Answer:
bbbbbbbb its bbbb
Explanation:
Which of the following statements about sensory transduction by hair cells is false? O A. Bending of the cilia toward the longest cilium produces depolarization O B. Bending of the cilia is caused by the sliding of the basilar membrane across the surface of the hair cells O c. Tip links are stretched by cilia deflection directly opening cation-selective ion channels O D. Hair cells are extremely sensitive since very slight movements in the clila cause a change in membrane potential
Sensory transduction by hair cells is an essential process in the human auditory and vestibular systems. Among the statements provided, option B is false. Bending of the cilia is not caused by the sliding of the basilar membrane across the surface of the hair cells. Instead, it is caused by the movement of the basilar membrane and the tectorial membrane, which results in a shearing force that bends the cilia.
In option A, it is true that bending of the cilia towards the longest cilium produces depolarization. This depolarization results in the opening of mechanically-gated ion channels, allowing an influx of cations, and thus increasing the membrane potential.
Option C is also accurate, as tip links, which connect adjacent cilia, are stretched by cilia deflection, directly opening cation-selective ion channels. This process further contributes to the changes in membrane potential.
Lastly, option D correctly states that hair cells are extremely sensitive. Indeed, very slight movements in the cilia can cause a change in membrane potential, allowing hair cells to detect even minute vibrations or sound frequencies. This sensitivity plays a crucial role in our ability to hear and maintain balance.
Thus, Option B is correct.
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how can alternative rna splicing result in different proteins derived from the same initial rna transcription
The alternative RNA splicing results in different proteins derived from the same initial RNA transcription introns/exons being read and chosen differently. Original RNA is spliced differently producing varying mRNA.
Alternative splicing can provide evolutionary flexibility. Single point mutations can result in the occasional exclusion or inclusion of specific exons from transcripts during splicing, allowing new protein isoforms to be generated without loss of the original protein.
Alternative splicing is a cellular process in which exons of the same gene are combined in different combinations to produce different but related mRNA transcripts. These mRNAs can be translated to produce different proteins with different structures and functions all from a single gene. Alternative splicing is a molecular mechanism that alters the pre-mRNA structure prior to translation.
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Question 2
In this part of the experiment, you’ll prepare and test three milk solutions: milk and water, milk and lactase enzyme, and milk and heated lactase enzyme.
Prepare
Use masking tape to label the three beakers: “milk,” “lactase solution,” and “heated lactase solution.”
Measure 60 milliliters of milk in the graduated cylinder (or ¼ cup of in a measuring cup. Pour it into the beaker labeled “milk.”
In the beaker labeled “lactase solution,” add one lactase tablet and 100 milliliters (or 1/2 cup) of cool or room-temperature water. Use the stirrer to dissolve the lactase tablet in the water.
Add 100 milliliters (or 1/2 cup) of milk to the microwaveable container. Dissolve a lactase pill in the container. Put the solution in the microwave and heat it to boiling (about 2 minutes). Use oven mitts to remove the container. Pour the heated lactase solution into the beaker labeled “heated lactase solution” and let it cool.
Stay safe! Be careful while handling the boiled mixture to avoid spilling it on your hands.
Test
While you’re waiting for the lactase solution to cool, read the directions on the test strips. The test strips in the Edmentum lab kit will react to glucose within a few seconds. If you use different strips, the reaction time may vary. Now follow these steps to test the solutions. Record your data in the answer space.
Milk and water solution: Fill the first test tube one fourth full of milk. Fill the small graduated cylinder with water and gently add it to the milk in the test tube until the test tube is half full. Use the stirrer to thoroughly mix the solution. Then insert the test strip for 10 to 20 seconds. Look at the test strip, and record whether it changed color. Wash the stirrer.
Milk and lactase enzyme solution: Fill the second test tube one fourth with milk and one fourth with the lactase solution. Use the stirrer to thoroughly mix the solution. Insert the test strip for 10 to 20 seconds, and record whether it changed color. Wash the stirrer.
Milk and heated lactase enzyme solution: Fill the third test tube with one fourth milk and one fourth of the heated lactase solution. Use the stirrer to mix the solution. Insert the test strip for 10 to 20 seconds, and record whether it changed color.
Note: Keep the lactase and heated lactase solutions for the next part of the experiment.
Wash the “milk” beaker, the test tubes, and the stirrer. If you used paper cups as an alternative, throw them away.
Answer:
The bacterium should stop production of lactase.
Explanation:
This is because the E. coli bacteria can degrade lactose, but lactose is not preferred as source of fuel or energy to glucose. If glucose is present, E. coli would preferably employ it over lactose as Glucose needs little process and minimal energy to degrade when compared to lactose. Although, if lactose is the major sugar that is present, the E. coli will have no option than to employ it as it's source of fuel or energy. The formation of lactase enzyme utilizes energy, which cannot be utilised in the presence of high level glucose.
This appears to be a set of instructions for a lab experiment involving testing different milk solutions with lactase enzyme.
What is lactase enzyme?Lactase is an enzyme that breaks down lactose, a sugar found in milk and dairy products, into glucose and galactose. People who are lactose intolerant have insufficient levels of lactase, which can lead to symptoms such as bloating, gas, and diarrhea when they consume dairy products.
The experiment involves preparing three different solutions (milk and water, milk and lactase enzyme, and milk and heated lactase enzyme) and then testing each solution with a test strip to see if it changes color, indicating the presence of glucose.
The instructions also include safety precautions, such as being careful while handling the heated lactase solution. Finally, the instructions remind the reader to wash all equipment used in the experiment.
Some people may have lactose intolerance, which means that they do not produce enough lactase to break down lactose in their bodies, resulting in discomfort and digestive problems.
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If both enzymes have to function together, what pH range would produce the highest activity?
A: pH 9.0 - 9.5
B: pH 7.5 - 8.5
C: pH 7.0 - 7.5
D: pH 8.5 - 9.5
Answer:
b is the answer
Explanation:
ph 7.5-8.5
Which of the following undermines our ability to apply parsimony to a trait to infer phylogenetic history? Note that multiple answers may be correct. A) An anatomical structure evolves independently on many lineages (e.g. wings or eyes). B) A trait defines a synapomorphic group within a clade. C) A nucleotide position evolves sufficiently rapidly that multiple changes are likely. D) A trait is likely to have changed only once in the relevant time interval
A) An anatomical structure evolves independently on many lineages (e.g. wings or eyes).
C) A nucleotide position evolves sufficiently rapidly that multiple changes are likely.
D) A trait is likely to have changed only once in the relevant time interval.
These factors undermine our ability to apply parsimony to a trait to infer phylogenetic history.
Parsimony is a principle used in phylogenetic analysis that suggests the simplest explanation or the fewest evolutionary changes is the most likely explanation for the observed data.
However, when these factors are present, they introduce complications that make it challenging to apply parsimony:
A) An anatomical structure evolving independently on many lineages means that the trait has evolved convergently multiple times, leading to a complex evolutionary history that cannot be explained by a single simple ancestral change.
C) A nucleotide position evolving rapidly with multiple changes likely means that multiple mutations have occurred at that position independently in different lineages.
This makes it difficult to determine the exact evolutionary history of the trait based solely on parsimony.
D) If a trait is likely to have changed only once in the relevant time interval, it suggests a single evolutionary event.
In such cases, applying parsimony may be straightforward, as there is a clear ancestral state and a single evolutionary change.
However, if the trait has changed multiple times within the time interval of interest, it becomes more complex, and the application of parsimony becomes less reliable.
B) A trait defining a synapomorphic group within a clade does not necessarily undermine the application of parsimony.
Synapomorphies are shared derived traits that help identify monophyletic groups or clades.
They can actually aid in inferring phylogenetic relationships based on parsimony, as they provide evidence of common ancestry and shared evolutionary changes.
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Which of the following events takes place during a single rotation of Earth?
A.
a complete day-night cycle
B.
a complete cycle of four seasons
C.
the passing of 12 hours
D.
a total of 365 revolutions
Answer:
A
Explanation:
Answer is option one, or A
Somoene please help and explain how to read the graph!!
Answer:
Whats going on is that you have a specific amount of enzyme that can do a specific amount of work. Think of it like a hotdog stand. the enzyme would be you and your substrats are your hot dogs and buns. You can only go as fast as the materials you have. If you have more materials than hands you can only go as fast as you can make them. To many substrates and you enzyme rate will cap out at maximum speed.
Explanation:
The plot shows rate of reaction of an enzyme as a function of the concentration of substrate. At stage B, the rate flattens and becomes constant, which means there's too much substrate for the enzyme to handle. So the fourth statement is false - the rate of reaction cannot be increasing.
How does the ozone layer affect life on Earth?
a.
causes respiratory illness
b.
increases photosynthesis in plants
c.
causes damage to living tissue
d.
filters out harmful ultraviolet radiation
(30 points!)
Answer:
(D) filters out harmful ultraviolet radiation
Explanation:
edg 2020
Mark ALL OF the statements that are true for DNA
a
bases held together by hydrogen bonds
b
bases are A, C, G and U
c
unzipped by helicase during replication
d
backbone of sugar molecule ribose and phosphate groups
e
found in the nucleus ofprokaryote cells
f
double helix
g
carries amino acids to ribosomes
h
bases are A, C, G and T
i
single helix
j
backbone of sugar molecule deoxyribose and phosphate groups
DNA is a double-helix molecule composed of two coiled strands. Each strand is formed by a sequence of nucleotides that carry genetic information. Options A, C, F, H, and J are TRUE.
What is DNA?DNA is a nucleic acid composed of two coiled strands, the double helix. This polymer is made of several nucleotides joined by their extremes that store genetic information.
Nucleotides are monomers containing a nitrogenated base, a phosphate group, and a sugar (deoxyribose).
Nitrogenated bases are purines (adenine and guanine) and pyrimidines (cytosine and thymine).
Adenine pair thymine through two hydrogen bonds, Guanine pairs cytosine through three hydrogen bonds.
During its replication, helicase is in charge of breaking hydrogen bonds and separating the two original strands, topoisomerase releases the tension between the strands, and DNA polymerase adds the correct nucleotides.
a) bases held together by hydrogen bonds ⇒ TRUE
b) bases are A, C, G and U ⇒ FALSE, bases are A, T, G, C
c) unzipped by helicase during replication ⇒ TRUE
d) backbone of sugar molecule ribose and phosphate groups ⇒ FALSE, the sugar is deoxyribose.
e) found in the nucleus of prokaryote cells ⇒ FALSE, prokaryote cells have DNA but not a nucleus
f) double helix ⇒ TRUE
g) carries amino acids to ribosomes ⇒ FALSE, tRNA carries amino acids to ribosomes during protein synthesis
h) bases are A, C, G and T ⇒ TRUE
i) single helix ⇒ FALSE, DNA is a double helix molecule.
j) backbone of sugar molecule deoxyribose and phosphate groups ⇒ TRUE
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What ocean did the Roman Empire border at its height
21. Sickle Cell Anemia is condition that shows codominance. The genotype for normal blood cells is NN. Sickle
cell trait is the heterozygous condition (NS) and contains both normal and sickle shaped blood cells. Sickle cell
disease (SS) is when all of the blood cells are sickle-shaped and has lifelong medical implications. A man with
sickle cell trait has a child with a woman with sickle cell trait.
Complete the Punnett Square.
What is the probability of having a child with sickle cell trait?
m. What is the probability of having a normal offspring?
What is the probability of having a child with sickle cell disease?
k.
1.
n.
One of the genetic diseases known as sickle cell disease is sickle cell anemia. Red blood cells, which deliver oxygen to every region of the body, are affected in terms of structure.
What causes sickle cell anemia?Sickle cell is a hereditary condition brought on by a gene deficiency. Only when two genes—one from the mother and one from the father—are inherited together will a person be born with sickle cell disease.
The Punnett square for the cross between a man with sickle cell trait (NS) and a woman with sickle cell trait (NS) is attached below.
The probabilities for each possible offspring are:
25% chance of having a child with normal blood cells (NN)50% chance of having a child with sickle cell trait (NS)25% chance of having a child with sickle cell disease (SS)Therefore, the probability of having a child with sickle cell trait is 50%, the probability of having a normal offspring is 25%, and the probability of having a child with sickle cell disease is 25%.
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Give five example molecules
These are the five examples for the molecules with the clear definition.
The following is a short section of a DNA template before replication:
3’ TACGGCTTAACC 5’
Based on your knowledge of DNA replication, which sequence would be the correct complementary strand synthesized by DNA polymerase?
5’ AUGCCGTTUUGG 3’
5’ GGTTAAGCCGTA 3’
5’ ATGCCGAATTGG 3’
5’ CCAATTCGGCAT 3’
3’ TACGGCTTAACC 5’ based on knowledge of DNA replication, the correct complementary strand synthesized by DNA polymerase is
5’ ATGCCGAATTGG 3’ (option -c) is correct answer.
What is the procedure for DNA replication?There are three main steps in the replication process: priming the template strand, assembly of the new DNA segment, and opening of the double helix and separation of the DNA strands. The DNA double helix's two strands uncoil during separation at a specific spot known as the origin.
Additionally, in vitro DNA replication (DNA amplification) is possible (artificially, outside a cell). Starting DNA synthesis at recognized sequences in a template DNA molecule is possible using artificial DNA primers and cell-isolated DNA polymerases.
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a researcher wanted to create a loss of function mutation for studying type i diabetes in a model animal. what mutation might mimic the effects of type i diabetes?
A loss-of-function mutation is a genetic mutation that makes a gene product less effective or entirely ineffective. In order to analyze the pathological mechanisms underlying type 1 diabetes, scientists have attempted to develop disease models utilizing mutant mice that harbor spontaneous and targeted gene mutations.
A loss-of-function mutation is a genetic mutation that makes a gene product less effective or entirely ineffective. In order to analyze the pathological mechanisms underlying type 1 diabetes, scientists have attempted to develop disease models utilizing mutant mice that harbor spontaneous and targeted gene mutations. The best mutant mice are those that exhibit diabetes, and these mice have been critical to gaining a better understanding of the molecular mechanisms underlying the disease.
To mimic the effects of type 1 diabetes, researchers would likely create a loss-of-function mutation in a gene that is critical for insulin synthesis or insulin release. Insulin is a hormone that regulates blood sugar levels, and people with type 1 diabetes have insufficient insulin production due to the destruction of the insulin-producing beta cells of the pancreas. Mutations in genes such as insulin, glucagon, and somatostatin, which are critical for the regulation of blood sugar, are frequently used to generate diabetic mouse models.
In summary, a loss-of-function mutation in a gene critical for insulin synthesis or insulin release would likely mimic the effects of type 1 diabetes. The exact gene and the nature of the mutation will vary based on the particular research objectives.
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Indicate whether the given structure is located in the outer, middle, or inner ear. Middle : Inner :
Outer :
- Tensor tympani muscle - Cochlea - Basilar membrane - Pinna - Semicircular canals - Saccule - Vestibule - Stapes
The given structures can be classified as follows:
- Middle ear: Tensor tympani muscle, Stapes
- Inner ear: Cochlea, Basilar membrane, Semicircular canals, Saccule, Vestibule
- Outer ear: Pinna
The tensor tympani muscle is located in the middle ear. Its function is to control the tension of the tympanic membrane (eardrum) in response to loud sounds. The stapes, also known as the stirrup, is the smallest bone in the human body and is found in the middle ear. It transmits sound vibrations from the tympanic membrane to the inner ear.
The cochlea, basilar membrane, semicircular canals, saccule, and vestibule are all part of the inner ear. The cochlea is responsible for converting sound vibrations into electrical signals that can be interpreted by the brain.
The basilar membrane is a thin, flexible structure within the cochlea that vibrates in response to different frequencies of sound. The semicircular canals are involved in detecting head movements and maintaining balance. The saccule and vestibule are also involved in balance and orientation.
The pinna, also known as the auricle, is the visible external part of the ear and is considered part of the outer ear. It helps to collect and direct sound waves into the ear canal.
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Describe three adaptations
of exchange surfaces?
Answer:
large surface area, high concentration gradient to the substance and thin barriers to minimize diffusion distance
Explanation:
Match the given symbol or molecular formula to the term that best describes it.
SO2
K
Cl2
C6H6
element
arrowRight
organic compound
arrowRight
inorganic elemental molecule
arrowRight
inorganic compound
The given symbols or molecular formula can be matched with the term that best describes it as follows: SO2 - inorganic elemental molecule KCl - inorganic compoundC6H6 - organic compound.
Inorganic elemental molecules are elements that are composed of the same type of atoms. They are often found in nature as gases. SO2 is a gaseous compound that is composed of sulfur and oxygen. It is an inorganic elemental molecule because it contains only two types of atoms.
Inorganic compounds are substances that do not contain carbon and hydrogen bonds. KCl is an inorganic compound that is composed of potassium and chlorine. It is an inorganic compound because it does not contain any carbon or hydrogen.
Organic compounds are substances that contain carbon and hydrogen bonds. C6H6 is an organic compound that is composed of six carbon atoms and six hydrogen atoms. It is an organic compound because it contains carbon and hydrogen.
In conclusion, SO2 is an inorganic elemental molecule, KCl is an inorganic compound, and C6H6 is an organic compound.
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When plants are closer to sunlight photosynthesis occurs
Answer:
Yeahh
Explanation:
When plants is around the place where the sunlight falls then plants can make their food
HELP pls will mark you the brainliest
Answer:
cell division
Explanation:
I hope this answer finds you well. In an effort to assist your understanding of the topic, I would like to provide you with an explanation. Please let me know if you have any questions or if there is anything else I can do to help.
Cell division can occur through two main processes: mitosis and meiosis. Mitosis is the process by which a single cell divides into two identical daughter cells, each with the same number of chromosomes as the parent cell. This process is important for the growth and repair of tissues in the body. Meiosis, on the other hand, is a specialized form of cell division that occurs in reproductive cells, such as eggs and sperm. During meiosis, the cell divides into four daughter cells, each with half the number of chromosomes as the parent cell. This process is necessary for sexual reproduction and the creation of genetically diverse offspring.
The other options you provided, metaphase, interphase, and G1 phase, are all stages of the cell cycle, which refers to the series of events that occur in a cell leading up to cell division. Metaphase is a stage of mitosis during which the chromosomes line up in the middle of the cell. Interphase is the stage of the cell cycle between cell divisions, during which the cell grows and prepares for division. The G1 phase is the first stage of interphase, during which the cell performs its normal functions and grows in size.
the older adult endocrine system undergoes age-related changes, one of which is prolonged hyperglycemia levels. what statement best describes why this occurs?
The older adult endocrine system undergoes age-related changes, one of which is prolonged hyperglycemia levels just because the pancreatic beta cells release insulin more slowly and inadequately.
Diabetes patients have hyperglycemia, often known as high blood sugar. Patients with diabetes who have hyperglycemia may be affected by a variety of circumstances. They include things like diet and exercise, as well as illnesses and medications that have nothing to do with diabetes. Skipping dosages, not taking enough insulin, or utilizing other blood sugar-lowering drugs insufficiently are additional causes of hyperglycemia. Treatment for hyperglycemia is crucial. If left untreated, hyperglycemia can worsen and lead to major health issues including a diabetic coma, which call for emergency care. Even modest persistent hyperglycemia can lead to issues with the heart, kidneys, eyes, nerves, and neurological system. Blood sugar levels exceeding 180 to 200 mg/dL, or 10 to 11.1 mmol/L, are known as hyperglycemia. However, symptoms often don't appear until the level is high. The signs and symptoms of hyperglycemia progressively appear over the period of days or weeks. The symptoms could be more severe the longer blood sugar levels are elevated.
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Compare and contrast a fossil with a trace fossil. hlep please 13 points
Answer:
see below:
Explanation:
a fossil is usually the animal/insect/whatever, whereas the trace fossil is something like a footprint or excrament! ^.^
Which element is the main component of all organic molecules?
A) Aluminum
B) Carbon
c) Iron
D) Silicon
the human immunodeficiency virus (hiv) is an example of a retrovirus. in order to prevent incorporation of the viral dna into the host genome, antiretroviral drugs could group of answer choices inhibit the viral reverse transcriptase and polyadenine removal. none of these are good antiretroviral strategies. inhibit the viral reverse transcriptase. inhibit polyadenine removal. inhibit retro-rna polymerase.
One way to prevent incorporation of the HIV viral DNA into the host genome is to inhibit the viral reverse transcriptase enzyme.
This is because HIV is a retrovirus, which means it uses reverse transcriptase to convert its RNA genome into DNA that can then be integrated into the host cell's DNA. By inhibiting this enzyme, the virus is unable to complete this process and its DNA cannot be integrated into the host genome.
To explain in more detail, antiretroviral drugs are a class of medications used to treat HIV infection. There are several different types of antiretroviral drugs that work in different ways to inhibit the virus. One common strategy is to target the viral enzymes that are essential for the replication of the virus, such as the reverse transcriptase enzyme.
Reverse transcriptase is an enzyme that converts the viral RNA genome into DNA, which can then be integrated into the host cell's DNA. By inhibiting this enzyme, the virus is unable to complete this process and its DNA cannot be incorporated into the host genome. This prevents the virus from replicating and spreading to other cells.
In summary, inhibiting the viral reverse transcriptase enzyme is a key strategy for preventing the incorporation of HIV viral DNA into the host genome, and is a commonly used approach in antiretroviral therapy.
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Students are asked to pair a plant structure to its physiological function and give evidence for why that structure is important for the process they describe. Anne : The root hairs are used to pull in sugars available in the soil that give the plant energy for growth and reproduction. Jamil : The phloem is needed for cellular respiration as it transports the carbon dioxide to the leaves in order to perform cellular respiration. Shantal : The cambium is important for photosynthesis because it allows water into the plant which is needed as a reactant for photosynthesis. Enrique The guard cells are needed during transpiration because they open and close to regulate the gases that enter and leave the plant Which student correctly described a plant structure and its physiological function? Anne O Jamil Shantal O Enrique
Answer:
Enrique
Explanation:
I literally taking the test rn
All student's answers are correct and well thought out, Each of the structures they mentioned play key role in the physiological process of plants.
Why is everyone's answer correct?Anne correctly identified that root hairs play a role in the plant's uptake of sugars from the soil, which are important for energy and growth. Root hairs are small outgrowths from the surface of the root that increase the root's surface area, allowing for more efficient absorption of water and minerals from the soil.
Jamil correctly identified that the phloem is involved in transport, specifically for sugars produced through photosynthesis in the leaves to other parts of the plant for storage or metabolic use. Phloem is a complex tissue that moves water, mineral nutrients and sugars through the stem and leaves.
Shantal correctly identified that the cambium is a meristematic tissue that allows the plant to grow in width. Cambium is a layer of actively dividing cells between the bark and wood of a stem or root, responsible for secondary growth in woody plants.
Enrique correctly identified that the guard cells are responsible for regulating the gases that enter and leave the plant through the process of transpiration. Guard cells are a special type of epidermal cell surrounding the stomata, the opening that allows for gas exchange and water loss from the plant.
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The color of a mineral in powdered form is the mineral's
Classify the phrases. Does each phrase describe a kinase, a phosphatase, neither, or both?.
Each sentence refers to a kinase or phosphatase that controls the activity of another protein.
A phosphatase is a sort of enzyme, right?The hydrolytic enzymes known as phosphomonoesterases or phosphohatases break the ester link that exists between the phosphate group and the organic residue in organic phosphates. Phosphatases come in two varieties: acid and alkaline, depending on the pH that is best for the action.
Do phosphatases count as enzymes?Dephosphorylation is catalyzed by phosphatases. These enzymes, known as hydrolases, swap out a phosphate group on the substrate for a hydroxyl group from water. Phosphotransferases work in a manner that is completely opposite from that of kinases and phosphorylases.
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When Charles Darwin published his theory in 1858, many people tried to discredit him. The following drawing of Charles Darwin appeared in a magazine, ‘The Hornet’, on 22 March 1871. Why was Darwin represented in this way?
1. Darwin’s theory was considered primitive.
2. Darwin’s theory was about humans descending from apes.
3. Darwin’s theory challenged religious and societal beliefs of his time.
4. Darwin’s theory was about an animal that was half ape and half human.
Answer:
Choice 3.
Explanation:
Choice 1 should be immediately deducted, since his theory was original and never mentioned before (atleast on a widescale), thus it could not be considered “primitive”. Choice 2 sounds correct, but is actually wrong due to his theory never mentioning humans descending from apes, rather sharing a common ancestor. Choice 4 is incorrect because Darwin never mentioned any “hybrid animal” that was half-monkey, half-human. Thus, we are left with choice 3.
EDIT: If per say, I am incorrect, the next best answer is probably 2. Some schools have a different evolution education, so perhaps in one’s school they teach choice 2 as fact?
TWO predator adaptations used to help them capture their prey are *
camouflage
outrunning prey
mimicry
fleeing
Answer:
Under the pressure of natural selection, predators have evolved a variety of physical adaptations for detecting, catching, killing, and digesting prey. These include speed, agility, stealth, sharp senses, claws, teeth, filters, and suitable digestive systems.
Explanation:
hope this helps mah dude.
Is a vacuole prokaryotic or eukaryotic
Answer:
what
Explanation:
4- Las plantas que viven en sitios húmedos y con poco sol tienen raíces poco profundas. Las que viven en sitios de mucho calor tienen raíces muy profundas. ¿Por qué?
Esto es debido a que en los lugares húmedos las plantas absorven el agua con las hojas y no necesitan hacer crecer la raíz para buscar agua. En cambio, las plantas que viven en lugares calurosos necesitan agua por lo que expanden las raíces hacia abajo para poder encontrar líquido e hidratarse.