red-flowering snapdragons are homozygous for allele R1. White-flowering snapdragons are homozygous for allele R2. Heterozygous plants (R1R2) bear pink flowers. What phenotypes should appear among F1 offspring of the crosses listed below? What are the expected probabilities for each phenotype?
Answer:
a) 1/2 red, 1/2 pink; b) all pink; c) 1/4 red, 1/2 pink, 1/4 white; d) 1/2 white, 1/2 pink
Explanation:
The flower color trait in snapdragons shows incomplete dominance: the heterozygous genotype produces an intermediate phenotype between the two different homozygous genoytpes.
The possible genotypes and phenotypes are:
R1R1 : redR1R2: pinkR2R2: whitea. R1R1 X R1R2The R1R1 individual only produces R1 gametes. The R1R2 parent produces 1/2 R1 gametes and 1/2 R2 gametes.
For that reason, the F1 will be:
1/2 R1R1 (red)1/2 R1R2 (pink)b. R1R1 X R2R2The R1R1 individual only produces R1 gametes. The R2R2 parent only produces R2 gametes.
For that reason, the F1 will be :
100% R1R2 (pink)c. R1R2 X R1R2Both parents are heterozygous. This is a monohybrid cross, and from Mendel's Laws we expect the following offspring:
1/4 R1R1 (red)2/4 R1R2 (pink)1/4 R2R2 (white)d. R1R2 X R2R2The R1R2 parent produces 1/2 R1 gametes and 1/2 R2 gametes. The R2R2 individual only produces R2 gametes.
For that reason, the F1 will be:
1/2 R2R2 (white)1/2 R1R2 (pink)Choose the key terms listed below with the phrase that is the best match for it.a. Reinforcement b. Cryptic species c. Biological species concept d. Morphospecies concept e. Species f. Phylogenetic species concepti. The smallest evolutionarily independent unit. ii. Distinguishing species based on their phenotypes. iii. Distinguishing species based on whether or not they interbreed regularly with each other and are reproductively isolated from other such groups iv. Distinguishing species by finding the smallest monophyletic group on a phylogeny v. Species that are evolutionarily independent from each other, but ARE NOT distinguishable based on morphology vi. Selection for reproductive isolation by means of reduced fitness of hybrids
Answer:
a) Reinforcement - Selection for reproductive isolation by means of reduced fitness of hybrids.
b) Cryptic species - Species that are evolutionarily independent from each other, but ARE NOT distinguishable based on morphology.
c) Biological species concept - Distinguishing species based on whether or not they interbreed regularly with each other and are reproductively isolated from other such groups.
d) Morphospecies concept - Distinguishing species based on their phenotypes.
e) Species - The smallest evolutionarily independent unit.
f) Phylogenetic species concept - Distinguishing species by finding the smallest monophyletic group on a phylogeny.
Explanation:
Species is a group of organisms which are closely related and very similar to each other. It is the smallest evolutionarily independent unit that are able to interbreed and produce fertile offspring.
The inability of related species to breed due to behavioral, geographical, morphological or genetic differences is known as reproductive isolation. When two populations have come back into contact after being separated, the complete reproductive isolation between them results in speciation and the incomplete reproductive isolation causes the production of hybrids, which are of low fitness or sterile.
The process of speciation in which natural selection increases the reproductive isolation between two populations of species by acting against the production of hybrids of low fitness is known as reinforcement.
Cryptic species are the species which are evolutionarily independent from each other, but are morphologically indistinguishable and cannot interbreed.
The different species concept are; biological species concept, ecological species concept, morphospecies concept, phylogenetic species concept, evolutionary species concept etc.
Biological species concept is a concept of species in which the individuals of a group can interbreed and create fertile offspring, but cannot breed with other groups (reproductively isolated).
Morphospecies concept or morphological species concept distinguish species based on the differences in their phenotypes (morphological characters).
Phylogenetic species concept distinguish species by finding the smallest monophyletic group (group of all organisms that are the descendants of a common ancestor) on a phylogeny (evolutionary history of genetically related group of organisms or species). This group have a shared and unique evolutionary history because they are descended from a common ancestor.
HIV is classified as a retrovirus because _____.
(A) it reverts to an inactive form when it infects B lymphocytes
(B) this virus is composed of two cells surrounded by a lipoprotein coat
(C) it makes a DNA copy of its RNA once inside the host cell
(D) it infects only cells with a CD4 receptor
(E) it causes the production of HIV antibodies
Answer: C
Explanation:
HIV is classified as a retrovirus because its genetic material is composed of single-stranded RNA nucleotide. And also Retroviruses have the enzyme reverse transcriptase, which is capable of making a DNA copy of its RNA once inside the infected host cell
Answer:
The answer is C; it makes a DNA copy of its RNA once inside the host cell
Explanation:
HIV is called a retrovirus because common to all retroviruses, they store their genetic information using RNA instead of DNA, and they need to ‘make’ or convert their RNA to DNA when they enter a human cell in order to make new copies of themselves.
How does the balance of populations in the ecosystem change when the rattlesnake population increases? Choose the two correct answers. A. The bison population increases. B. The grasshopper population increases. C. The grasshopper population decreases. D. The blue jay population increases. E. The prairie dog population decreases.
Answer: The correct options are C and E.
Explanation: When the rattlesnake population increases, the population of grasshopper and prairie dog are also decreases.
Both grasshopper and prairie dog are herbivores and eaten by rattlesnake. When rattlesnake population increases they feed on grasshopper and prairie dog and decrease occurs in their population.
Indicate at which step of the replication-transcription-translation process each type of RNA first plays a role.During which step of the replication-transcription-translation process does each type of RNA first play a role?replication?translation?transcription/ rna processing?
Answer:
Ribosomal RNA (rRNA) is used in the translation process
Messenger RNA (mRNA) is produced during the transcription process and is used in the translation process
Transport RNA (tRNA) is used in the translation process
and if you count the RNA produced by RNA primase than that is used in the replication process.
Answer:
that answer is correct
Explanation:
You cross a true breeding tomato plant with large fruits to a true breeding plant that has small fruits. All of the F1 offspring have medium sized fruits, with a variance in fruit size of 0.06 cm2 . When you create F2 plants, the variance in fruit size is 0.60 cm2 . What is the broad sense heritability (H2 ) of fruit size?
Answer:
H² = 0.9
Explanation:
Phenotypic variance (VP) = genetic variance (VG) + environmental variance (VE)
The observed VP in the F1 is composed only of VE, because the parents were true breeding (thus VG=0).
The F2 variance is due to both genetic and environmental variation. If the tomato plants are in the same environment, the VE wil be the same, and then we can calculate VG as:
VG = VP - VE
VG = 0.60 cm² - 0.06 cm²
VG= 0.54 cm²
Broad sense heritability is calculated as: H² = VG/VP. It is used to describe how much of the observed phenotypic variation is due to genetic variation (comprised by additive, dominance and epistasis variance).
In this case,
H² = VG/VP
H² = 0.54 cm² / 0.60 cm²
H² = 0.9
The broad sense heritability (H^2) of fruit size in the cross breeding experiment involving tomato plants with large fruits and small fruits is 0.90.
Explanation:The question is about determining the broad sense heritability (H2) of fruit size in a cross breeding experiment involving tomato plants with large fruits and small fruits. Broad sense heritability (H2) is a statistic used in genetics to estimate the proportion of the phenotypic variance in a population that is due to the genetic variance. It is given by the formula:
H2 = VG/VP
Where, VG is the genetic variance, and VP is the phenotypical variance. The phenotypic variance is made up of genetic variance (VG), environmental variance (VE), and possibly the interaction between genetic and environmental variance (VGxE).
In this case, we need to find out the VG and VP to find H2. Since the F2 variance (VF2) is provided (0.60 cm2) and the F1 variance (VF1) is provided (0.06 cm2), we can assume that the F1 variance represents the environmental variance (VE) and the F2 variance minus the F1 variance gives the genetic variance (VG) since F1 was produced by crossing two true breeding plants. Hence:
VG = VF2 - VF1
VG = 0.60 cm2 - 0.06 cm2 = 0.54 cm2
Substituting the values in the formula:
H2 = VG/VP = 0.54 cm2/0.60 cm2 = 0.90
So, the broad sense heritability of fruit size in this experiment is 0.90.
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In guinea pigs, the gene for black fur, B, is dominant over the gene for white fur, b. Complete the Punnett square below to show the results of one possible cross between two guinea pigs with black fur.
Answer:
In guinea pigs, the gene for black fur, B, is dominant over the gene for white fur, b. Complete the Punnett square below to show the results of one possible cross between two guinea pigs with black fur.
BB X Bb = BB, Bb, BB, Bb
Explanation:
From the analogy shown, it is obvious that both parents are gene with black fur. During such crossing, the offspring are two homozygous dominant black fur and two heterozygous dominant black fur
Day-to-day choices can help reduce the risk of heart disease. One of the major risk factors for development of heart disease is elevated LDL, which can be affected by the types and amounts of dietary fat consumed as well as other dietary factors. Read the statements below and select all of the correct statements regarding how various dietary fats affect LDL cholesterol levels. Select all that apply. a. Typically, the higher your consumption of unsaturated fats, the higher the LDL cholesterol levels in your blood. b. Trans fats are worse for heart health than saturated fats because they raise LDL cholesterol and lower HDL cholesterol. c. Increasing intake of plant foods may be one of the easiest ways to decrease LDL cholesterol. d. Typically, the higher your consumption of saturated fats, the higher the LDL cholesterol levels in your blood. e. Dietary cholesterol does not affect blood cholesterol levels.
Answer:
b. Trans fats are worse for heart health than saturated fats because they raise LDL cholesterol and lower HDL cholesterol.
c. Increasing intake of plant foods may be one of the easiest ways to decrease LDL cholesterol.
d. Typically, the higher your consumption of saturated fats, the higher the LDL cholesterol levels in your blood.
Explanation:
Generally, the consumption of saturated fat increases the amount of LDL cholesterol in the blood system while the consumption of unsaturated fat can reduce the amount of LDL cholesterol in the blood system. In addition, the consumption of plant food can reduce the amount of LDL in the blood system. Trans fats are more harmful to one's heart than saturated fat because they can increase the amount of cholesterol in the blood system.
Answer:b,c,d
Explanation:a. Unsaturated fats are mostly beneficial to health, e.g- omega 3 fatty acids, and are not known causes of LDL increase.
b. Trans fats are known to increase LDL cholesterol while also decreasing HDL cholesterol levels. We do not find this in saturated fats which are known to only increase LDL cholesterol.
c. Plants possess "soluble fiber" that drastically help to reduce cholesterol levels. They are known to be very low in saturated fats, and also free of cholesterol.
d. Saturated fats are common causes of elevated LDL in blood stream.
e. Dietary cholesterol as opposed to serum cholesterol are fats found in our diets(food) that may contain high levels of cholesterol that increase LDL levels in blood.
The major advantage of using artificial chromosomes such as YACs and BACs for cloning genes is that
plasmids are unable to replicate in cells.
only one copy of a plasmid can be present in any given cell, whereas many copies of a YAC or BAC can coexist in a single cell.
YACs and BACs can carry much larger DNA fragments than ordinary plasmids can.
YACs and BACs can be used to express proteins encoded by inserted genes, but plasmids cannot.
all of the above
H. Simply inserting an entire eukaryotic gene into a prokaryotic expression system will most likely not work for the following reason.
Prokaryotes lack promotors
Prokaryotes lack introns
Prokaryotes lack polymerase
Prokaryotes lack repressors
Question is a multiple choice question.
Answer:
Question 1. The major advantage of using artificial chromosomes such as YACs and BACs for cloning genes is that:
Answer:
(C)
Explanation:
YACs and BACs can carry much larger DNA fragments than ordinary plasmids can.
Question 2. Simply inserting an entire eukaryotic gene into a prokaryotic expression system will most likely not work for the following reason.
Answer:
(B)
Explanation:
Prokaryotes lack introns
Human hair color is a classic, if oversimplified, example of recessive epistasis. Red hair is caused by a recessive allele r. However, if an individual has a dominant R allele, they may have either brown or blonde hair depending on whether they have a dominant B allele (which causes brown hair), or are homozygous for a recessive b allele (which causes blonde hair). The B and R loci are on located on different chromosomes. If a couple with genotypes Bb; Rr and bb; rr have children, what hair colors and in what proportions are expected among their children?
Answer:
1 Brown: 1 Blonde: 2 Red
Explanation:
According to the given information, the recessive allele "r" gives red color to hair but is epistatic to alleles B and b. Therefore, the genotype with two copies of the "r" allele would have red-colored hair. The genotypes with at least one copy of "B" and “R" alleles each would have brown hair while the "R" allele would give blond hair in presence of allele "b".
Therefore, a cross between BbRr and bbrr would produce progeny in following phenotype ratio= 1 Brown: 1 Blonde: 2 Red
Human hair color is a polygenic trait influenced by multiple genes. For a couple with genotypes Bb; Rr and bb; rr, a Punnett square predicts their children will have an equal chance of brown, blonde, or red hair in a 1:1:1:1 ratio.
Explanation:Hair color in humans is an example of a polygenic trait influenced by multiple genes. Individuals with certain combinations of alleles for these genes will display various hair colors such as black, brown, blonde, or red. In the scenario provided, where one parent has the genotype Bb; Rr (capable of brown or blonde hair with a possibility of red) and the other parent has the genotype bb; rr (red hair), we can predict the possible hair colors of their offspring using a Punnett square to combine their genotypes.
Since red hair requires two recessive r alleles and the second parent provides only recessive r alleles, all the offspring must carry at least one r allele, ensuring the possibility of red hair. Also, as brown hair is dominant over blonde and requires at least one B allele, which only one parent carries, we can expect some children to inherit brown or blonde hair depending on whether they inherit the B allele from the first parent.
Combining these alleles from both parents would yield a 1:1:1:1 ratio of the following phenotypes: Brown (Bb; Rr), Blonde (bb; Rr), Red (Bb; rr), and Red (bb; rr).
Using the patch clamp technique, a researcher measures a single channel in a membrane. The receptor channel passes 3 pA (picoamperes) of ionic current over a period of 8 msec at -60 mV; 1 pA = 10 -12 ampere.Based on your knowledge of nerve transmission, this movement of ions willA. stimulate a change in a postsynaptic membrane if 50 channel events are combined.B. stimulate a change in a postsynaptic membrane if three channel events are combined.C. not stimulate a change in a postsynaptic membrane.D. generate a current resulting in membrane depolarization.E. stimulate a change in a postsynaptic membrane
Answer:
(C)
Explanation:
Not stimulate a change in a postsynaptic membrane because it is preset in inner membrane.
Which accessory eye structures function to produce the tears that cleanse and protect the eye?
a. lacrimal glandsb. conjunctivac. Meibomian glandsd. medial canthi
Answer:
a. lacrimal glands
Explanation:
Lacrimal glands are paired, almond-shaped exocrine glands, situated at each eye. Their main function is to secrete the aqueous layer of the tear film.This lacrimal gland produce the tears that flow into canals that connect to the lacrimal sac. and also cleanse and protect the eye. They can be located at the upper lateral region of each orbit, in the lacrimal fossa of the orbit formed by the frontal bone.
The major function of the conjuctiva is to Produce mucus to prevent the eyes from drying out.
Meibomian glands are holocrine type exocrine glands, along the rims of the eyelid inside the tarsal plate. They produce meibum, (an oily substance).This oily substance prevents evaporation of the eye's tear film.
Medial canthi is the corner of the eye where the upper and lower eyelids meet.
This basic explanation clearly depicts that the Lacrimal glands is the correct answer.
You are interested in two traits in rabbits, each of which is controlled by a separate gene with two alleles:
a) coat color (brown, B, is completely dominant to white,
b) and tail (tailed, T, is completely dominant to tail-less,
c) You cross a brown, tailed rabbit that is heterozygous at both loci with a white, tail-less rabbit and produce a large number of offspring.
Among the offspring you find only two phenotypes in equal proportions: brown, tailed and white, tail-less.
Answer:the scenario here with the brown tailed rabbit RrTt
The white tailess rabbit is rt.
Crossing both yields the below offsprings possibilities;
For colour: Rr or rr
For tail: Tt or tt
That means they would produce equal numbers of Brown tailed & white tailess rabbits.
Explanation:
The presence of circular dna in mitochondria and chloroplasts suggests that these organelles evolved from ___.
Answer:
Bacteria.
Explanation:
These are DNA that formed closes loops.
They have no ends.
A typical example is plasmids of bacteria., circular bacteria chromosomes,
In viruses example are ccc DNA (circular closed circular DNAf) formed by viruses.
Mitochondria and chloroplasts evolved from prokaryotes through endosymbiosis.
Explanation:The presence of circular DNA in mitochondria and chloroplasts suggests that these organelles evolved from prokaryotes. Circular DNA is a characteristic feature of prokaryotic cells, such as bacteria, and is not typically found in the nucleus of eukaryotic cells. This indicates that mitochondria and chloroplasts were likely once free-living prokaryotic organisms that were engulfed by a larger host cell through a process called endosymbiosis. Over time, these organelles became integrated within the host cell and established a mutually beneficial relationship.
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Which of the following statements about viruses is FALSE?a. Vaccines are available to prevent several diseases caused by viruses. b. If a virus mutates, the immune system may not recognize the virus. c.The widespread use of antibiotics has led to resistant strains of viruses. d. People can get colds and flus again because the viruses are constantly mutating.
Answer: option C - The widespread use of antibiotics has led to resistant strains of viruses.
Explanation:
Antibiotics are substances (usually drugs) that can destroy or inhibit the growth of BACTERIA and similar microorganisms.
Do note that viral infection DO NOT respond to antibiotic treatment.
So, it is FALSE to say that the widespread use of antibiotics has led to resistant strains of viruses
The statement C- 'The widespread use of antibiotics has led to resistant strains of viruses.' is false. Antibiotics are used for bacterial infections, not viral ones.
Explanation:The statement that is FALSE among those given about viruses is: c. The widespread use of antibiotics has led to resistant strains of viruses. Antibiotics are used to treat bacterial infections, not viral infections. While bacteria can indeed develop resistance to antibiotics, viruses do not. This is due to the different nature of bacteria and viruses. Vaccines do help prevent diseases caused by viruses (a), and it's also true that the immune system may not recognize a mutated virus (b), and that constant mutation is the reason why people can get colds and flus more than once (d).
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During DNA replication
A.each strand can act as a template for the replication of the moleculeB.the two strands of the original DNA molecule must be separatedC.enzymes facilitate all the steps involvedD.each of the original two strands of DNA will wind up in a different chromatid
Answer:
The answer is A: Each strand can act as a template for the replication of the molecule
Explanation:
During DNA replication, each of the two strands that make up the double helix serves as a template from which new strands are copied. The new strand will be complementary to the template or parent strand.
Small vesicles containing pigment inside of pigmented fish epidermal cells aggregate or disperse in response to treatment with certain chemicals. When nocodazole is added to cells in which the pigment granules have been induced to aggregate, the granules cannot disperse again. Small vesicles containing pigment inside of pigmented fish epidermal cells aggregate or disperse in response to treatment with certain chemicals. When nocodazole is added to cells in which the pigment granules have been induced to aggregate, the granules cannot disperse again. Pigment granules can either aggregate or disperse. Once aggregated the granules cannot disperse again, and vice versa. Intermediate filaments have a large effect on the pigment granule dispersal process and can stabilize resulting aggregates. Pigment granule dispersal is a microtubule-dependent process.
Answer:
Pigment granule dispersal is a microtubule-dependent process.
The question discusses how nocodazole, by disrupting microtubules, prevents the dispersion of pigment granules in fish epidermal cells, demonstrating the role of microtubules and intermediate filaments in pigment granule movement.
Explanation:The question pertains to the behavior of pigment granules in pigmented fish epidermal cells in response to chemical treatment. Nocodazole, a chemical inhibitor that disrupts microtubules, prevents the dispersion of pigment granules once they have aggregated. This indicates that pigment granule dispersal is a microtubule-dependent process. Furthermore, the question notes that intermediate filaments significantly impact the dispersal and stabilization of these pigment granules.
This process is akin to those observed in eukaryotic cells, where microtubules, actin filaments, and intermediate filaments form the cytoskeleton and play a critical role in various cellular functions, including intracellular transport and cell division. Observations like this are often visualized using fluorescent dyes, highlighting specific cellular structures under a fluorescence microscope.
_______________ occurs when coal is burned and sulfur oxides are released into the atmosphere, causing acid-forming particles to accumulate.
Answer:
Acid rain
Explanation:
So2 +H2O = HSO3+
Two large populations of horses are being systematically crossed (mares from one population bred to stallions of the other and vice versa). Coat color is not a factor in determining which animals are selected and which individual matings are made (random matings). Frequencies of coat color genes at the C locus for population 1 are.85 for Cand.15 for c. Frequencies for Care.6 and care.4 for population 2. Given these values, what are the gene and genotypic frequencies of the F1?
a. p=0.725, q = 0.275: P=0.51, H=0.43, Q=0.06
b. p=0.725.q = 0.275; P=0.06. H=0.56, Q=0.51
c. p=0.4.q = 0.6: P=0.12. H=0.56, Q=0.32
d. p=0.725.q = 0.275: P=0.34. H=0.57. Q=0.09
Answer:
a. p=0.725, q = 0.275: P=0.51, H=0.43, Q=0.06
Explanation:
Let state our given parameters from the question:
Frequencies of coat color genes at the C locus for population 1 are .85 for C
This implies that the Allelic frequency C for population p1 =0.85
Frequencies of coat color genes at the c locus for population 1 are .15 for c
This implies that the Allelic frequency c for population q1 = 0.15
Frequencies for Care .6 i.e p2= 0.6
Frequencies for care .4 i.e, let that be q2= 0.4
The table below shows a diagrammatic representation of the above expression:
Alllelic Frequency C c
Population 1 (p1) 0.85 (q1) 0.15
Population 2 (p2) 0.6 (q2) 0.4
Now, from above: let think of the table as a punnet square and then cross it together;
(p1) = 0.85 (q1) = 0.15
p2 = 0.6 p1p2 p2q1
= 0.6 × 0.85 = 0.15 × 0.6
= 0.51 (P) = 0.09 (H)
q2 = 0.4 p1q2 q1p2
= 0.85 × 0.4 = 0.4 × 0.15
=0.34 (H) = 0.06 (Q)
From the above table, the heterozygous are represented by (H)
∴ Frequency of heterozygous can be calculated as:
= 0.09 + 0.34
= 0.43
Thus, we can conclude that the progeny F1 genotypic frequencies are:
P= 0.51
H= 0.43
Q= 0.06
Now, let us calculate the allelic frequencies, p and q in F1
p = P + 1/2 × (H)
= 0.51 + (1/2 × 0.43)
= 0.51 + 0.215
= 0.725
q = Q + 1/2 × (H)
= 0.06 + (1/2 × 0.43)
= 0.06 × 0.215
= 0.275
Hence, p=0.725, q = 0.275: P=0.51, H=0.43, Q=0.06 , This makes option a the correct answer.
U937D cells express high levels of creatine kinase (CK‑B) mRNA but do not translate the mRNA into protein. Ribosomes bind the 5' end of the CK‑B mRNA; however, translation into protein is repressed in these cells. U937D cells synthesize the CK‑B enzyme when researchers introduce numerous short segments of RNA containing 3' UTR consensus sequences into the cells. The total amount of CK‑B mRNA does not change after adding RNA containing 3' UTR sequences. Introducing short RNA segments without the 3' UTR consensus sequences does not stimulate CK‑B synthesis. Which of the statements explains how the introduction of short RNA containing 3' UTR sequences allows CK‑B translation in U937D cells?.Explain how the introduction of short segments of RNA containing the 3'UTR sequences might remove the inhibition.
Answer:
Basically the translation of CK-B protein is inhibited in the U937D cells, despite the fact that the CK-B protein is bounded to the ribosomes. This is because(mechanisms) the translation is inhibited by the binding of translational repressors to the 3’UTR of the CK-B mRNA rather than the actual CK-B mRNA 3'UTR.
Furthermore, the soluble protein inhibitions is due to the reaction of the U937 cells to the short RNA sequences with the 3’UTR.
The introduction of these sequences(shot segement of RNA) into the U937D cells leads to CK-B synthesis. This makes 3’UTR sequences to bind to the translational repressor proteins, thus preventing them from binding to the CK-B mRNA .
COMPLETED QUE.
A common feature of many eukaryotic mRNAS is the presence of a rather long 3' UTR, which often contains consensus sequences. Creatine kinase B (CK-B) is an enzyme important in cellular metabolism. Certain cells—termed U937D cells—have lots of CK-B mRNA, but no CK- B enzyme is present. In these cells the 5’ end of the CK-B mRNA is bound to ribosomes, the mRNA is apparently not translated. Something inhibits the translation of the CK-B mRNA in these cells. Researchers introduced numerous short segments of RNA containing only 3’UTR sequences into U937D cells. As result, the U937D cells began to synthesize the CK-B enzyme, but the total amount of CK-B mRNA did not increase. The introduction of short segments of other RNA sequences did not stimulate the synthesis of CK-B; only the 3’UTR sequences turned on the translation of the enzyme. Based on these results, purpose a mechanism for how CK-B translation is inhibited in U937D cells. Explain how the introduction of short segments of RNA containing the 3'UTR sequences might remove the inhibition.
Explanation:
What organisms are responsible for decomposing organic matter? Why are they essential? A. heterotrophs; they return biotic material to the abiotic component of the Earth B. social insects; they carry out fertilization of many crops C. decomposers; they return biotic material to the abiotic component of the Earth
Answer: Option C - decomposers; they return biotic material to the abiotic component of the Earth
Explanation:
Decomposers are also known as SAPROPHYTES. They feed on and break down dead and decaying remains of plants and animals, thus enabling the release of certain compounds like ammonia, methane gas etc into the environment that then enrich/improve the soil fertility.
Final answer:
Correct option is C. Decomposers, mainly bacteria and fungi, are responsible for decomposing organic matter and are essential for recycling nutrients back into the ecosystem, supporting the stability of food webs.
Explanation:
Organisms responsible for decomposing organic matter are known as decomposers, which include bacteria and fungi. These decomposers are essential because they are involved in the process of breaking down dead materials and waste products, thereby recycling nutrients such as nitrogen and phosphorus back into the environment. Without decomposers, nutrients would remain locked within dead organic matter, preventing them from being accessible to living organisms that require them for growth.
Decomposers carry out the crucial task of returning biotic material to the abiotic component of the Earth, which ensures the stability of ecosystems. They break down organic substances, which are then termed as biodegradable, into inorganic nutrients that can be utilized by primary producers like plants. This cycle is fundamental to the maintenance of the food web and the overall health of the ecosystem.
If the normal nucleotide sequence was TACGGCATG, what type of gene mutation is present if the resulting sequence becomes TAGGCATG?A. additionB. deletionC. substitutionD. inversion
Answer:
B. Deletion
Explanation:
the 3rd letter (C) is missing (deleted)
In the normal nucleotide sequence C is missing. So, the type of mutation is deletion. Thus, option B is correct.
What is mutation?The term mutation is defined as a change in the normal sequence of DNA at a particular gene locus and mutations are seems to be harmful as it results into serious life taking disorders such as cancer.
Mutation also affect the function and division of cell that causes cancer and other serious disorders. Mutation in a single gene causes the body to produce thick and sticky white substance known as mucus that blocks lungs and digestive organs.
There are mainly three types of DNA mutations and these are deletions, insertions, and base substitutions.
By nature mutation is recessive but in some condition it can be dominant and these are harmful for the individual. It is random and recurrent event.
Therefore, the normal nucleotide sequence C is missing. So, the type of mutation is deletion. Thus, option B is correct.
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If Chargaff's equivalence rule is valid, then hypothetically we could extrapolate this to the combined genomes of all species on Earth (as if there were one huge Earth genome).
In other words, the total amount of A in every genome on Earth should equal the total amount of T in every genome on Earth.
Likewise, the total amount of G in every genome on Earth should equal the total amount of C in every genome on Earth.Calculate the average percentage for each base in your completed table.
Do Chargaff's equivalence rules still hold true when you consider those six species together?
Answer:
yes . A approximately equals to T and G approximately equals C in average.
Explanation:
according to Chargaff's equivalence rule
the number and concentration of adenine is similar to the number of thymine and number of guanine is similar to that of cytosine in the DNA.
so according to conditions given in the question i.e average amount of A should be equal to average amount of T and total average amount of G in every genome on Earth should equal the total average amount of C . so Chargaff's equivalence rules still hold true when you consider those six species together.
An outbreak of salmonellosis occurred after an epidemiology department luncheon, which was attended by 485 faculty and staff. Assume everyone ate the same food items. Sixty-five people had fever and diarrhea, five of these people were severely affected. Subsequent laboratory tests on everyone who attended the luncheon revealed an additional 72 cases.
The ratio of severe cases to other clinically apparent cases was:
A. 65/485
B. 5/60
C. 72/485
D. 65/72
E. 5/65
Answer:
72/485
Explanation:
this question was just on my bio test i took
Which is the class of cnidarian that is characterized by having a dominant medusa stage and four gastric pouches?
Answer:scyphozoa
Explanation:cniderians are aquatic animals.they have radial symmetry, that is they can be divided into two halves through any plane.
Cniderians may have nematocyst,which are stinging cells.
They have two forms,which are polyp and Medusa .
There are four classes of cniderians;hydrozoa, scyphozoa,cubozoa and anthozoa.
Scyphozoa are predominantly medusa,which is umbrella shaped.they are jelly fishes.
They have four or eight oral arms.
They possess four gonads and four gastric pouches .an example is the Aurelia.
Why do you believe you need, or should receive, financial assistance, such as a Houston Livestock Show and Rodeo Scholarship to attend the college/university of your choice? This should include details about family situations such as:
a. care of family members,
b. medical situations,
c. family financial burdens and/or how important it will be for you to work while in college.
Houston Livestock Show and Rodeo Scholarship to attend the college/university of your choice because it is and will be.
What is Houston Livestock?One of the most lucrative professional rodeo events for the regular season is included. Since 2003, it has been held at NRG Stadium in Houston, Texas, with the exception of 2021 because of the COVID-19 pandemic's effects.
The Astrodome served as the previous venue. Similar to Mardi Gras in New Orleans, the Texas State Fair in Dallas, Comic-Con in San Diego, and New Year's Eve in Times Square in New York City, it is regarded as the city's "signature event."
A record-breaking 2,611,176 attendees and 33,000 volunteers showed up in 2017. The rodeo was dubbed "the year of the volunteer" in 2007.
Therefore, Houston Livestock Show and Rodeo Scholarship to attend the college/university of your choice because it is and will be.
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Ringing in the ears is called:
a. otitis externa.
b. Meniere's syndrome.
c. tinnitus.
d. otosclerosis.
Answer:
c. tinnitus
Explanation:
Tinnitus (a hearing impairment) is a form of hearing of sound when no external sound is present Which is often described as a Ringing,Tinnitus may also sound like a clicking or roaring. i.e usually unclear voices or music are heard. The sound may be soft or loud, low or high pitched, and appear to be coming from one or both ears. Ranging from one person to another, the sound may causes depression or anxiety and can interfere with concentration.
Otitis externa is a condition that causes inflammation , the inflammation also covers (redness and swelling) of the external ear canal, i.e the tube between the outer ear and eardrum.
Meniere's disease popularly called (MD), is a disorder of the inner ear that is characterized by episodes of feeling like the world is spinning (vertigo), hearing loss, and a fullness in the ear. The cause of MD involves both genetic and environmental factors. Some of the factors include constrictions in blood vessels, viral infections, and autoimmune reactions.
Otosclerosis is a condition where one or more foci of irregularly laid spongy bone replace part of normally dense enchondral layer of bony otic capsule in the bony labyrinth. This condition affects one of the ossicles (the stapes) resulting in hearing loss, vertigo or a combination of symptoms.
Therefore from the foregoing we can conclude that Tinnitus is the correct anwser.
Answer: option C
Explanation:
Ear ringing and strange ear noises is medically called tinnitus. People who have tinnitus hear buzzing and roaring sounds which are actually absent.
Tinnitus can be caused by problems in any of the four areas which are responsible for hearing: the outer ear, the middle ear, the inner ear, or in the brain.
Integrated pest management requires Select one:
a. a complete genetic make-up of pests.
b. a complete knowledge of the pest's life history.
c. extensive application of pesticides.
d. All of these are correct.
Answer:
Option B
Explanation:
Integrated pest management is a method to control the pest through some common sense practices.
In this method, in depth study of pest’s life cycle along with interaction of pest with the environment must be done. This knowledge is then combined with the available knowledge of pest control methodologies in order to deal with pests in a most economically and environment friendly way.
Hence, option B is correct