Answer: D
Explanation:osmosis is the movement of molecules from a region of lower concentration to a region of higher concentration through a semi permeable membrane.
Final answer:
Water will move from inside the cell to the outside by osmosis because the concentration of solute X is higher outside the cell, making the outside a region of lower water concentration.
Explanation:
If solute X is more concentrated outside the cell than inside, and the cell's membrane is permeable to water but impermeable to solute X, water will move by osmosis from inside the cell to outside. This is because osmosis is the diffusion of water across a membrane from an area of higher water concentration (or lower solute concentration) to an area of lower water concentration (or higher solute concentration). Therefore, the correct answer is D. Water will leave the cell.
Unequal crossing over results in A. an exchange between nonhomologous chromosomes. B. a loss of genetic material. C. a repair of UV-induced damage. D. a production of eggs containing Y chromosomes. E. a creation of deletions and duplications.
Answer: OPTION E
Explanation:unequal crossover usually leads to duplication or deletion of chromosome. In this case,. A DNa strand is deleted and replace usually by another DNA strand which is mostly a duplicate from a sister chromatid and this process leads to Gene families been produced beause one is deleted and again and again duplicate is produced on the same place (2 product formation). It is a form of chromosomal crossing over that exists between homologous sequence which were initially not paired together. In Gene duplication and mutation in organism, unequal crossover is said to be the pioneer or chief cause of it with Gene conversion beside it.
One of your lab partners has followed the recommended procedure of running Gram-positive and Gram-negative control organisms on her Gram stain of an unknown species. Her choices of controls were Escherichia coli and Bacillus subtilis. She tries several times and each time concludes she is decolorizing too long because both controls have pink cells (one more than the other). What might you suggest she try and why?
Answer:
Reduced in holding time of decolrization step and also used less Alcohol because decolrization step is important in Gram's staining.The decolorization step must be performed carefully. Otherwise over-decolorization may occur. This step is critical and must be timed correctly otherwise the CV stain will be removed from the Gram-positive cells. If the decolorizing agent is applied on the cell for too long time, the Gram-positive organisms to appear Gram-negative..
Explanation:
Gram' staining is a technique used in microbiology labs to differentiate between Gram's positive and negative
Gram-positive bacteria :Stain dark purple due to retaining the primary dye called CV in the cell wall.
:Gram-negative bacteria Stain red or pink due to retaining the counter staining dye called Safranin or neutral red.
There are four basic step in Gram" staining
1) Application of the Primary Stain to a Heat-Fixed Smear of Bacterial Culture
2)Addition of Gram's Iodine
3)Decolorization with 95% Ethyl Alcohol:Alcohol or acetone dissolves the lipid outer membrane of Gram-negative bacteria, thus leaving the peptidoglycan layer exposed and increases the porosity of the cell wall. The CV-I complex is then washed away from the thin peptidoglycan layer, leaving Gram-negative bacteria colorless.
On the other hand, alcohol has a dehydrating effect on the cell walls of Gram-positive bacteria that causes the pores of the cell wall to shrink. The CV-I complex gets tightly bound into the multi-layered, highly cross-linked Gram-positive cell wall thus staining the cells purple.
The decolorization step must be performed carefully. Otherwise over-decolorization may occur. This step is critical and must be timed correctly otherwise the CV stain will be removed from the Gram-positive cells. If the decolorizing agent is applied on the cell for too long time, the Gram-positive organisms to appear Gram-negative. Under-decolorization occurs when the alcohol is not left on long enough to wash out the CV-I complex from the Gram-negative cells, resulting in Gram-negative bacteria to appear Gram-positive.
Which three statements may correctly explain why the population size increases after time point C?a. Bacteria that acquired a mutation that conferred drug- resistance had a growth advantage over non-resistant bacteria.b. The population increase just after time point C indicates that antibiotic use was discontinued.c. Between time points C and D, drug-resistant bacteria were reproducing faster than non-drug resistant bacteria were dying.d. The few drug-resistant bacteria in the population reproduced, quickly leading to a large drug- resistant population.e. Because the population grew more rapidly after time point C, the bacteria must have acquired a second drug- resistance mutation.
Answer:
A,C, D
Explanation:
Bacteria that acquired a mutation that conferred drug- resistance had a growth advantage over non-resistant bacteria.
Between time points C and D, drug-resistant bacteria were reproducing faster than non-drug resistant bacteria were dying.
The few drug-resistant bacteria in the population reproduced, quickly leading to a large drug- resistant population.
There is no indication that a second antibiotic was used, so it is not possible to assume a second drug-resistance mutation occurred. There is also no indication that antibiotics were discontinued.
Resistant bacteria are the one which fight against the drug or immune system. It is the capacity of bacteria to withstand the effect of antibiotic which are intended to kill them.
The correct statements are :
a. Bacteria that acquired a mutation that conferred drug- resistance had a growth advantage over non-resistant bacteria.
c. Between time points C and D, drug-resistant bacteria were reproducing faster than non-drug resistant bacteria were dying.
d. The few drug-resistant bacteria in the population reproduced, quickly leading to a large drug- resistant population
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HIV does not have any enzymes to ensure that the replication of its genome is error free. Use this information to propose an explanation as to why it has been difficult for scientists to create a vaccine to HIV?
Answer:
The human immunodeficiency viruses (HIV) belong to the category of retrovirus and leads to fatal condition to the affected individual due to its action on the immune system.
Explanation:
This virus is known to contain its own replication machinery that lacks the property of proof reading. But despite vaccine against its is not yet formulated. This is due to the ability of higher mutation rate of the associated virus. This genetic variability allows the spread of the virus all throughout the body but with specific or unique genetic composition.
As the genetic composition of the virus is different, it adds to the difficulty level in creation of the antiviral drug as the target for the drug will not be specified.
The production of a continuous new strand of DNA using the many separate Okazaki fragments (in other words, the joining of the already made fragments) found on the lagging strand requires all of the following except which one?
A. nuclease
B. ATP
C. repair polymerase
D. DNA primase
E. DNA ligase
Answer:B
Explanation:
because if we strand dna we wouldnt have genes
Production of continuous new strand of DNA using Okazaki fragments found on lagging strand requires following components: ATP, repair polymerase, DNA primase, and DNA ligase,thus correct options are all except A.
1. ATP: ATP provides the energy necessary for the synthesis of DNA. It is required during the formation of phosphodiester bonds between nucleotides, which link them together to form the new DNA strand.
2. Repair polymerase: Repair polymerase, also known as DNA polymerase I, is responsible for replacing the RNA primers used in DNA replication with DNA nucleotides. It removes the RNA primers and fills in the gaps with complementary DNA nucleotides.
3. DNA primase: DNA primase synthesizes short RNA primers that provide a starting point for DNA synthesis. These primers are required for DNA polymerase to initiate DNA synthesis.
4. DNA ligase: DNA ligase is an enzyme that seals the gaps between the Okazaki fragments on the lagging strand. It catalyzes the formation of phosphodiester bonds between adjacent nucleotides, joining the fragments together to form a continuous DNA strand.
Based on this information, the correct answer is A. nuclease. Nuclease is not required for the production of a continuous new strand of DNA using the Okazaki fragments. Nucleases are enzymes that break down DNA or RNA molecules by hydrolyzing the phosphodiester bonds between nucleotides. In the context of DNA replication, nuclease activity is not involved in joining the Okazaki fragments.
Thus, production of continuous new strand of DNA using Okazaki fragments found on lagging strand requires following components: ATP, repair polymerase, DNA primase, and DNA ligase,thus correct options are all except A.
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MHC class II molecules expressed on the surface of thymic cortical epithelial cells normally have a wide repertoire of different peptides bound to them. By engineering a construct that fuses the MHC class II protein to a single peptide sequence, and expressing this construct in thymic cortical epithelial cells that have their endogenous MHC class II genes knocked out, it is possible to generate a mouse line where all MHC class II proteins expressed on all thymic cortical epithelial cells are bound to the same peptide. These mice are often referred to as ‘single-peptide’ mice. Examination of the T cell developing in these single peptide mice would likely show:
A. A significant reduction in the numbers of mature CD4 T cells
B. No change in the numbers of mature CD4 T cells
C. A block in T cell development at the CD4+CD8+double-positive stage
D. A repertoire of T-cell receptors on mature CD4 T cells restricted to a single Vbeta
E. A block in T cell development at the CD4-CD8-double-negative stage
Answer:
A. A significant reduction in the numbers of mature CD4 T cells
Explanation:
In the given problem, there is an engineering of the MHC class II protein with the sequence of a single peptide. In addition, it was expressed in the epithelial cells of the thymic cortical. Based on the result obtained from the engineering construction, it is obvious that there would be a large decreases in the CD4 T cells numbers.
ou are studying an animal and inject fluorescein, a fluorescent dye, into a single cell on the surface epithelium of the animal. After a brief period of time, the dye spreads to cells neighboring the injected cell. What do you concludea. The cells are connected by gap junctions. b. The cells are connected by zonulae adherens. c. The cells are connected by tight junctions. d. The cells are connected by plasmodesmata.
Answer:
a. The cells are connected by gap junctions.
Explanation:
Gap junctions are the points of connection between the adjacent cells of animal tissues. These structures are formed by proteins which in turn form narrow channels in the plasma membranes of the neighboring cells. These channels allow the small molecules to pass from the cytoplasm of the one cell to that of another. The spread of dye to the neighboring animal cells would have occurred by gap junctions present between these cells.
Scientists theorize that all living things are fundamentally alike at the cellular and molecular level. These fundamental similarities are the basis of evolutionary theory: all life shares a common ancestor. If we compare the two cell types, prokaryotic and eukaryotic, there is evidence of these similarities and that evidence includes all EXCEPT:
A) the presence of DNA containing the same four nitrogen bases.
B) the presence of ribosomes capable of preforming protein synthesis.
C) the compartmentalization of the cell due to the presence of organelles.
D) that all life is composed of one or more cells, although the organization of cells does vary.
Prokaryotic and eukaryotic cells share fundamental similarities at the cellular and molecular level. The evidence of these similarities is seen in the presence of DNA, ribosomes, and the fact that all life is composed of cells. The exception is the compartmentalization of cells due to organelles, which is a feature exclusive to eukaryotic cells.
Explanation:The question asks for evidence of fundamental similarities between prokaryotic and eukaryotic cells, with the exception of one option. The options are:
The presence of DNA containing the same four nitrogen basesThe presence of ribosomes capable of performing protein synthesisThe compartmentalization of the cell due to the presence of organellesThat all life is composed of one or more cells, although the organization of cells does varyThe correct answer is C) the compartmentalization of the cell due to the presence of organelles because this is a characteristic unique to eukaryotic cells and not found in prokaryotic cells.
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Two gene loci, A and B, assort independently, and alleles A and B are dominant over alleles a and b. Indicate the probability of producing an AB gamete from an AaBb individual.
Explanation:
Step 1. The two quality loci, A and B, assort autonomously. The alleles A and B are predominant over the alleles an and b. In this way, when a cross happens between AaBb X AaBb, the subsequent gametes would be AB, Ab, aB, and ab.Step 2.The offsprings which have in any event one A and B allele, will show AB phenotype. Along these lines, AABB, AaBb, AABb, AaBB, will all have AB phenotype.If two gene loci, A and B, assort independently, the probability of producing an AB gamete from AaBb would be 1/4 or 25%
Since both A and B are independently inherited, Aa and Bb can be crossed just we would have it in a monohybrid cross following Mendelian pattern.
Thus:
Aa x Bb
AB Ab aB ab
The gametes and their respective probabilities of appearing would be:
AB - 1/4Ab - 1/4aB - 1/4ab - 1/4In other words, all 4 gametes have an equal chance of being produced during gametogenesis.
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An enzyme that follows Michaelis-Menten kinetics has a KM value of 20.0 μM and a kcat value of 231 s−1. At an initial enzyme concentration of 0.0100 μM, the initial reaction velocity was found to be 1.07×10−6 μM/s. What was the initial concentration of the substrate, [S], used in the reaction ?
The initial substrate concentration was 0.52 μM.
We can apply the Michaelis-Menten equation to solve for the initial substrate concentration ([S]):
v = (Vmax * [S]) / (KM + [S])
where:
v is the initial reaction velocity (1.07 × 10^-6 μM/s)
Vmax is the maximum reaction velocity (determined by kcat and enzyme concentration)
KM is the Michaelis-Menten constant (20.0 μM)
[S] is the initial substrate concentration (unknown)
Step 1: Calculate Vmax based on kcat and enzyme concentration:
Vmax = kcat * [E] = 231 s^-1 * 0.0100 μM = 2.31 μM/s
Step 2: Rearrange the Michaelis-Menten equation to solve for [S]:
[S] = (v * KM) / (Vmax - v)
Step 3: Substitute known values and solve for [S]:
[S] = (1.07 × 10^-6 μM/s * 20.0 μM) / (2.31 μM/s - 1.07 × 10^-6 μM/s) ≈ 0.52 μM
Therefore, the initial concentration of the substrate ([S]) used in the reaction was approximately 0.52 μM.
Please help with number 8,9and 10
If the killer whale obtained 26000 joules of energy by eating crab-eater seal, how much energy was available at each of the following trophic levels
Answer:
it is not very specific im sorry if i could see the paragraph i could help
Explanation:
"Suppose two independently assorting genes are involved in the pathway that determines fruit color in squash. These genes interact with each other to produce the squash colors seen in the grocery store. At the first locus, the W allele codes for a dominant white phenotype, whereas the w allele codes for a colored squash. At the second locus, the allele Y codes for a dominant yellow phenotype, and the allele y codes for a recessive green phenotype. The phenotypes from the first locus will always mask the phenotype produced by the second locus if the dominant allele (W) is present at the first locus. This masking pattern is known as dominant epistasis. A dihybrid squash, Ww Yy, is selfed and produces 320 offspring. How many offspring are expected to have the white, yellow, and green phenotypes
Answer:
White:120 Yellow:30 Green: 10Explanation:
Given;
W codes for a dominant white phenotype;
w codes for a colored squash;
the Y codes for a dominant yellow phenotype;
the y codes for a recessive green phenotype.
The phenotypes from the first locus mask the phenotype produced by the second locus.
A dihybrid squash, of Ww Yy, is self crossed
WwYy * WwYy
F1 are:
WWYY,WWYy, WWyy, WwYy, Wwyy,wwYy, Wwyy and wwyy etc
Due to dominant epistasis, the progeny are:
White:120
Yellow:30
Green: 10
These many offspring are expected to have the white, yellow, and green phenotypes
White: 160, Yellow: 160, Green: 0
The given scenario involves the interaction of two independently assorting genes in determining fruit color in squash. At the first locus, the W allele confers a dominant white phenotype, while the w allele codes for a colored squash. At the second locus, the Y allele results in a dominant yellow phenotype, and the y allele produces a recessive green phenotype. The key aspect is the dominance relationship between the alleles: W masks the expression of w, and Y masks the expression of y.
In the dihybrid squash Ww Yy, the alleles segregate independently, and the possible combinations are WWYY, WWYy, WwYY, WwYy, and so on. However, due to the dominant epistasis described, the presence of at least one dominant allele (W) at the first locus masks the effect of the alleles at the second locus. Therefore, individuals with the genotype Ww Yy and WW Yy will both exhibit the white phenotype, while those with ww Yy and ww yy will display the yellow and green phenotypes, respectively.
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If your sample does not grow on the plate in the anaerobic chamber but does grow in the presence of oxygen, it produces bubbles with the addition of hydrogen peroxide, and the dextrose tube remains orange, you can conclude:
Answer:
it is an obligate (or strict) aerobe
Explanation:
By whether an organism requires oxygen or not for respiration. We can classify it as either aerobic or anaerobic.
Aerobic organisms require oxygen and anaerobic do not.
For aerobes, it can be facultative or obligate. Facaltative aerobes require oxygen but they can however switch to fermentation when oxygen is not available.
Obligate aerobes are aerobes require oxygen for cellular metabolism.
In the test the dextrose tube remained yellow because fermentation had not taken place.
Therefore we can conclude the sample contained an obligate aerobe which was catalase positive since it produces bubbles when Hydrogen peroxide was added.
In some circumstances, when two different carbon sources are available, growth will occur first using one carbon source, then after a short lag period, growth will resume using the second carbon growth source.
a. This process is called _____ growth.
Answer: Diauxic growth
Explanation:
The diauxic growth or diphasic growth is a bacterial growth which is characterized by the growth in two phases depending upon the source of carbon used.
The preferred carbon source is consumed first, this leads to enhance the growth in the bacteria, followed by the lag phase. During the process of lag phase the bacteria start to metabolize the second carbon source and the growth resumes.
For example, a colony of E.coli bacteria was cultured in a medium containing the glucose, and lactose sugars. At the initial level the bacteria was capable of using the glucose sugar and growth became rapid this is followed by a lag phase. In the lag phase the bacteria started to utilize the lactose sugar and again the growth resumes.
Final answer:
Diauxic growth is when a cell uses one carbon source and then switches to another after the first is depleted. In the given E. coli growth curve example, glucose is used first, leading to rapid growth, followed by a lag phase and then slower growth on xylose once glucose is exhausted.
Explanation:
The process described in the student's question, where a cell uses one carbon source for growth and then switches to another after the first is depleted, is known as diauxic growth. Tackling Monod's research, we can clarify what is occurring at different points (A-D) in the growth curve. At point A, the E. coli is primarily using glucose as its carbon source for growth, causing a rapid increase in population. The xylose-use operon is not being expressed at this point because the presence of glucose typically represses the expression of enzymes for the metabolism of other sugars.
After glucose is exhausted, demonstrated by a leveling off of the growth curve at point B, the cells enter a short lag phase as they begin to express the necessary enzymes for the uptake and metabolism of xylose. At point C, we observe the resumption of growth as the bacteria start to use xylose. As these enzymes are less efficient or the substrates are utilized less favorably, the second phase of growth is slower compared to the first phase. Finally, at point D, the growth rate declines again as the bacterial population exhausts the xylose.
What is the genus of plant that is easily identified by having no true leaves, dichotomous branching as well as lobed sporangia (usually yellow in color). a. Marchantia b. Cardiospermum c. Psilotum Ginkgo d. Equisetum e. None of these
Answer:
Option C.
Psilotum.
Explanation:
Psilotum is a genus of whisk ferns. They are vascular plants. They lack true stem and true leaves, their stems is the organ that contain the conducting tissues. They have stems with many branches and synangium with three lobes. The synangia is is fused sporangia that produce spores. They have rhizome which form rhizoids and help to anchor the psilophyte sporophyte.
Answer: The Genus is Psilotum.
Explanation: Psilotum is the genus of fern-like vascular plants, commonly known as whisk ferns. It is one of two genera in the family Psilotaceae. They are easily identified by having no true leaves, dichotomous branching as well as lobed sporangia.
Image of a Psilotum attached below.
Wild-type bacteria can grow on minimal medium. Four mutants that cannot grow on minimal medium but can grow on minimal medium supplemented with the nutrient "H" are isolated. It is suspected that metabolites T, P, and A are in the biochemical pathway for synthesis of H, so each mutant is tested for the ability to grow on minimal medium supplemented with these metabolites:
A. Mutant 1: can grow on minimal medium supplemented with T, but not P or A
B. Mutant 2: is unable to grow on minimal medium supplemented with T, P, or A
C. Mutant 3: is able to grow on minimal medium supplemented with A or T, but not P
D. Mutant 4: can grow on minimal medium supplemented with T, P, or A.
Answer:
Mutant 4: can grow on minimal medium supplemented with T, P, or A.Explanation:
In minimal medium, those bacteria who lacks some enzyme which are necessary for the synthesis of all their required compounds can not grow. They require these metabolites supplemented in minimal medium. In this case , if any of these 3 metabolites T,P,A which are required for the synthesis of H are absent in minimal medium,then there will be no bacteria growth.
The use of selective media to test growth of mutants with specific nutritional requirements helps to understand the biosynthetic pathways and elucidate the genetic defects in metabolic genes. This resembles Beadle and Tatum's experiments and illuminates the concept of auxotrophs and the one gene-one enzyme hypothesis.
Explanation:The question revolves around the principles of biochemical genetics, where wild-type bacteria that can grow on minimal medium are contrasted with four mutant strains that require a nutrient (H) supplementation for growth. The growth behavior of these mutants on media supplemented with metabolites T, P, and A allows us to deduce the order of steps in the metabolic pathway for H synthesis. Mutants that grow on media with a specific supplement likely have a functional gene for the metabolic step before the supplemented compound but have a mutation in the gene responsible for utilizing the compound to make the subsequent metabolite in the pathway.
For instance, based on the given data:
Mutant 1: Can grow on minimal medium with T suggests it has a downstream block after T.Mutant 2: Unable to grow on any supplemented medium implies a mutation early in the pathway before T.Mutant 3: Can grow on A or T, but not P, indicating a block just before P in the pathway.Mutant 4: Able to grow on all supplements, so the block is after A in the pathway.This assessment of mutant growth on selective media is reminiscent of Beadle and Tatum's experiments, which contributed to the one gene-one enzyme hypothesis. Such experiments are essential for understanding metabolic pathways and identifying specific genetic defects in auxotrophs, organisms that have lost the ability to synthesize certain compounds due to mutations.
Indicate true (T) and false (F) statements below regarding cytokinesis in animal cells. Choose the correct answer represented by a four-letter string composed of letters T and F only, e.g. TFFF.( ) The force for cytokinesis is generated by kinesin motors on microtubule bundles that form the contractile ring.( ) As the contractile ring constricts, its thickness increases to keep a constant volume.( ) The midbody forms from bundles of actin and myosin II.( ) Local activation of Ran GTPase triggers the assembly and contraction of the contractile ring.A FTFTB FFTFC FTFFD FFFTE FFFF
Explanation of true and false statements regarding cytokinesis in animal cells. A FTFT is the correct option.
FTFT
The correct answer sequence is FTFT:
True: The force for cytokinesis in animal cells is generated by kinesin motors on microtubule bundles that form the contractile ring.
False: As the contractile ring constricts, its thickness does not increase to keep a constant volume.
True: The midbody forms from bundles of actin and myosin II.
False: Local activation of Ran GTPase does not trigger the assembly and contraction of the contractile ring.
The answer is option D: FFFF, as explained in the detailed response regarding the statements related to cytokinesis in animal cells.
The correct answer is option D: FFFF.
Let's break down each statement:
(F) The force for cytokinesis is generated by kinesin motors on microtubule bundles that form the contractile ring. This statement is false because the force is mainly generated by myosin II motors on actin filaments.(F) As the contractile ring constricts, its thickness increases to keep a constant volume. This statement is false; the thickness of the ring decreases as it contracts.(F) The midbody forms from bundles of actin and myosin II. This statement is false; the midbody contains microtubules and other proteins but not actin and myosin II.(F) Local activation of Ran GTPase triggers the assembly and contraction of the contractile ring. This statement is false; Rho GTPase, not Ran GTPase, plays a crucial role in activating the contractile ring.Complete question is-
Indicate true (T) and false (F) statements below regarding cytokinesis in animal cells. Choose the correct answer represented by a four-letter string composed of letters T and F only, e.g. TFFF.
( ) The force for cytokinesis is generated by kinesin motors on microtubule bundles that form the contractile ring.
( ) As the contractile ring constricts, its thickness increases to keep a constant volume.
( ) The midbody forms from bundles of actin and myosin II.
( ) Local activation of Ran GTPase triggers the assembly and contraction of the contractile ring.
A. FTFT
B. FFTF
C. FTFF
D. FFFF
3. The peppered moth provides an example of , as the predominant color of moths
changed over time as pollution dictated which color was best camouflaged for protection from
predators.
Answer:
Natural selection
Explanation:
When there was no pollution, the light color of tree bark provided excellent camouflage for the peppered moth, which was predominantly light in color. Some black peppered moths did exist, although they were rare.
During the industrial revolution when there was lots of soot around, the trees became a darker color. This means that now they were a better camouflage for the black moths. As a result, the black moths had a survival advantage, and were much more likely to survive than the lighter colored moths.
As a result, they were more likely to reach reproductive age and pass their genes on to the next generation, leading to an increase in the frequency of black moths. This is an example of natural selection, and illustrates how a species can adapt to its environment,
gHow would an inhibitor of cAMP phosphodiesterase affect glucose mobilization in muscle? It would reduce cAMP levels and inhibit glucose mobilization. It would maintain high cAMP levels and elevate glucose mobilization. It would increase AMP concentration, thereby increasing glucose mobilization. It would increase cAMP levels, which would inhibit glucose mobilization.
Explanation:
It would maintain high cAMP level and elevate glucose mobilization
Phosphodiesterase is an effector enzyme which degrades secondary messenger cAMP(cyclic adenosine monophosphate)Here in this case an inhibitor is inhibiting the phosphodiesterase therefore cAMP level will increaseAs cAMP level rise it activates a protein called protein kinase A which phosphorylates phosphorylase kinase and activates it Phosphorylase kinase becomes active that phosphorylates glycogen phosphorylase and makes it active,glycogen phosphorylase catalyse breakdown of glycogen(in liver and muscle cells)In liver cells breakdown of glycogen occurs and glucose 1 phosphate gets converted into glucose and supplied to whole body through blood
What would happen if someone stabbed your leg with a syringe full of calcium and injected the calcium directly into your muscle?
a.The actin active sites would stay covered by tropomyosin.
b.Cross-bridges would form in the absence of an action potential from a motor neuron.
c.Tropomyosin would bind the calcium and change the conformation of troponin.
d.Myosin would be unable to hydrolyze ATP.
Answer:
Cross-bridges would form in the absence of an action potential from a motor neuron.
Explanation:
The injected calcium ions would bind to troponin. Troponin would make tropomyosin move away from the myosin-binding sites on actin. The presence of free binding sites on the actin would be followed by the contraction cycle. This would include hydrolysis of ATP to energize myosin heads and binding of these heads to actin to form cross-bridges. Therefore, cross-bridge formation would occur without any action potential if calcium ions are injected directly into the muscle.
Many bacteria and fungi that are parasitic of plants face the daunting task of finding and infecting a new host by airborne spore dispersal followed by germinating upon and then penetrating the leaf surface of their host. What are some of the specific adaptations possessed by some of these parasites to gain access to leaf tissue by entering through stomata thereby evading the plant leaf cuticle? (Hint: search on rust fungi, Uromyces.)
Answer:
Explanation:
The parasites are those organisms which invade the host cell machinery to obtain nutrition, shelter and to reproduce their own progeny.
The fungi are the saprophytic organisms which obtain their food from the dead and decaying organic matter. But the fungi can also be parasitic in nature. The rust fungi produce spores which get attached to the plant surface. These spores germinate over the plant surface. They produce their germ tubes in the stomata of the leaves.
Some of the fungal strains damage the cuticle, and enter into the cells of the host plant. Some of the fungal strain also produce special structures called as haustoria which helps in deriving the nutrients from the plants.
Final answer:
Parasitic fungi and bacteria possess adaptations such as specialized structures called haustoria and appressoria to penetrate plant tissues and evade plant defenses. They infect through natural openings, wounds, and directly through plant cuticles, utilizing the plant's nutrients for their own survival.
Explanation:
Specific Adaptations of Parasitic Fungi and Bacteria
Bacteria and fungi have evolved various adaptations to parasitize and infect plant hosts. Notably, parasitic fungi like rust fungi produce specialized structures known as haustoria which penetrate the cell walls of their host plants. These haustoria make it possible for the fungi to access the cytosol materials such as sugars and amino acids, essential for their survival. The Uromyces species is a prime example of rust fungi that utilizes such mechanisms. Furthermore, these organisms can enter plant tissue through natural openings, wounds, or even directly through the cuticle. In addition to haustoria, pathogens and herbivores may have specialized cells called appressoria which help the fungi to attach and breach the epidermal cells of the plant. Additionally, certain bacteria enter through natural plant openings like stomata or wounds and multiply rapidly within the plant tissues causing cell death. Parasitic adaptation is an evolutionary process, which allows these organisms to resist the host's defense mechanisms and thrive within their living tissues.
Match the # in the diagram with the correct structure/term.
Column A Column B
1. ____ 1 A. codon
2. ____ 2 B. amino acid
3._____ 3 C. tRNA (anit-codon)
4._____ 4 D. Polypeptide chain (protein)
Answer:
1 A. amino acid
2 B. Polypeptide chain (protein)
3 C. tRNA (anit-codon)
4 D. codon
Explanation:
The whole diagram explains protein synthesis from transcription to translation.
- The codon, GGU is a codon with a triplet nature coding for glycine, and it consists of three nucleotides
- The amino acid, phenylanine is encoded from the codon, UUU found on the mRNA molecule
- transfer RNA (tRNA) helps to combines covalently with a specific amino acid, threonine and transfers the amino acid to the ribosomes to join the polypeptide chain
- polypeptide chain is a pentapeptide (with five amino acids) and is formed in the final stage of protein synthesis.
Which of the following pairs of microbe classification terms and optimal growth temperatures is mismatched? Which of the following pairs of microbe classification terms and optimal growth temperatures is mismatched? hyperthermophiles growth at 95°C psychrotroph growth at 22°C psychrophile growth at 37°C mesophile growth at 37°C
Psychrophile growth at 37°C is a mismatch.
Explanation:
Psychrophiles are also known as cryophyles.These organisms are extremophilic because they inhabit extremely cold places whose temperatures are at about 10 degree Celsius to -20 degree Celsius.These places include polar belts, high snow covered mountains, deep sea beds and other regions with permafrost etc.Examples include many bacterial genus like Polaromonas, Psychrobacter, Arthrobacter etc.The mismatched pair of microbe classification terms and optimal growth temperatures is psychrophile and growth at 37°C.
Explanation:The pair of microbe classification terms and optimal growth temperatures that is mismatched is psychrophile and growth at 37°C. A psychrophile is a type of microorganism that grows best at very low temperatures, typically around 0-20°C. On the other hand, mesophiles are microbes that grow best at moderate temperatures, between roughly 20-45°C. The other pairs in the question are correctly matched: hyperthermophiles grow at extreme high temperatures (around 80-113°C), psychrotrophs grow at cold temperatures but have an optimum growth temperature around 20-30°C, and mesophiles grow at moderate temperatures around 20-45°C.
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Bacteria are grouped into two catgories reflecting structural features of their cells. Bacteria are classified as Gram-positive or Gram-negative based on whether or not they retain the crystal violet dye used in the Gram stain procedure. Cell wall structure determines the ability to retain the dye, thus cell wall structure is the basis of categorization into Gram-postive (G+) and Gram-negative (G-).
Which of the following occurs only in Gram-negative bacteria?
(A) Peptidoglycan
(B) Integral proteins
(C) Lipoteichoic acid
(D) Lipopolysaccharide
(E) Phospholipids
Answer:
A) Peptidoglycan
D) Lipopolysaccharide
Explanation:
Gram-negative bacteria have a cytoplasmic membrane, a thin peptidoglycan layer, and an outer membrane containing lipopolysaccharide. There is a space between the cytoplasmic membrane and the outer membrane called the periplasmic space or periplasm.
Final answer:
Lipopolysaccharide (LPS)(D) is the compound that occurs only in Gram-negative bacteria, indicating the difference in their cell wall structure compared to Gram-positive bacteria.
Explanation:
Bacteria are classified as Gram-positive or Gram-negative based on their cell wall structure which determines their reaction to the Gram staining procedure. The distinction lies in the presence or absence of certain cellular components. Specifically, lipopolysaccharide (LPS) is a compound that is found only in the outer membrane of Gram-negative bacteria, distinguishing them from Gram-positive bacteria which do not have an outer membrane containing LPS.
Imagine you are cutting a bagel (one of the most common household injuries) and you get a cut. The cut heals. How do the new cells compare to the original (pre-cut) cells
Answer:
The new cells are the same as the previous ones, since they are the result of the mitosis process.
Explanation:
When we cut our skin, our brain sends information to millions of cells to take action and prevent this cut from putting us in danger. At that moment, the blood cells begin their work, supplying enough oxygen to stop possible bleeding and start the healing process. Then another group of cells swap out possible bacteria that may be trying to get into the wound. Last but not least, skin cells enter cell division and undergo mitosis, to generate new cells and create a new skin layer.
New cells are the same as old cells, as they are the result of mitosis. Mitosis is the process of cell division where one cell gives rise to two cells exactly the same as it.
The new cells should be the same as the previous ones, because they are the result of the mitosis process.
What is the mitosis process ?
Mitosis refers to the process where a eukaryotic cell nucleus divides in two, followed by the division of the parent cell into two daughter cells. Also, it is the process of cell division where one cell gives rise to two cells exactly the similar it is.
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In most normal human somatic cells, telomeres shorten with each division. In stem cells and in cancer cells, however, telomere length is maintained. In the synthesis of telomeres: Telomerase, a ribonucleoprotein, provides both the RNA and the polymerase needed for synthesis. the direction of synthesis is 3'~5'. the polymerase of telomerase is a DNA-directed DNA polymerase. the RNA of telomerase serves as a primer. the shorter, 3'~5' strand gets extended.
Answer:
Provides both the RNA and the polymerase needed for synthesis
Explanation:
Telomerase is an enzyme which extends the telomere sequences present at the end of the chromosomes. The telomerase enzyme acts as DNA polymerase as well as provides the RNA which serves as a template strand rather than the primer.
The polymerase acts as a reverse transcriptase enzyme and synthesizes the DNA strand from the RNA template. Since the telomerase provides RNA and acts as DNA polymerase, therefore, it is known as the ribonucleoprotein molecule.
Thus, the selected option is correct.
Sarah is a sprinter who specializes in quick and powerful bursts of speed followed by periods of rest. Priya is a marathon runner who specializes in long, steady runs. Compared to Priya, Sarah is likely to have_____________.
Answer:
The correct answer is : legs with a larger diameter.
Explanation:
Priya is a marathon runner who is better in long and steady runs and have more long skinny legs in compare to the Sarah is sprinter who specializes in quick and powerful speed and then a period of rest.
In Sarah the legs would have larger diameter as in marathon runners you will develop special muscles over the period of time. However, the type of running do make leaner muscles and sprinting adds bulk.
Thus, the correct answer is : legs with a larger diameter.
One of the first diagnostic tools used at the hospital was an electrocardiogram (EKG or ECG), which reflects the electrical activity of the cardiac muscle. We know that the atria contract first (the P wave) and then, after a brief delay, the ventricles contract (the QRS complex). Given that the heart does not have any nerves to stimulate the cardiac muscle cells, how is the timing of contraction coordinated? How do action potentials get from muscle cell to muscle cell? If the EKG shows a long delay between the P wave and the QRS complex, which type of cardiac tissue might have been damaged?
Answer:
The heart has an intrinsic conduction system that causes electrical activity in the heart muscles causing them to contract. The intrinsic conduction system is made up specialized cells, that contain nerve and muscular characteristics.The muscle cells in the heart are linked together by gap junctions, allowing cardiac action potentials to travel from one muscle cell to another.Atrioventricular (AV) node. The damage to the AV node causes the electrical signals traveling from the upper chambers to the lower chambers to be impaired causing an AV block.Explanation:
Discuss the effects of population size on both theeventual fate of an allele (fixation, loss) and the time to fixation. Describe evidence from your simulations, citing specific examples.
Explanation:
In populace hereditary qualities, obsession is the adjustment in a genetic supply from a circumstance where there exists in any event two variations of a specific quality (allele) in an offered populace to a circumstance where just one of the alleles remains. Genetic drift is an arbitrary procedure that can prompt large changes in populations over a brief time frame. It is produced the repeating little populace sizes by the Random drift , serious decreases in populace size called "bottlenecks" and founder events where another populace begins from few people. There are two significant sorts of hereditary float populace bottlenecks and the organizer impact. A populace bottleneck is the point at which a populace's size turns out to be little rapidly.Vascular plants have vascular tissues called xylem and phloem that transport materials throughout the plant. Classify these materials according to whether they are transported by the xylem or the phloem. Xylem Phloem
Answer:
Xylem transports water from roots and stems to leaves and Phloem transports food produced from photosynthesis from leaves to roots and stems.
Explanation:
The xylem and the phloem make up the vascular tissue of a plant and transports water, sugars, and other important substances around a plant.Phloem and xylem are closely associated and are usually found right next to one another. One xylem and one phloem are known as a ‘vascular bundle’ and most plants have multiple vascular bundles running the length of their leaves, stems, and roots.
The phloem carries important sugars, organic compounds, and minerals from the leaves to the non photosynthesized part of plant such as roots and stems . The phloem is made from cells two cells
1.sieve-tube members
2.companion cells.
The xylem is responsible for keeping a plant hydrated, it transports water from stems and roots to leaves. Two different types of cells are known to form the xylem
1. tracheids
2. vessel elements.
Xylem transports water and minerals, while phloem transports organic nutrients (such as sugars).
Xylem and phloem are specialized vascular tissues in vascular plants responsible for the transportation of essential substances throughout the plant. Xylem primarily functions in the upward transport of water and minerals absorbed by the roots from the soil. This process, known as transpiration, is driven by evaporation from the leaves, creating a negative pressure that pulls water upward. Xylem vessels, composed of specialized cells called tracheids and vessel elements, facilitate this transport, providing mechanical support to the plant as well.
On the other hand, phloem is responsible for the transport of organic nutrients, primarily sugars produced through photosynthesis in the leaves, to other parts of the plant. This process, called translocation, occurs bidirectionally, allowing for the distribution of sugars from sources (usually leaves) to sinks (areas of active growth or storage). Phloem consists of sieve tubes and companion cells, forming a continuous network throughout the plant.
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