Answer:
the answer would be that the frequency of the allele Rh + = 0.6
Explanation:
the frequency of the Rh value is incorrect, the value must be expressed in an integer that can have decimals but not a meaningless numerical sequence(0.01.0.01?), however, an exercise is performed taking the frequency value 0.16 as an example for the development of subsequent
we can start from the formula taking into account that the population is in Hardy-Weinberg equilibrium
p2 + 2pq + q2 = 1.0
the above is based on the fact that
p2 is considered as the dominant homozygous ratio
q2 as the proportion of recessive homozygotes
2pq as the proportion of heterozygous individuals
knowing that the Rh phenotype is recessive and has a value of 0.16, we replace obtaining
q2 = 0.16 or q = 0.4
where solving the equation
p + q = 1
p = 1- 0.4 = 0.6
the answer would be that the frequency of the allele Rh + = 0.6
Write a complete step-by-step plan for dealing with broken glassware on the floor or lab table. Write your plan in list form in the order in which steps should be taken.
Answer:
Notify the teacher and follow his or her instructions.
Wash the cut with soap and water.
Treat the cut with antiseptic from the first-aid kit.
Put a bandage over the cut.
Explanation:
The step-by-step plan for dealing with broken glassware on the floor or lab table has been stated below:
1 tell the person in charge or the teacher
2 make sure u did not get cut
3 get the large pieces and put them into glass waist container
4 get broom or brush weep area and get small pieces depose in same place
5 take a wet wipe and cover area to pick up anything that as missed.
What is glassware?Glassware has been refer to the any laboratory apparatus made of glass. When you are done using a glassware, wipe down the glassware to remove any cleaning solvent. They are often washed with a solvent and particles of the cleaning solvent adheres to the glassware.
A lot of our laboratory apparatus are made of glass e.g beakers, conical flask, measuring cylinders etc. These glassware must be properly cleaned after use.
These particles of the solvent must be wiped to remove these particles of cleaning solvent that could contaminate a sample placed in the glassware.
Therefore, The step-by-step plan for dealing with broken glassware on the floor or lab table has been stated below:
1 tell the person in charge or the teacher
2 make sure u did not get cut.
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Describe all the forces acting on the ice cream in the root beer float relate these forces to the fact that the ice cream is floating
Answer: The density of the ice cream and buoyant.
Explanation:
The density of the ice cream plus buoyant will cause the carbonation in root bear will cause gravity to push down on the ice cream and buoyant pushes the ice cream upwards, the ice cream floats because the buoyant force is greater than the weight of the ice cream.
Answer: The root beer exerts a force on all sides of the ice cream. These forces are all equal and cancel each other out. Gravity pushes downward on the ice cream. This can also be called the weight of the ice cream. Buoyant force pushes the ice cream upward. The ice cream floats because the buoyant force is greater than the weight of the ice cream.
Helpppppppp with Q2 Q3 Q4 and Q5.
Answer:
Q2 is D and Q3 is C Q4 is A and Q5 is A
Explanation:
a flood carries soil and rock with it which proves 2 c is the only physical change which proves 3 waves beak stones they don't combine 4 and more rainfall means more erosion so more sediment proves 5 the other answers on 5 is contradictory
Imagine that you could microinject cytochrome c into the cytosol of both wild-type cells and cells that were lacking both Bax and Bak, which are apoptosis-promoting members of the Bcl-2 family of proteins. Would you expect one, both, or neither of the cell lines to undergo apoptosis?
Answer:
In the given case, if the cytochrome c was administered within the cytosol of both the cells that were defective doubly for Bax and Bak and the wild type cells then the cell lines would go through the process of apoptosis. Both the Bax and Bak refer to the proteins that come under the Bcl-2 family and plays an essential role in triggering the process of apoptosis.
The cytochrome c administered in the process combines with an adapter protein that further stimulates a procaspase. Thus, due to the combination, the beginning of the caspase cascade takes place, which eventually results in apoptosis.
Though the cells that were devoid of both Bax and Bak fails to discharge cytochrome c from the mitochondria even after attaining the upstream signals, however, due to no issue in the pathway's downstream section because of the stimulation done by the cytosolic cytochrome c, the microinjection of cytochrome c nulls the effect of the doubly defective cells, and starts the process of apoptosis.
It is expected that the wild-type cells would undergo apoptosis, while the cells lacking both Bax and Bak would not.
Cytochrome c is a key component in the induction of apoptosis, particularly in the intrinsic pathway. When released from the mitochondria into the cytosol, cytochrome c binds to Apaf-1 and procaspase-9 to form the apoptosome, leading to the activation of caspase-9 and subsequent activation of effector caspases, such as caspase-3, which execute the apoptotic process.
In wild-type cells, the injection of cytochrome c into the cytosol would bypass the normal mitochondrial control mechanisms and directly trigger the formation of the apoptosome, thus initiating apoptosis. This is because all the necessary components for apoptosis execution are present and functional in wild-type cells.
On the other hand, Bax and Bak are pro-apoptotic proteins that are essential for the release of cytochrome c from the mitochondria into the cytosol during apoptosis. Cells lacking both Bax and Bak have a defect in their ability to release cytochrome c from the mitochondria in response to apoptotic stimuli. Therefore, even if cytochrome c is artificially introduced into the cytosol of these double-knockout cells, the absence of Bax and Bak would mean that the cells are unable to initiate the apoptotic program effectively, as the signal for apoptosome formation would not be properly transmitted.
In summary, microinjecting cytochrome c into the cytosol of wild-type cells would lead to apoptosis because the downstream apoptotic machinery is intact and functional. However, in cells lacking both Bax and Bak, the apoptotic pathway is disrupted, and these cells would be resistant to apoptosis induction by cytochrome c microinjection.
Which type of relationship is formed when a bristle worm lives on a White Sea urchin
The relationship between a bristle worm and a White Sea urchin is an example of commensalism, where the bristle worm benefits without harming or benefiting the urchin.
Explanation:The relationship between a bristle worm and a White Sea urchin is an example of commensalism.
In this type of relationship, one species (the bristle worm) benefits by living on or near another species (the White Sea urchin) without causing harm or benefiting it. The bristle worm uses the White Sea urchin as a source of protection, while the urchin remains unaffected.
This relationship is similar to when a bird builds a nest in a tree, benefiting the bird while not harming or benefiting the tree.
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Why is the sickle cell allele a nuisance trait rather than an advantage in most of the modern world?
Answer:
Sickle cell anaemia is a generic disorder which affects the red blood cells. The red blood cells assumes a sickle shape and contains lesser amount of the protein hemoglobin. Hemoglobin helps in the necessary transport of oxygen to other parts of the body.
The sickle cell allele is known to be advantageous to mainly people who stay in developing/underdeveloped counties who malaria endemic . People with the sickle cell allele have a higher chance of survival against malaria than people without this allele.
In modern countries malaria isn’t prevalent there due to the lack of vectors, poverty and environmental conditions which tends to support the survival of the mosquito. This means the sickle cell allele isn’t advantageous in this region.
A population of 150 fish have two alleles for the size of a black spot on their sides: large spot alleles and small spot alleles. The large spot allele is dominant.
For each situation listed below, identify the mechanism of evolution at work. Select whether it is an example of natural selection, genetic drift, or gene flow.
1. A fish net captures twenty fish, all who happened to have large spots.
2. Small spotted fish escape from predators better.
3. 15 large spotted fish move into this population.
4. Fish-eating birds catch large spotted fish more easily
Options:
A. Gene flow
B. Genetic Drift
C. Natural Selection
Answer:
1. genetic drift; 2. natural selection; 3. gene flow; 4. natural selection
Explanation:
1. A fish net captures twenty fish, all who happened to have large spots:
The mechanism of evolution that is playing out in this situation is most likely genetic drift. Genetic drift occurs when a random change occurs, resulting in the selection of a number of individuals by chance, rather than based on their level of fitness. The fish with large spots allele were selected against by chance, not necessarily because the fish with the small spot alleles were better fitted.
2. Small spotted fish escape from predators better:
The situation here is natural selection. The small spotted fish possess a greater fitness as they become well adapted to escaping from predators better than the large spotted fish, which gives confers on them a greater fitness to be selected for against the large spotted fish.
3. 15 large spotted fish move into this population:
The movement or introduction of new individuals i nto a population is what is referred to as gene flow in evolution, which is another mechanism of evolution that changes the allele frequency of the original population.
4. Fish-eating birds catch large spotted fish more easily:
This is another example of natural selection. The small spotted fish possess a greater fitness which makes them not easily preyed upon by Fish-eating bird easily, compared to large spotted fish. Natural selection favors the fish with small spotted alleles against the large spotted fish, and as a result, more small spotted fish would survive and reproduce more offspring with the small spotted alleles compared to those with large spotted alleles.
Compare biomes of North Africa and iceland
Final answer:
The biomes of North Africa, mainly characterized by the Sahara Desert, and Iceland, known for its volcanic and icy terrain, offer a stark contrast in climate, geography, and ecological diversity. These differences influence local flora, fauna, and human cultures.
Each biome has uniquely adapted to its environmental conditions, reflecting the planet's wide range of ecosystems.
Explanation:
The biomes of North Africa and Iceland represent a compelling study of contrast within the Earth's ecological diversity. North Africa is predominantly known for the Sahara Desert, the world's largest hot desert, characterized by its sand dunes, sparse rainfall, and extreme temperatures.
This biome is in stark contrast to the cold and volcanic landscape of Iceland, which is primarily recognized for its tundra and glacial regions that are influenced by its position near the Arctic Circle and its volcanic activity.
The biome in North Africa, beyond the Sahara, transitions into the Sahel and Mediterranean regions with more vegetation and milder climates suitable for agriculture, mirroring the adaptation of cultures and agricultural practices through history. Conversely, Iceland's geography, marked by its ice caps, geothermal hot springs, and volcanic terrains, creates unique ecosystems with specialized flora and fauna adapted to cold weather and geological activity.
Moreover, the cultural and historical impacts of these biomes on local civilizations are profound. The Sahara and its surrounding regions have facilitated trade routes and cultural exchange across continents, while Iceland's isolated and rugged terrain has shaped a society resilient and innovative in harnessing geological resources.
Final answer:
The Sahara Desert in North Africa and the glacier-covered landscapes of Iceland are examples of extreme ecosystems on Earth, with the former being a hot desert with adapted flora and fauna, and the latter a cold tundra with its unique biodiversity.
Explanation:
The biomes of North Africa and Iceland represent two extremes in the Earth's ecosystem. North Africa is home to the Sahara Desert, the world's largest hot desert, characterized by its sand dunes, rock formations, and highly arid conditions. In contrast, Iceland features a tundra biome and is known for its glacier-covered landscapes, such as the Vatnajökull ice cap, one of the largest in Europe. Both regions are remarkable for their distinct environmental features and adaptations of flora and fauna.
While the Sahara Desert in North Africa experiences extremely low precipitation and high temperatures, making survival for most plants and animals challenging, the area does support a variety of life adapted to these harsh conditions, including reptiles, small mammals, and desert-adapted plants. Conversely, Iceland's cold climate and short growing seasons result in a biome where mosses, lichens, and small shrubs predominate, and the animal life includes arctic foxes and various bird species.
The contrasts between these biomes highlight the incredible diversity of ecosystems on Earth. It underscores the importance of understanding and preserving these unique environments amidst global environmental changes.
The accumulation of high-energy protons creates an H gradient between the Energy released blank is used to move H ions into the blank. The accumulation of high-energy protons creates an H gradient between the blank and the blank. and the Energy released blank is used to move H ions into the blank. The accumulation of high-energy protons creates an H gradient between the ___ and the ___
Answer:
This is refers to the sequence of events which occurs during chemiosmosis for ATPs Synthesis.
1. the released during electron transport along the cytochrome carries. as proton motive force(PMF), is used to move H+ into the intramembrane space from the matrix, The accumulation of high -energy protons creates an H gradient (electochemical gradients )between the Intramembrane space and the mitochrondrial matrix.
The elctrochemical gradients produced from the Proton accumulation produced energy for the influx of H+ back into the matrix, and this energy (produced from the gradient) is used by ATPase synthase to synthesis ATPs from ADP and Pi.
High-energy protons create an H+ gradient, and the energy released moves these ions from a higher concentration area to a lower one. This ATP synthesis from potential energy stored in a gradient is termed chemiosmosis and relies on ATP synthase.
Explanation:The accumulation of high-energy protons creates an H+ gradient between the thylakoid space (in cellular respiration, between the mitochondrial matrix and the intermembrane space) and the stroma of the chloroplast. Energy released is used to move H+ ions from a high concentration area to a low concentration area. This movement, termed as chemiosmosis, relies on ATP synthase, a transmembrane enzyme that helps to convert potential energy stored in this gradient into chemical energy, stored in ATP molecules. Because of H+ ions positive charge and their higher concentration on one side of the membrane, an electrochemical gradient is established which powers ATP synthesis. The current of hydrogen ions powers the catalytic action of ATP synthase, which phosphorylates ADP, thereby producing ATP.
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11. Fill in the blanks to describe the similarities between transcription and DNA
replication.
Both transcription and DNA replication produce nucleic acids which are polymers of
(C, G, A, and T or U). Both processes depend on a
enzyme which adds one husketide at a time. To
determine which nucleotide is added next, both transcription and DNA replication
use the nucleotide sequence in a
strand and the
rules.
Transcription and DNA replication both produce nucleic acids, using a polymerase enzyme to add one nucleotide at a time. They follow a template strand and base-pairing rules to determine which nucleotide is added next.
Explanation:Both transcription and DNA replication produce nucleic acids which are polymers of nucleotides, namely C (cytosine), G (guanine), A (adenine), and either T (thymine) for DNA or U (uracil) for RNA. Both processes depend on a polymerase enzyme, which adds one nucleotide at a time. To determine which nucleotide is added next, both transcription and DNA replication use the nucleotide sequence in a template strand and follow base-pairing rules- A with T or U, and C with G.
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Both transcription and DNA replication produce nucleic acids which are polymers of nucleotides: cytosine (C), guanine (G), adenine (A), and in DNA replication, thymine (T); in transcription, uracil (U) instead of thymine. Both processes depend on a polymerase enzyme which adds nucleotides one at a time. To determine which nucleotide is added next, both transcription and DNA replication use the nucleotide sequence in a template strand and the base-pairing rules.
Transcription and DNA replication are both processes that produce nucleic acids, using polymerase enzymes and template strands to add nucleotides based on the base-pairing rules.
Nucleic acids, which are polymers of nucleotides, are produced by both transcription and DNA replication. In transcription, uracil (U) is produced in place of thymine, while in DNA replication, cytosine (C), guanine (G), and adenine (A) are produced. A polymerase enzyme that adds nucleotides one at a time is necessary for both processes. Both transcription and DNA replication employ base-pairing rules and the nucleotide sequence in a template strand to decide which nucleotide is inserted next.
The similarities illustrate that both processes involve the synthesis of new strands of nucleic acids, using a template to ensure that the new strand is complementary to the original. The differences, however, are impactful; replication results in an identical copy of DNA, while transcription translates DNA into mRNA, which carries the genetic message to the ribosome for protein synthesis. It is important to note that DNA replication employs DNA polymerase, involves both strands of the DNA, and uses thymine, whereas transcription uses RNA polymerase, generally involves only one strand of DNA, and incorporates uracil in the RNA.
complete question:
Fill in the blanks to describe the similarities between transcription and DNA replication.
Both transcription and DNA replication produce nucleic acids which are polymers of ...................(C, G, A, and T or U). Both processes depend on a........................... enzyme which adds ............................. at a time. To determine which nucleotide is added next, both transcription and DNA replication use the nucleotide sequence in a .................strand and the.................... rules.
Which material is a part of bed rock? Silk plants wood water
Answer:
I think you meant 'silt'
The slit is a part of the bedrock.
What is bedrock?The deposit of solid rock that is essentially buried beneath the soil and other fractured or segregated substances is referred to as bedrock.
The parent substance (the source of rock and mineral particles) for soil and regolith is the bedrock, which is made up of igneous, metamorphic, or sedimentary rock.
Bedrock is a source of nitrogen in the Earth's nitrogen cycle. An outcrop is a bedrock accumulation that occurs at the Earth's surface.
Thus, we can conclude that option A, the slit is correct.
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Edward Jenner is considered the founder of vaccines in the western world. In 1796, he inoculated a 13 year-old-boy with the virus that caused cowpox. Since cowpox and smallpox are closely related, the result was a demonstrated immunity to smallpox. Later, the first smallpox vaccine was developed. During the next two centuries, systematic implementation of mass smallpox immunizations culminated in its global eradication in 1979. Using the cowpox virus to create a virus for smallpox ... A) ...allowed only expectant mothers to pass the immunity on to their unborn children. B) ... induced a secondary response because the body immediately recognized cowpox as smallpox due to the similarities in their genetic blueprints. C) ...induced a primary immune response and immunological memory. Because of this, an encounter with the pathogen would trigger a rapid and strong secondary response. D) ...allowed the virus to lie dormant within the cells of the person. When introduced to the virus again, the body could use the dormant pathogen as a point of reference. Eliminate
Answer:
c
Explanation:
i got it right!
Final answer:
Using the cowpox virus to create a vaccine for smallpox induced a primary immune response and immunological memory in the individual, enabling a strong secondary response upon exposure to smallpox.
Explanation:
The use of the cowpox virus to create a vaccine for smallpox by Edward Jenner in 1796 represents a crucial development in the field of immunology and vaccine development. When Jenner inoculated a 13-year-old boy with the cowpox virus, it induced a primary immune response and the subsequent development of immunological memory. As such, upon later exposure to the smallpox pathogen, the body was able to mobilize a rapid and robust secondary immune response, protecting the individual from getting sick with smallpox. The correct answer to the question is Option C, as exposure to cowpox antigens gave rise to a primary response and memory cells ready to recognize and respond to similar antigens presented by the smallpox viruses upon future encounters.
What are traits? A)small hairs that grow on you’re forehead. B)characteristics passed from parents to children. C)people that are unpatriotic.
Answer:
b
Explanation:
Answer:
characteristics passed from parents to children.Answer B
a land manager proposes planting shrubs to help restore land damaged by erosion. describe a study or experiment that you could carry out to evaluate whether this proposal will work
Answer:
Yes, this proposal which was presented by a land manager will work.
Explanation:
Land with no vegetation has high rate of erosion as compared to those having more vegetative cover. This is because grasses, shrubs and trees have large root system in which soil particles stick to the roots which prevent erosion. If more vegetation is present in an area so these is very low chance of soil erosion due to wind and water. So the proposal which is presented by the land manager will work perfectly.
2. Why are there so many small faults in the San Francisco Bay Area?
Large stars with high mass also have high luminosity
O True
O False
Answer:
false
Explanation:
because the gas is not strong enough to be
done right
Suggest reasons why the information represented in the pyramid of numbers of animals of one of the ecosystems you studied may not truly represent that ecosystem. 2. According to your data, what is the ratio of third-order consumers to producers? Explain your answer. 3. Compare and contrast two of the ecosystems you studied. How is the energy conversion efficiency similar or different? 4. Does the population size increase or decrease at higher trophic levels in the pyramid of numbers of an ecosystem consisting of a tree, insects (that are herbivores) and birds feeding on the insects? Explain your answer. 5. What might happen to an ecological pyramid of numbers in a forest ecosystem if most of the deer were killed due to hunting by people and disease? 6. What would happen to an ecosystem if the decomposers disappeared? 7. Could there be a food chain without herbivores and carnivores?
Answer:
1. Because the ecosystem is changing every day that affects the numbers of animals. The numbers would also be estimated and not the true number of the animals in the ecosystem.
2.100:0.1 is the ratio of the third order consumer because only 10% of energy is transfered as the producer are eaten by the first order consumer. When the first order consumer eaten by second order, it only gets 1% of the total energy, and so on; therefor; the ratio is 100:1.
3. Lets consider two ecosystems water and forest an both are similar because both the ecosystem has more numbers of producers than that of consumers. But the number of products and consumption is different.
4. Yes, the populaton size decreases at higher trophic levels because the population of producers decline as the first orders producer require to eat lot of plants for their survival, so that the first order would soon consume all the producers and die of nutrition shortages.
5. Death of deer in the forest ecosystem will disturb the food chain as it will reduces the number of second order animals due to starving and die. It will also result in teh inrease the producers population becuase deer are not present to balance that.
6. If the decomposers will get disaaperaed thatn the dead plants and animals will not be able to decompose and affect the soil fertility. The soil will not get enough nutrition to grow new plants an remaisn of dead bodies can produce harmful gases which can affect the health of other living organisms.
7. No, because food chain represent the flow of energy between different levels and herbivores and carnivores are two important pillars of food chain.
Identify the following as biotic or abiotic.
Sunlight and heat:
Insects:
Rotting log:
Bacteria:
Water:
Rocks
Answer:
Answer:
sunlight and heat: Abiotic
insects: Biotic
rotting log: Biotic
bacteria: Biotic
water: Abiotic
rocks: Abiotic
Explanation:
Explanation:
The categories of biotic or abiotic things are:
Sunlight and heat:
Insects: biotic factor
Rotting log: biotic factor
Bacteria: biotic factor
Water: abiotic factor
Rocks: abiotic factor
What are biotic and abiotic factors?Biotic factors are those that are living, growing, and reproducing. These are plants, animals, fungi, etc.
Abiotic factors are those which are nonliving. Like a rock, or water.
Thus, the categories are:
Sunlight and heat: abiotic factor
Insects: biotic factor
Rotting log: biotic factor
Bacteria: biotic factor
Water: abiotic factor
Rocks: abiotic factor
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The information in pedigrees that is useful for mapping disease genes is sometimes limited. This pedigree shows segregation of neurofibromatosis, which is a fully-penetrant, autosomal dominant condition, and a single nucleotide polymorphism, SNP1, which has alleles G or T. Why is the cross in generation I not useful for determining the degree of linkage between the disease-causing gene NF and SNP1?
Answer:
Since neurofibromatosis is an autosomal dominant condition, the cross between individuals with genotypes carrying the same SNP type is not useful to show the degree of linkage between the disease-causing gene NF and SNP1. This occurs because it is necessary to observe segregation for the trait in order to determine its effects
The cross in generation I is not useful for determining the degree of linkage between neurofibromatosis and SNP1 because it's lacking the information about recombination frequency. Recombination frequency is required to determine relative gene locations on chromosomes and predict if the disease and the trait will be inherited together.
Explanation:The cross in generation I is not useful for determining the degree of linkage between the disease-causing gene NF and SNP1 because it does not provide information about recombination frequency.
Recombination frequency is a key element in determining linkage, allowing geneticists to map relative gene locations on chromosomes.
In the case of neurofibromatosis, a fully penetrant autosomal dominant condition, individuals with one copy of the gene (Nn) will express the trait. However, without a full understanding of the recombination rate between NF and SNP1, it is challenging to predict if the disease and the SPT1 trait will be inherited together.
Furthermore, autosomal dominant traits are expressed by the presence of just one gene, meaning that an individual with the disorder will know that they have at least one faulty gene. Nevertheless, the expression of the disease may manifest later in life, after the childbearing years, which further complicates the analysis.
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Which Are characteristics of natural selection
From your experiences with the lawn and the butterfly garden, you can affirm that ________. higher biodiversity of plants means a lower biodiversity of animals pesticides need to be used to maintain a diversity of native species once natural habitats are altered or destroyed they can never be restored, even in part native plants are more likely than exotic species to support local herbivores native plant landscaping can be done so that no predators or parasites are present
Answer:
native plants are more likely than exotic species to support local herbivores native plant landscaping can be done so that no predators or parasites are present
native plants are more likely to support local herbivorous animals than are exotic plant species
Explanation:
Exotic species are known to perform well and survive/thrive due to the fact that they lack natural enemies in their new environment making it obvious that the native species will likely support more of local herbivores
The graph shows the petroleum reserves that remain in five regions of Europe. Arrange the regions by decreasing amount of petroleum reserves
Answer:
B>A>C>D>E
Explanation:
The graph in the question shows the amount of petroleum reserves (on y-axis) in five regions of Europe (on x-axis). To arrange the amount of reserves in a decreasing order, this means we arrange from the most amount to the least amount of petroleum reserves (million of barrels of oil equivalent).
The height of the the bar graph represents the amount of petroleum reserves in each region. Hence, the tallest and shortest bar represents the most and least respectively.
According to the graph in the image, Region B has the most amount of petroleum reserves. Region A is next to it. Region C follows A. Region D follows and the least amount is Region E. Based on this, the descending or decreasing order is:
B>A>C>D>E
Evidence supporting E.coli DNA polymerase III having the major role in nucleotide incorporation during replication includes:
a) A point mutation can inactivate DNA pol III and inhibit replication.
b) DNA polymerase III has a high Vmax.
c) Replication is blocked at high temperatures in temperature sensitive mutants with a thermolabile form of DNA polymerase III.
d) There are ten or fewer molecules of enzyme per cell.
e) All of the above.
Answer:
c
Explanation:
For which of the following pairs does the molecule given as the first term on the left contribute to the synthesis of
the molecule that is the second term on the right?
Select one:
a fatty acid - starch
b. amino acid - protein
C. nucleotides - glycogen
d glucose - fat
e urea - DNA
Answer: B) amino acid - protein
Explanation:
Systemic lupus erythematosus (SLE or lupus) is distinct from other autoimmune disorders because Systemic lupus erythematosus (SLE or lupus) is distinct from other autoimmune disorders because there is a genetic predisposition. it is also a hypersensitivity disorder. multiple organs and tissues are affected. it involves an antibody response. it may be triggered by an infection.
Answer:
All these responses are correct
Explanation:
The systemic lupus erythematosus (SLE) is an autoimmune disorder that may be linked to both environmental and genetic factors. This disease is caused by the own immune system (i.e., antibodies) that is directed against the body's tissues in a healthy person.
SLE is a systemic autoimmune disease characterized by its widespread impact on the body, presence of autoantibodies against the individual's own DNA/proteins, and resulting type III hypersensitivity reactions leading to varied organ dysfunction.
Explanation:Systemic lupus erythematosus (SLE or lupus) is distinctive among autoimmune disorders due to several characteristics. SLE is a systemic autoimmune disease that can affect multiple organs and tissues across the body, including the muscular, skeletal, integumentary, respiratory, and cardiovascular systems. Unlike diseases such as type I diabetes, which target specific cell types, SLE has a broad impact on the body. The autoimmune mechanisms in SLE include type III hypersensitivity reactions, and a hallmark of the disease is the presence of autoantibodies against nuclear and cytoplasmic proteins, such as anti-nuclear antibodies (ANAs) - found in over 95% of patients. Two distinct autoantibodies, anti-double-stranded DNA (ds-DNA) and anti-Sm antibodies (antibodies to small nuclear ribonucleoprotein), are particularly associated with SLE and are used as part of the classification criteria. These antibodies lead to the formation of immune complexes and subsequent tissue damage.
The variety of symptoms associated with SLE reflects its systemic nature; these can include psychological fatigue, loss of appetite, a characteristic butterfly rash, joint arthritis, and other issues dependent on which areas of the body are affected. While the symptoms of SLE can often be managed with drugs or other treatments, there is no cure for the underlying autoimmune process.
You know that the narrow-sense heritability of milk production in Ayrshire cattle is 0.587. You perform an experiment where you take a parental population of cattle with a mean milk production of 5.1L/hr and selectively breed only the top 10% of cows in this population, who have a mean milk production of 8.9L/hr, what is the response to selection (R) in your experiment
Answer:
Explanation:
Narrow sense heritability - h2
selection differential - S
Selection differential is calculated from the difference between the population average and the parental population.
Breeder's equation:
Response to selection - R = h2S
Mean milk production of 10% cows for experiment = 8.9L/hr
Mean milk production of parental population = 5.1L/hr
Selection differential S= 8.9 - 5.1 = 3.8 L/hr
Response to selection = 0.587 × 3.8 = 2.23
When would gluconeogenesis be most active, after a meal or when you wake up in the morning? Explain your reasoning. Match the words in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer.
Answer:
Gluconeogenesis would be more active when you wake up in the morning.
Explanation:
The words and the sentences to match are missing but the reason why gluconeogenesis is more active when we wake up in the morning is that gluconeogenesis is the process by which our body produces glucose, the main fuel of the body, from non-carbohydrate carbon elements, such as proteins and lipids, when there are no carbohydrate carbon elements present in the body, since a long time has passed since our last meal, and the body has already consumed the glucose that was obtained from the ingested food, the system produces glucose from other elements.
Answer: Gluconeogenesis would be more active when you wake up in the morning.
Explanation:
Gluconeogenesis is the metabolic process by which living organisms produce sugars (namely glucose) for catabolic reactions from non-carbohydrate precursors (non carbohydrate elements). This is why gluconeogenesis is more active when we wake up in the morning due to the fact that a long time has passed since our last meal, and the body has already consumed the glucose that was obtained from the ingested food hours ago, the system tends to produces glucose from other elements available in the body.
The words and the sentences to match are missing.
Which of these modes of transmission require a bodily opening either natural or artificial?
Answer:
Answer:
Vectors-These are diseases transmission which involves the piercing of the skin-vectors penetration e.g by mosquito. ticks.
Air borne which involves entry through the nasal passages, mouth, eyes to reach the body the respiratory tracts and the blood.
Blood borne diseases are transmitted from e.g piercing of the skin by contaminated objects needles, broken glass. If there is an opening in the skin these transmission is also possible.
Sexual can be transmitted through open sores at the genitals.or use of contaminated needle by infected subject. Sores in the mouth during kissing may be another route
food borne or water borne needs to enter through natural opening of the mouth to reach the body fluids and the blood through the GIT.
Contact diseases does not necessarily need a direct opening for transmission . rather there should be a direct contact between the infected individual and the other. Sometimes a direct contact between the two is not need. e.g common cold, chicken pox , influenza , etc.
Explanation:
Which of the following address(es) the problems posed by chromatin during transcription?
a) In some instances, protein factors can interfere with chromatin structure at specific locations, opening sensitive sites.
b) The action of chromatin remodeling factors is involved, which require ATP hydrolysis to allow the promoter regions to accept the complex.
c) In some cases, the chromatin structure rearranges at the time of replication.
d) All of the listed statements are true.
Answer:
Option D All of the listed statements are true
Explanation:
the major problem posed by chromatin during transcription includes its state at that moment in time. The chromatin might exist in a state that prevent the transcription machinery from accessing the DNA to carry out transcription and bring about gene expression. thus, its state can either allow for gene expression or not.
Thus, some factors/processes have been put in place to regulate this. This includes chromatin remodeling complexes that utilize the energy supplied by ATP hydrolysis to affect nucleosomes causing structure alterations, covalent histone modification that allows for unwinding of the DNA from nucleosome to allow for transcription to take lace.
Spiders and lobsters?!? Juan was surprised that spiders and lobsters were similar in at least two ways. Juan also noted that all insects were similar to each other because they all had A) wings. B) 6 legs. Eliminate C) 8 legs. D) two body segments
Answer:
B
Explanation:
Answer:
The correct answer would be B. 6 legs
Explanation:
Insects are similar because they all have 6 legs