Answer:
I would go with A maybe,am not really sure though.
The original name for Asian Homo erectus was Group of answer choices Homo rudolfensis. Australopithecus. Paranthropus. Pithecanthropus.
Answer: Pithecanthropus erectus.
Explanation:
Between 1891 and 1892 Eugène Dubois believed he had found the "missing link", hypothesized by Ernst Haeckel, when he discovered some loose teeth, a skull cap and a femur - very similar to that of modern man - in the excavations he was carrying out in Trinil, located on the island of Java, Indonesia. Homo erectus erectus was the first specimen of Homo erectus to be discovered. Dubois first named it Anthropopithecus erectus and then renamed it Pithecanthropus erectus. The name Homo erectus means in Latin "erect man", wich means, "standing man", whereas Pithecantropus erectus means "standing ape-man".
So, Dubois published these findings as Pithecanthropus erectus in 1894, more popularly known as "Java Man" or "Trinil Man". In the 1930s the German palaeontologist Ralpf von Koenigswald obtained new fossils, both from Trinil and from new locations such as Sangiran and in 1938 von Koenigswald identified a magnificent Sangiran skull as "Pithecanthropus". It was not until 1940 that Mayr attributed all these remains to the genus Homo (Homo erectus erectus).
Could an object, tool, or structure be made without the use of a natural resource? Explain your answer.
What does ingestion and degradation of extracellular antigens and their subsequent presentation by MHC class II molecules lead to
Answer:
"Activation of CD4+ helper T cells" is the correct choice.
Explanation:
The major complicated of histocompatibility II seems to be the chemical compound for something like the T cells that presets antibodies. Antigens collected from those in the pathogens become transformed as well as inserted into a large histocompatibility system that's also distributed on either the cellular cell surface membrane representing antigen as well as the trigger is perceived by complicated systems of T cells as well as MHC II but rather contained throughout CD4 + support T cells.A/An ______________________________ is a specialized radiographic examination of the uterus and fallopian tubes. Question 94 options:
Answer:
Hysterosalpinography
Explanation:
It is Hysterosalpingography because it is an x-ray examination of uterus and fallopian tubes that uses a specific type of x-ray called fluoroscopy and a radiopaque material. It is use to examine women who find it difficult in getting pregnant to examine the shape of the uterus.
PLz answer my scientific question
Answer:
the independent variable is time.the independent variable is time in the second graph because time is the only independent variable.Explanation:
hope this will help :)
Answer:
I would have to say D but I might be wrong.
Explanation:
A decrease in muscular activity or damage to neurons that attach to skeletal muscle can lead to a reduction in the size of muscle called
Answer:
The correct answer is atrophy
A purebred tall pea plant is cross-pollinated with a tall, can you please help
heterozygous pea plant. Use a Punnett square to determine the probability the offspring inherita
recessive short allele. (I point)
75%
25%
0%
50%
Answer:
50%
Explanation:
because if you use the box method you see that half receive a short alele
T t
T TT Tt
T TT Tt
During cellular movement which filaments will be the ones that are responsible for attaching and pulling the other filaments along?
Answer:
Myosin
Explanation:
Myosin are the major components of both muscle and non muscle cells in striated muscles which is responsible for contraction or movement that convert chemical energy (ATP) into mechanical work thereby generating force and movement.
During cellular movement , myosin filament will be the ones that are responsible for attaching and pulling the other filaments along because Myosin binds to actin at a binding site on the globular actin protein,it also have ATP binding sites which hydrolyses ADP to ATP , ATP binding causes myosin to release actin, which make actin and myosin to detach from each other thereby causing attachment and contraction of other filaments
Circle
1
2
Circle the most reactive metal in each set.
1) Magnesium / Potassium
2) Aluminum / Gold
3) Cobalt / Cesium / Calcium
4) Iron / Titanium / Potassium
5) Francium / Lithium / Beryllium
لی
Answer:
Explanation:
1) potassium
2)Aluminium
3) Cesium
4) potassium
5) Beryllium
For this task, you will imagine that you are a reporter for a scientific magazine. Your task is to explain the process of protein synthesis to someone who does NOT have a science background. Therefore, the explanation needs to be in simple enough terms for anyone to understand. You will organize your article in the following way: Structure and Function of DNA and RNA Transcription and RNA processing Translation Protein modification (general) You must also include the following terms: Double helix Helicase Codon Polymerase 5’ cap Poly (A) tail Introns Exons Splicesomes rRNA, tRNA, Mrna Ribosomes Anticodons E site, P site, A site Initiation Elongation Termination
Answer:
Please find the explanation below
Explanation:
Our nucleus is made up of a genetic material called deoxyribonucleic acid (DNA), which is a double-helical structure that stores the genetic information needed for the optimal functioning of any organism. DNA, alongside RNA are nucleic acids that are composed of NUCLEOTIDES subunits. The nucleotide consists of a pentose sugar (deoxyribose in DNA and ribose in RNA), nitrogenous base and a phosphate group.
However, the genetic information stored in the DNA molecule needs to be expressed in order to form useful products (proteins). This genetic expression is done in two stages viz: transcription and translation. Transcription, which is catalyzed by an enzyme called RNA polymerase is the process whereby the information stored in the DNA is used to synthesize a mRNA molecule. However, this mRNA molecule is considered pre-mature until it is processed. RNA processing occurs in three stages viz: 5' capping, polyadenylation, and splicing.
5' capping involves adding a 5' cap to the marks molecule. Polyadenylation involves adding a poly(A) tail to the mRNA molecule while splicing is the removal of introns (non coding regions) with the aid of Spliceosomes and joining of the exons (coding region). After processing, the mRNA becomes matured and ready to be translated.
Translation is the process whereby the mRNA transcript is used to synthesize a protein molecule. It occurs in the ribosomes (organelles for protein synthesis, a complex of rRNA and proteins) where the mRNA is read in a group of three nucleotides called CODON. The reading is done by the Anticodon of a transfer RNA (tRNA), which is complementary to the codon.
Translation occurs in three stages: initiation, elongation and termination. The mRNA attaches to the P site of the ribosomes (initiation) where it is attached to by a tRNA's anticodon complementary to it. The anticodon carries the amino acid corresponding the codon and shifts to the A-site. The addition of amino acid to the polypeptide chain continues (elongation) until a stop codon is encountered, which signals the end of the translation process i.e. termination. This causes the polypeptide (protein) to be released from the E-site.
The synthesized protein undergoes packaging and modification in the Golgi apparatus.
Some people use the terms "cell cycle" and "mitosis" as if they were the same thing. Which of these best describes why they are not the same?
a. the cell cycle is just a stage of mitosis
b. mitosis and the cell cycle are different types of cell division that occur in different of cells in a multicellular organism
c. mitosis is a sequence of events within the cell cycle that separates the genetic material
d. mitosis is a type of cell division that occurs in eukaryots and the cell cycle takes place in prokaryots
Answer: C. Cells in meiosis have unique genetic information
Explanation: Meiosis is the cell division that forms four daughter cells from one parent cell. It includes two successive divisions called as meiosis I and meiosis II. Crossing over during prophase I of meiosis I imparts new gene combinations to the daughter cells of meiosis. Hence, daughter cells formed by the end of meiosis have some new gene combinations, that is, unique genetic information.
(06.03 LC) Which of the following is an example of how HIV can be transmitted from one person to another? Contact between infected blood and a mucus membrane Contact between infected saliva and an open wound Contact between infected blood and skin Contact between infected saliva and lining of mouth
Answer:
i believe the answer is contact between a infected blood and a mucus membrane
Explanation:
What are two theories proposed to explain Primate Origins and how do they relate to traits we see in modern primates
Answer:
The level of specialization is given by the function.
The function of the organ makes the cell as it becomes mature, its more specific function becomes.
First, a stem cell differentiates itself into a hepatic or neuronal cell, thus generating that its functions are limited compared to the rest of the stem cells.
After this process, throughout the cell development and maturation of the organ, its cells become more specific in a single activity.
Explanation:
The differentiation of the stem cells is regulated by the genetic code, the migration of these during the embryonic stage of the mouse and the external stimulation that these stem cells perceive.
basic question for fun . HOW TO FIGHT WITH TYPES OF VIRUS IN THE WORLD AND HOW GET RID FROM IT EXAMPLES: CORONA VIRUS, INFLUENZA, TYPHOID, HEPHATITISAND CHLOREA?
Answer:
Boost immunity power( follow these):
1. Have enougn sleep.
2. Be happy...overcome stress.
3.Do exercises frequently.
4. Have regular check ups ;these may help to diagnosis on early stage.
5. Consume the foods which boosts immunity.
6. Be mentally strong.
Explanation:
There isn't proved ideas ...however, we can fight by making our immune system strong...
HIT LIKE
Use the following scenario to answer the next following question(s):
You and your friends go to the beach for vacation. You all walk down to the beach to go swimming. When you get there, you see the water is murky and green, and there are algae blooms floating on top.
Reference: Ref 6-3
If it is excess nutrients which are feeding the algae blooms and lowering the oxygen content in the water, that process is called:
A. nutrient cycling.
B. nitrification.
C. eutrophication.
D. hypoxia.
Answer:
eutrophication
Explanation:
Eutrophication refers to a situation in which the aquatic environment becomes excessively enriched with nutrients. This leads to algal blooms in aquatic habitats such as lakes. These nutrients come from Fertilisers used in farming, which find their way into water bodies through run-off thereby increasing nutrient levels.
Excess nutrients causes phytoplankton to grow and reproduce at an alarming rate resulting in algal blooms. This bloom disrupts the balance in the ecosystem leading to many problems.
The algae may end up using all the oxygen in the water, causing oxygen shortage for aquatic life. Some of the algae may die, their decay may lead to further oxygen depletion. As oxygen is depleted, aquatic organisms may also begin to die.
PLS HELP ME OUT WITH THIS ANSWER........ WHAT IS DINOFLAGELLATES ?
Answer:
These are single-celled MESOKARYOTES constituting the phylum DINOFLAGELLATA
Explanation:
Their populations are distributed depending on sea surface temperature,e.t.c.
Hope it helps.
• They are basically unicellular, motile, biflagellate and photosynthetic protists. Predominate colour is golden brown but yellow, green, red and even blue also exists.Some Dinoflagellates like Gymnodinium and Gonyaulax grow in large number in the sea and make the water look red and cause the so called “red tide”.
Which structure is found in the cytoplasm of a prokaryotic cell, but is not found in the cytoplasm of a eukaryotic cell?
DNA
ribosomes
nucleus
mitochondria
Answer:
DNA
Explanation:
cytoplasm of a prokaryotic cell, but is not found in the cytoplasm of a eukaryotic cell.
Answer:
A- DNA
Explanation:
"Acidic" is an appropriate description for four of the following. Which one is the exception?
A). Soup and ammonia
B). HCI
C). Excess hydrogen ions
D). The contents of the stomach
E). A pH less then 7
Answer:
E) A pH less than 7
Explanation:
if a solution has a higher concentration of hydronium ions than pure water , it has a pH lower than 7
Answer: e
Explanation:
I did the test
identify the components ( parts) of DNA
Answer:
Phosphorus group, a sugar(like deoxyribose or ribose), and a nitrogenous base
Nucleus is where dna is stored and where rna transcription occurs
Answer:
Yes that is correct
Explanation:
The ____ ring is built onto ribose-5-phosphate of PRPP for its de-novo nucleotide biosynthesis, while the ring structure of the ______ bases are synthesized separately and then coupled to ribose-5-phosphate via the C-N glycosidic bond.
Answer:
The purine ring is built onto ribose-5-phosphate of PRPP for its de-novo nucleotide biosynthesis, while the ring structure of the pyrimidine bases are synthesized separately and then coupled to ribose-5-phosphate via the C-N glycosidic bond.
Explanation:
In the de novo synthesis of nucleotides, their metabolic precursors such as aminoacids, ribose-5-phosphate, CO₂ and NH₃ are used as starting materials.
In purine nucleotide synthesis, the ring structure is built up on ribose-5-phosphate of PRPP by addition of one or a few atoms one at a time starting with the amino group donated by glutamine until the first intermediate inosinate is synthesized.
In pyrimidine ring synthesis, orotate is first synthesized from carbamoyl phosphate and aspartate, and then is attached to ribose-5-phosphate of PRPP, before it is then converted to the common pyrimidine nucleotides starting from uridylate.
why does mosquito net easily form lather with hard water
Answer:
Hard water is defined as the type of water that does not readily give a good lather with soap.
Mosquito net do not form easily lather with hard water because hard water is made up of magnesium and calcium ions which precipitate the soap and unable to form lather.
The end result of a chemical reaction is always:
A. the formation of new kinds of elements.
B. the production of water molecules.
c. a substance that was not ope of the reactants.
ak
s
D. a molecule that does not have an electric charge.
Answer:
The formation of new kinds of elements.
When a chemical reaction is finished, a new type of element is made. because the chemicals couldn't react to eachother without making something new.
Which location is least likely to experience a volcanic eruption? Α. an island hot spot, such as the island of Hawaii B. Hamilton County on the plains of central Texas с. a convergent boundary, as in the Ring of Fire D a volcanic island arc, such as the Aleutian Arc in Alaska
Answer:
i think that the answer is B. Hamilton County on the plains of central Texas i took the test
Explanation:
Hamilton County on the plains of central Texas is least likely to experience a volcanic eruption. Therefore, option (B) is correct.
What are volcanoes?Molten rock and gases stored under the surface erupt through a volcano, generating a hill or mountain.
Active, inactive, or extinct volcanoes. Active volcanoes are likely to erupt again. Dormant volcanoes may erupt again. Extinct volcanoes won't erupt.Magma collects inside active volcanoes. The magma chamber's pressure forces it through rock channels and onto the planet's surface.
Volcanic eruptions can be violent or slow-moving. Volcanoes erupt through vents on the sides or a primary entrance at the top. The volcano's morphology depends on eruption rate and magma chemistry. Land and sea volcanoes exist. As lava cools and hardens, underwater volcanoes build mountains and ranges. When volcanoes rise above the ocean, they create islands.
Learn more about volcano, here:
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When the body cells are hypotonic to the blood plasma, water will move from intracellular fluid to extracellular fluid.
A. True
B. False
Answer:
True
Explanation:
What are the biological methods of controlling mosquito population?
Explanation:
Biological control or "biocontrol" is the use of natural enemies to manage mosquito populations. ... Effective biocontrol agents include predatory fish that feed on mosquito larvae
You continue this approach by designing two separate experiments. In the first experiment, you use glucose where "Coccupies position 1. In the second experiments, 14C occupies position 2 and 6 in the glucose molecules. For each experiment, you use 0.2 moles of radiolabeled glucose and you assume that all the pyruvate formed is converted to acetyl-CoA. What following statements are correct? (select 3)
a) When glucose is labeled on carbon #1, 0.2 mole of acetyl-CoA is radiolabeled
b) When glucose is labeled on carbon #1, 0.1 mole of acetyl-CoA is radiolabeled
c) When glucose is labeled on carbons #2 and #6, 0.2 mole of acetyl-CoA is radiolabeled
d) When glucose is labeled on carbons #2 and 6, 0.1 mole of acetyl-CoA is radiolabeled
e) When glucose is labeled on carbons #2 and 6, the carbonyl groups of half of the acetyl-CoA molecules are radiolabeled
f) When glucose is labeled on carbons #2 and 6, the methyl groups of half of the acetyl-CoA molecules are radiolabeled
Answer:
b) When glucose is labeled on carbon #1, 0.1 mole of acetyl-CoA is radiolabeled
c) When glucose is labeled on carbons #2 and #6, 0.2 mole of acetyl-CoA is radiolabeled
e) When glucose is labeled on carbons #2 and 6, the carbonyl groups of half of the acetyl-CoA molecules are radiolabeled
f) When glucose is labeled on carbons #2 and 6, the methyl groups of half of the acetyl-CoA molecules are radiolabeled
Explanation:
When glucose undergoes glycolysis, it is converted into two molecules of pyruvate with carbon 1 to 3 forming the first molecule of pyruvate and carbon 4 to 6 the second molecule of pyruvate. The C-1 and C-6 of the glucose molecule becomes the methyl groups of each of the two molecules of pyruvate. The C-2 and C-5 of the glucose molecule forms the carbonyl carbon of each of the two pyruvate molecules. Each of the two pyruvate molecules undergoes further oxidation to yield acetyl-CoA with the carbonyl and methyl groups of pyruvate retained in the acetyl-CoA molecules.
Thus when 0.2 moles of glucose are labelled at C-1 and then C-2 and C-6 in each of the two experiments the following results are obtained:
When glucose is labeled on carbon-1, 0.1 mole of acetyl-CoA is radiolabeled since half of the two pyruvate molecules are obtained from C-1
When glucose is labeled on carbon-2 and carbon-6, 0.2 mole of acetyl-CoA is radiolabeled since the C-2 and C-6 of the glucose molecules forms a part of one of each of the two pyruvate molecules.
When glucose is labeled on carbons #2 and 6, the carbonyl groups of half of the acetyl-CoA molecules are radiolabeled since one of the two carbonyl groups of the two pyruvate molecules is formed from C-2 of glucose.
When glucose is labeled on carbon-2 and carbon-6, the methyl groups of half of the acetyl-CoA molecules are radiolabeled since one of the two methyl groups of the two pyruvate molecules is formed from C-6 of glucose.
use the numbers 12345 to place the protien creations steps below in the correct order
Use the numbers 1, 2, 3, 4, and 5 to place the protein creation steps below in the correct order. Ribosome attaches to the mRNA. Information is transcribed in DNA to mRNA. tRNA anticodon carries an amino acid that compliments the mRNA codon. mRNA leaves the nucleus. The chain of amino acids forms a protein.
if it is helpfull please mark as brainlist
Answer:
please list numbers 12345 and then i will
Explanation:
Place the respiratory structures into the order that air would pass through them during a normal inspiration.
Outside of Body
1. nares
2. vestibule
3. nasal cavity
4. choanae
5. nasopharynx
6. oropharynx
7. laryngopharynx
8. larynx
Inside of Body
Answer:
Outside of the body, nares, vestibule, nasal cavity, choanae, nasopharynx, oropharynx, laryngopharynx, and larynx.
Explanation:
During inspiration, the air enters through the nares in the nose. From here, it goes up passing through the vestibule, the nasal cavity, and the choanae to start descending through the pharynx. The pharynx has three parts: the nasopharynx, which is in contact with the nose; the oropharynx that is in contact with the mouth; and the laryngopharynx, which is in contact with the larynx and esophagus. Between the oropharynx and laryngopharynx is the epiglottis, this is cartilage that prevents food from going to the lungs when we swallow food. After passing through the pharynx, the air flows through the larynx. From here, the air goes to the trachea, the bronchi, which enters the lungs giving the bronchioles, and these divide giving the alveolus, which is the place where the exchange of Oxygen and Carbon dioxide occurs.
Placing the respiratory structures in the correct order :
Outside of Body --> nares --> vestibule --> nasal cavity --> choanae --> nasopharynx --> oropharynx --> laryngopharynx --> larynx -- > inside of body
During respiration the air enters from outside of the body into the body through the nares down to the vestibule, the nasal cavity which will descend down through to the pharynx which is further divided into three ( 3 ) parts which are; nasopharynx, oropharynx, and laryngopharynx. The epiglottis is located between oropharynx and laryngopharynx.
Hence we can conclude that Placing the respiratory structures in the correct order ; Outside of Body --> nares --> vestibule --> nasal cavity --> choanae --> nasopharynx --> oropharynx --> laryngopharynx --> larynx
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A purebred tall pea plant is cross-pollinated with a tall, heterozygous pea plant. Use a Punnett square to determine the probability the offspring inherita
recessive short allele. (I point)
75%
25%
0%
50%
Answer:
The correct answer is - 25%.
Explanation:
A cross pollination between purebred tall pea plant and a heterozygous pea plant occurred as per the given condition. TT is the allele representation of the purebred plant and heterozygous plant is represented by Tt, so the gametes will be formed would be - T & T and T & t allele.
The Punnett square of this cross is attached with the answer,
So the probability of short allele would be:
= (2/8) *100 , (where 2 is the number of short allele in offspring where 8 is total number of alleles)
= (1/4) *100
= 25%
