8. Local and global winds on Earth are caused by the unequal heating of the atmosphere, land, and oceans.
Which of the following is a local wind pattern?
A trade winds
B. sea breeze
C polar easterlies
D. prevailing easterlies
Answers
It is because it comes in the local wind pattern
Answer:
B is right answer. sea breeze
Related Questions
A student isolates a sample of nucleic acids from a cell. which of these experiments can the student perform to distinguish whether the isolated compound is dna or rna?
Answers
A student can perform enzymatic digestion, gel electrophoresis, and reverse transcription experiments to distinguish whether the isolated compound from a cell is DNA or RNA. A student isolates a sample of nucleic acids from a cell and wants to determine whether the isolated compound is DNA or RNA.
There are a few experiments the student can perform to distinguish between the two:
1. Enzymatic Digestion: The student can use specific enzymes like DNase and RNase to digest the isolated compound. If the compound is DNA, it will be resistant to DNase digestion but susceptible to RNase digestion. Conversely, if the compound is RNA, it will be resistant to RNase digestion but susceptible to DNase digestion.
2. Gel Electrophoresis: The student can run the isolated compound on an agarose gel using gel electrophoresis. DNA and RNA have different migration rates due to their differences in size and charge. By comparing the migration of the isolated compound to known DNA and RNA markers, the student can determine whether it is DNA or RNA.
3. Reverse Transcription: If the student suspects the isolated compound might be RNA, they can perform reverse transcription. Reverse transcription is a process that converts RNA into complementary DNA (cDNA) using the enzyme reverse transcriptase. If the isolated compound can be converted into cDNA, it confirms the presence of RNA.
In conclusion, a student can perform enzymatic digestion, gel electrophoresis, and reverse transcription experiments to distinguish whether the isolated compound from a cell is DNA or RNA.
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Water is a polar molecule because
Answers
Answer:
Water is a polar molecule because of its unequal sharing of electrons.
Answer:
The reason why Water is considered a polar molecule is because of Unequal sharing of electrons.
Explanation:
This makes the oxygen end of the molecule slightly negative. Since the electrons are not near the hydrogen end as much, that end is slightly positive. When a covalent bonded molecule has more electrons in one area than another, it is called a polar molecule.
Which process provides plants with the glucose needed for cellular respiration?.
Answers
Answer:
Photosynthesis is the process that plants use in the presence of sunlight to convert carbon dioxide taken in by their leaves and water taken in through the roots to produce oxygen and sugar (glucose).
choose the correct makeup of a haploid set and the correct makeup of a diploid set. select all that apply. choose the correct makeup of a haploid set and the correct makeup of a diploid set.select all that apply. red chromosomes make up a diploid set. the chromosomes of one colour make up a haploid set. the chromosomes of one colour make up a diploid set. all red and blue chromosomes together make up a diploid set. blue chromosomes make up a diploid set. all red and blue chromosomes together make up a haploid set.
Answers
The correct makeup of a haploid set is that the chromosomes of one color make up a haploid set. This means that all the chromosomes that an organism receives from one parent are included in a haploid set. For example, if an organism has four chromosomes, and it receives two from each parent, then the haploid set would contain two chromosomes of one color.
The correct makeup of a diploid set is that red and blue chromosomes together make up a diploid set. This means that an organism has two complete sets of chromosomes, one from each parent. For example, if an organism has four chromosomes, and it receives two from each parent, then the diploid set would contain all four chromosomes, two of one color and two of the other color.
Therefore, the correct options are: the chromosomes of one color make up a haploid set and red and blue chromosomes together make up a diploid set. The other options are incorrect.
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Patient 2 has Type _______
blood.
Answers
Answer:
Patient 2has type B+ blood
for animal cells a ____ starts devloping from the outside of the cell and will eventually spluit the cell into two
Answers
for animal cells a cleavage furrow starts devloping from the outside of the cell and will eventually spluit the cell into two
Imagine you are working with a culture of yeast that is 2 X10^10 cells/ml. You expose the cells to 20
secs of UV light, which you are told results in a 20% survival rate
a. If you plate 0.1 ml from the 10^-6 dilution onto YEPD, how many colonies do you predict to see?
b. If you wanted to have 40 surviving colonies on the plate after the 20 sec UV exposure, how
many cells would you need to plate?
c. What dilution of the culture would you need to plate this number of cells (assuming you plate 0.1 ml)?
Answers
a. If you plate 0.1 ml from the 10^-6 dilution onto YEPD, 4 × 10^11 colonies/ml colonies are predicted to be seen.
b. If you wanted to have 40 surviving colonies on the plate after the 20 sec UV exposure, 1000 cells are needed to be plated.
c. 5 × 10^-9 dilution of the culture is needed to plate 1000 cells.
a. To calculate the predicted number of colonies when you plate 0.1 ml from the 10^-6 dilution onto YEPD, you can use the following formula:
N = (number of colonies)/(volume plated × dilution factor).
First, you need to calculate the total number of cells in the culture that were exposed to UV light:
N0 = 2 × 10^10 cells/ml × V, where,
V is the volume of the culture that was exposed to UV light.
Since we don't know the volume of the culture, we can't calculate N0 directly. However, we do know that the survival rate after the UV exposure was 20%, which means that only 20% of the cells were still alive:
N = 0.2 × N0
Now we can calculate the number of cells in the 10^-6 dilution that we plated:
N1 = 10^6 × N0 = 2 × 10^16 cells/ml.
Then we can calculate the number of surviving cells in the 10^-6 dilution that we plated:
N2 = 0.2 × N1 = 4 × 10^15 cells/ml
Finally, we can calculate the predicted number of colonies:
N = N2 × 0.1 ml/(10^-6) = 4 × 10^11 colonies/ml
b. To have 40 surviving colonies on the plate after the 20 sec UV exposure, you need to plate a number of cells that will give rise to 40/0.2 = 200 cells after the UV exposure. Let's call this number N3. We can calculate N3 as follows:
N3 = 200 cells/0.2 = 1000 cells
c. To calculate the dilution of the culture that we need to plate to get 1000 cells in 0.1 ml, we can use the following formula:
N4 = N3 × (volume plated × dilution factor) = 1000 cells
N5 = N0 × dilution factor
N6 = N5 × (volume plated × dilution factor)
= N4N5/N6 = dilution factor 2 × 10^10 cells/ml × dilution factor × 0.1 ml
= 1000 cells
dilution factor = 5 × 10^-9
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one initial suspect bacteria you considered was the common sepsis (bloodstream infection) organism staphylococcus aureus. why can we rule this organism out as the cause in the patient?
Answers
As Staphylococcus aureus are the cause of the patient's infection because the blood culture tests performed on the patient did not show the presence of this particular organism.
Blood culture tests are highly sensitive and specific for detecting the presence of bacteria in the bloodstream, and the absence of Staphylococcus aureus in the patient's blood cultures indicates that this organism is not the cause of their sepsis. Additionally, other clinical factors such as the patient's symptoms and medical history may have contributed to ruling out Staphylococcus aureus as a possible culprit.
Staphylococcus aureus as the cause of sepsis in the patient, we need to consider factors such as the patient's symptoms, lab results, and any other relevant clinical information. If these factors don't match the typical presentation of a Staphylococcus aureus infection, then it's likely not the cause and we can rule it out. Additionally, a negative blood culture result for Staphylococcus aureus would further support this conclusion.
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What are the 4 steps of mitosis and briefly explain them?
Answers
The 4 steps of mitosis are Prophase, Metaphase, Anaphase, Telophase.
The chromosomes shorten and thicken during prophase. Chromosomes align in the centre of the cell during metaphase. Chromatids split off at the centromere and migrate to opposing poles during anaphase. Telophase: Following the reorganisation of nuclear envelopes surrounding each cluster of chromosomes, two nuclei are created.
A centriole that is positioned outside the nucleus splits during the prophase, the first phase. The nuclear membrane vanishes when the lengthy, threadlike substance of the nucleus coils up into the visible chromosomes. Long, fine threads protrude from the centrioles in all directions. The spindle is made up of several of them from one centriole joining strands from the other.
Metaphase, the second stage of mitosis, sees the chromosomes travel onto the spindle's equatorial plane. The third phase, called anaphase, begins when the chromatids divide and move to the opposite ends of the cell. The chromosomes are the split chromatids. A whole pair of chromosomes moves in this way to reach each centriole.
In the telophase, the last step, the cell divides. Contrary to what happened during prophase, the chromosomes uncoil, fresh membranes form around the nuclei, and the spindle fibres disappear. Two identical cells have formed from the original one, and they are now ready to begin their initial growth phase.
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Which grouping(s) in an amino acid allows for hydrogen bonding to occur among amino acids?
N-H
C-N
C-H
C-0
0-H
Answers
Answer: 291
Explanation:
because i know
Answer:
C N i was having the same question in the high school and that's correct
Neck Pain: Changes in Strength & Endurance- decreased (concentric/isometric/eccentric) strength & endurance in...
- What are the 3 muscle groups??
Answers
The three muscle groups commonly affected by neck pain are the sternocleidomastoid, the scalene, and the upper trapezius.
The sternocleidomastoid is a long, superficial muscle located on both sides of the neck, stretching from the jaw to the collarbone and the base of the skull. The scalene muscles are three small muscles located on each side of the neck and connect the cervical vertebrae with the first rib.
The upper trapezius is a large muscle located on each side of the neck, stretching from the base of the skull to the middle of the back. When these muscle groups are affected by neck pain, they can experience decreased strength and endurance in concentric, isometric, and eccentric contractions.
Concentric contractions involve the muscles shortening, such as when you lift your arm up. Isometric contractions involve the muscle staying the same length, such as when you hold your arm in a certain position. Eccentric contractions involve the muscles lengthening, such as when you lower your arm down.
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Carbon dioxide is a waste product of the Krebs Cycle in cellular respiration.
True
False
Answers
Which are functions of small RNAS?
Select all that apply.
to build proteins using codons
to cause a protein complex to break down mRNA molecules
to carry all the information necessary to build proteins
to prevent mRNA molecules from being translated
Answers
Small RNAs serve a variety of purposes, like option B: to cause mRNA transcripts to be broken down by a protein complex, and option D: mRNA molecules from being translated.
By attaching to particular mRNAs and either preventing or triggering the destruction of their translation, small RNAs, such as microRNAs and siRNAs, can control the expression of genes. They neither carry the information required to construct proteins on their own, nor do they construct proteins using codons.
Numerous biological processes require RNA, also known as ribonucleic acid, which is a single-stranded nucleic acid. Its structure is comparable to that of DNA, except instead of deoxyribose and thymine, it contains the sugar ribose and the nucleotide uracil.
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how can i get my mom and dad to trust me ans to give me my phone and PlayStation back. ive been grounded for two months already and overtime i think of my friends i get in my feelings
Answers
which of the following describes the resting membrane potential of a neuron?
Answers
The resting membrane potential of a neuron refers to its stable electrical charge when not transmitting signals, typically around -70 millivolts (mV), maintained by ion concentration gradients and selective ion channels.
What is the resting membrane potential?The resting membrane potential of a neuron refers to the electrical charge difference across its cell membrane when the neuron is at rest, meaning it is not sending or receiving signals. Typically, the resting membrane potential of a neuron is around -70 millivolts (mV) inside the cell compared to the outside.
This polarization is maintained by the distribution of ions, primarily potassium (K+) and sodium (Na+), across the membrane. The resting potential is a result of the balance between passive ion diffusion and active ion pumping mechanisms.
The sodium-potassium pump helps maintain this potential by actively pumping out three Na+ ions for every two K+ ions it pumps in. The resting membrane potential plays a crucial role in neuronal excitability and the transmission of signals within the nervous system.
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Write a suggestion for your friend who wants to marry at early age.
Answers
Select the correct statement(s) about invertebrate taxa.
Select all that apply.
a) Rotifers are smaller than many protists.
b) Tapeworms lack a mouth and gastrovascular cavity.
c) Ectoprocts lack a distinct head.
Answers
The following are the correct statements about invertebrate taxa: Rotifers are smaller than many protists. Tapeworms lack a mouth and gastrovascular cavity. Ectoprocts lack a distinct head. The above statement (a, b and c) are correct about invertebrate taxa.
Invertebrates are a group of animals that lack a vertebral column or backbone. As a result, they do not have a spine. They are characterized as the most numerous and varied animal group on the planet. The majority of animals on Earth are invertebrates. They can be found in almost every environment, ranging from freshwater to saltwater, hot to cold, and terrestrial to marine. Invertebrates include a wide range of animals, including sponges, cnidarians, flatworms, roundworms, mollusks, arthropods, and echinoderms. These animals come in a variety of shapes and sizes, from tiny rotifers to giant squid. Invertebrates also include many of the world's most vital species, such as insects and other arthropods.
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Helpppp please pretty pretty plzzzzzzz characteristics that can be observed or measured without changing the matters identify called chemical properties. True Flase
Answers
Answer:
False
Explanation:
Those are physical changes. Changing the identity of matter means changing its chemical composition which is what happens in chemical changes.
The bleeding disorders Hemophilia A, B and C is a significant medical conditions. Compare and contrast the pathophysiology, pathways or factors which are disrupted, clinical outcomes, patient management and diagnostic tests utilized for these related bleeding disorders. Where do these conditions alter the coagulation pathway?
Answers
Hemophilia A, Hemophilia B, and Hemophilia C are all bleeding disorders that affect the coagulation pathway, but they differ in terms of their underlying causes, specific factor deficiencies, clinical outcomes, patient management, and diagnostic tests. Here's a comparison and contrast of these three conditions:
Hemophilia A:
Pathophysiology: Hemophilia A is caused by a deficiency or dysfunction of clotting factor VIII (FVIII), which plays a crucial role in the intrinsic coagulation pathway.
Pathways or Factors Disrupted: Hemophilia A affects the intrinsic pathway of the coagulation cascade, leading to impaired formation of fibrin clots.
Clinical Outcomes: Individuals with Hemophilia A experience prolonged bleeding, especially in response to trauma or injury. Spontaneous bleeding into joints and muscles is common.
Patient Management: Treatment involves the administration of exogenous factor VIII, either on-demand or as prophylaxis, to control bleeding episodes. Physical therapy and joint protection strategies are also important for long-term management.
Diagnostic Tests: Diagnosis is confirmed through laboratory tests that measure FVIII levels, such as the activated partial thromboplastin time (aPTT) and specific FVIII assays.
Hemophilia B:
Pathophysiology: Hemophilia B, also known as Christmas disease, is caused by a deficiency or dysfunction of clotting factor IX (FIX), which is involved in the intrinsic coagulation pathway.
Pathways or Factors Disrupted: Hemophilia B affects the intrinsic pathway of the coagulation cascade, similar to Hemophilia A.
Clinical Outcomes: Clinical manifestations of Hemophilia B are similar to Hemophilia A, with prolonged bleeding, joint and muscle bleeds, and increased risk of spontaneous bleeding.
Patient Management: Treatment involves the replacement of factor IX through factor IX concentrates. Prophylactic or on-demand treatment is used to prevent or control bleeding episodes.
Diagnostic Tests: Diagnosis is confirmed by measuring FIX levels through laboratory tests such as the aPTT and specific factor IX assays.
Hemophilia C:
Pathophysiology: Hemophilia C is caused by a deficiency or dysfunction of clotting factor XI (FXI), which is involved in the intrinsic pathway of the coagulation cascade.
Pathways or Factors Disrupted: Hemophilia C affects the intrinsic pathway of coagulation, similar to Hemophilia A and Hemophilia B.
Clinical Outcomes: Hemophilia C generally causes milder bleeding symptoms compared to Hemophilia A and B. The bleeding tendency is more pronounced in specific situations, such as surgery or trauma.
Patient Management: Treatment of Hemophilia C is generally not required unless the patient is undergoing surgery or experiences significant bleeding. In those cases, FXI concentrates or fresh frozen plasma (FFP) can be administered.
Diagnostic Tests: Diagnosis is confirmed through laboratory tests, including the aPTT and specific factor XI assays.
In summary, Hemophilia A, Hemophilia B, and Hemophilia C are all bleeding disorders that disrupt the intrinsic pathway of the coagulation cascade. Hemophilia A involves a deficiency of factor VIII, Hemophilia B involves a deficiency of factor IX, and Hemophilia C involves a deficiency of factor XI. The severity of bleeding symptoms and management strategies may vary among these disorders. Diagnostic tests, such as aPTT and specific factor assays, are utilized to diagnose and monitor these conditions.
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A malignant tumor is capable of spreading to other body tissues, while a benign one is not. true or false
Answers
The statement 'a malignant tumor is capable of spreading to other body tissues, while a benign one is not' is TRUE. This is the major difference between both types of tumors.
Cancer cells and malignant tumorsCancer cells are a specific type of cells that proliferate in an uncontrolled manner.
Malignant tumors are produced by cancer cells that are able to migrate to other tissues.
The major difference between both types of tumors (malignant and benignant) is the ability to produce metastasis.
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Sample Problem Given the DNA molecule below:
5' - TTCCATGTCAGTGCGTATACATACC -3'
3' - AAGGTACAGTCACGCATATGTATGG -5'
Transcription starts at the right most base pair and goes to the end of the molecule.
Which strand is the template strand (top or bottom)?
Give the sequence of the mRNA formed and label the ends.
Give the sequence of the protein formed and label the ends.
Answers
The template strand for transcription is the bottom strand (3' - AAGGTACAGTCACGCATATGTATGG -5').
Transcription involves the synthesis of an RNA molecule using the DNA template strand as a guide, and the complementary base pairing between the DNA template and the RNA transcript is crucial for this process.
The sequence of the mRNA formed is 5' - AUGUAUACGCACUGACAUUGGAA -3', with the start codon AUG at the 5' end and the stop codon UAA at the 3' end. The labeling of the ends refers to the addition of a 5' cap and a 3' poly(A) tail to the mRNA, which occur after transcription.
The sequence of the protein formed depends on the genetic code, which translates the mRNA sequence into a string of amino acids.
Using the standard genetic code, the protein formed from this mRNA sequence would have the sequence Met-Tyr-Thr-His-Asp-Leu, with the N-terminus (amino end) starting with the methionine (Met) residue and the C-terminus (carboxy end) ending with the leucine (Leu) residue.
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Egg is composed of the shell, egg white and egg yolk
yr own word please
Agree or disagree explain why
Answers
Two differences between the heart of a fetus and an adult heart.
Answers
As in an adult heart, the prenatal heart develops four chambers and four valves. But because the fetal lungs will not be used until after birth, blood must bypass the lungs. Two structures develop in the prenatal heart that allow the blood to be routed around the lungs: the foramen ovale and the ductus arteriosus.
Hope it helps...Both chimeric antigen receptors and BiTEs function to bring cytotoxic T cells into close proximity to cells bearing a target antigen in order to kill the target cell. What is one advantage of using BiTEs over CAR T cells?
CAR T cells lock directly onto the tumor cells using a receptor on their surfaces while BiTEs rely on an intermediate molecule, reducing the chance of attacking normal tissue cells. BiTEs can be designed to target many different types of antigen, while CAR T cells can only target a limited number of types of antigen. CAR T cells last a long time in the body so the effect lasts long after the initial injection without needing additional treatment. BiTEs do not last very long in the body so the treatment can be dosed to limit adverse effects
Answers
Option 2 is Correct. One benefit of employing BiTEs over CAR T cells is that they can be programmed to target a wide variety of antigens, as opposed to CAR T cells, which can only target a small variety of antigens.
By bringing cytotoxic T cells into close contact with cells expressing a target antigen, chimeric antigen receptors and BiTEs both work to destroy the target cell. The use of CAR (chimeric antigen receptor) T cells is one of the key ACT strategies.
By allowing T cells to bind target cell surface antigens via a single-chain variable fragment (scFv) recognition domain, CAR T cells facilitate MHC-unrestricted tumour cell killing. a particular class of immune cell that is capable of eliminating specific types of cells, such as virus-infected cells, cancer cells, and alien cells.
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Correct Question:
Both chimeric antigen receptors and BiTEs function to bring cytotoxic T cells into close proximity to cells bearing a target antigen in order to kill the target cell. What is one advantage of using BiTEs over CAR T cells?
1. CAR T cells lock directly onto the tumor cells using a receptor on their surfaces while BiTEs rely on an intermediate molecule, reducing the chance of attacking normal tissue cells.
2. BiTEs can be designed to target many different types of antigen, while CAR T cells can only target a limited number of types of antigen.
3. CAR T cells last a long time in the body so the effect lasts long after the initial injection without needing additional treatment.
4. BiTEs do not last very long in the body so the treatment can be dosed to limit adverse effects.
Please hurry. Which characteristic would most likely remain constant when a limestone cobble is subjected to extensive abrasion? Explanation
Answers
Answer: I think the answer is Composition.
Explanation: i hope this helps you
The aye-aye lemur is a mammal that feeds mostly on insect larvae that live inside trees. The aye-aye lemur has a specialized middle finger that is long and thin. The aye-aye lemur moves along a tree branch and taps the branch with its specialized finger. When the aye-aye lemur hears a difference in the echo, it will tear open the bark with its teeth until the insect tunnel is exposed. The aye-aye lemur then uses its specialized finger to reach the insect larvae and remove it.
Which selective pressure most likely resulted in the development of the aye-aye lemur's special adaptation?
A.Limited availability of water
B.Competing for mates
C.Large numbers of natural predators
D Food sources that are hard to find
Answers
Answer: D because the lemur sticks it’s middle finger in the tree to eat
Explanation:
Just took test
Adaptation is the process of survival to the environment and the surroundings. The development of the special adaptation is due to the food sources that are hard to find.
What are the adaptations?Adaptations are the specialized changes that an organism acquires for survival and reproduction in altered conditions so that their generations can be maintained.
The lemurs adapted to the thin and long middle fingers and teeth so that they can tear open and reach the insects living in the barks that are the food source for them.
Therefore, option D. hard sources of food led to the special adaptation.
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What are the letters that make up the 4 bases in dna?.
Answers
Answer:
acgt
Explanation:
adenine, cytosine, guanine, and thymine
Describe how a substrate interacts with an enzyme. Use the phrase lock and key in your description. Answer in 2-3 sentences.
Answers
A substrate interacts with an enzyme to generate an enzyme-substrate conformational change that resembles a key capable of matching to a particular lock in a door.
What is enzyme specificity?Enzyme specificity refers to the ability of an e3znyme to bind to a particular substrate in order to catalyze a chemical reaction. This binding lead to the formation of the enzyme-substrate complex where the substrate matches with the enzyme at the active site in a similar way to a lock and a key.
Therefore, with this data, we can see that enzyme and substrate are very specific and they interact in a similar way to a key and a lock in the door.
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How did sir francis bacon believe science should be determined
Answers
Sir Francis Bacon believed that science should be determined through the scientific method, which involves empirical observation, experimentation, and the use of inductive reasoning to derive general principles and laws.
Sir Francis Bacon was an English philosopher, statesman, and scientist who is credited with developing the scientific method. Bacon believed that the key to scientific knowledge was the use of empirical observation, experimentation, and the use of inductive reasoning to derive general principles and laws. He believed that scientific knowledge should be based on facts and evidence rather than on speculation or tradition. Bacon argued that scientific knowledge should be based on observation and experimentation rather than on speculation or authority. He believed that scientific knowledge should be tested and verified through empirical methods and that it should be based on facts and evidence. Bacon's scientific method involved four steps: observation, hypothesis, experimentation, and conclusion.
In conclusion, Sir Francis Bacon believed that science should be determined through the scientific method, which involves empirical observation, experimentation, and the use of inductive reasoning to derive general principles and laws. He believed that scientific knowledge should be based on facts and evidence rather than on speculation or tradition, and that it should be tested and verified through empirical methods.
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You learned about how an individual’s genes express various physical and behavioral traits. The study of genes can shed light on various aspects of human development. The Human Genome Project is one such initiative that set out to map the genome of various random individuals. Research and write a short report about the Human Genome Project, the people involved in the program, the project’s goal, and some of its key findings.
Answers
The HGP mapped the entire human genome, involving scientists from around the world, launched in 1990. The HGP aimed to map all human genes, identify DNA sequences, find genetic variations causing diseases, and create new genetic study tools.
What is the genes about?The Human Genome Project (HGP) was a big project led by scientists at NIH, involving thousands of people. Francis Collins later became the NIH director. Scientists mapped 20,000+ human genes during the project.
Over the course of the project, chemists were smart to recognize and plan as well 20,000 human genes. The HGP again aided to recognize the historical mutations that help differing afflictions, containing cystic fibrosis, Huntington's affliction, and feelings malignancy. By recognizing these historical differences, physicists have happened capable to cultivate new situations and medicines for these environments.
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