Answer:
D. Oxygen
Explanation:
Oxygen is the most abundant element in the bodies of living things. For example, oxygen is about 65% of the human body.
Oxygen in the body is mostly in the form of water, since the bodies of living things need water to create energy needed every day.
In this diagram of a long bone, which type of bone marking is circled? anatomical drawing of a long bone showing the internal structure, with the point where the blood vessels enter circled
Answer:
the answer is osteon
Explanation:
HELP ASAP PLEASE HELP FAST
Answer:
D. the basic unit of life
Explanation:
A cell is the basic fundamental unit of a living organism. There two types of organisms which are unicellular (single cell organism) and multicellular (many cells)
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:
0%
Explanation:
This question involves a gene coding for height in pea plants. The allele for tallness (T) is dominant over the allele for shortness (t). This means that allele T will be expressed over allele t in an heterozygous state.
A purebred tall plant will possess genotype: TT while a heterozygous tall plant will possess genotype: Tt. The two parents will produce the following gametes:
TT- T and T
Tt- T and t
Using these gametes in a punnet square (see attached image), the following offsprings with genotypes: TT and Tt in a ratio 1:1 will be produced.
TT offsprings are purebreed tall while Tt offsprings are heterozygous tall. Hence, based on the question, no offsprings of this cross will possess the recessive genotype (tt). This means that 0% of the offsprings of this cross will be short.
What are the features of Nuclear Receptor Screening Services?
Nuclear receptors (NRs) are a class of proteins responsible for thyroid hormones and sensing steroid and certain other molecules. There are 48 genes in the human genome that code for the NRs. As the largest family of eukaryotic transcription factors, NRs control numerous processes involved in development, cell cycle, and important metabolic pathways.
https://dda.creative-bioarray.com/nuclear-receptor-screening.html
Answer:
The nuclear receptor superfamily comprises a large group of transcription factors that play a key regulatory role in development and homeostasis of multicellular organisms. A special feature of nuclear receptors is their ability to bind to condensed chromatin templates, which makes them important initiators of gene transcription. Moreover, the ability of nuclear receptors to sequentially recruit a variety of transcription factors and coregulators to target promoters and to orchestrate the whole process of gene transcription confirms their biological significance and stimulates intensive research and a high level of scientific interest in this field. In this review, we summarise current knowledge regarding the structure and function of nuclear receptors as principal regulators of gene expression. Emphasis is given to the molecular mechanisms of nuclear receptor-mediated transcriptional activation and repression including recent progress made in this area.
Fermentation converts organic material into a fuel whose primary component is
A. methane
B. sugar
C. yeast
D. alcohol
Answer:
D. alcohol
Explanation:
Fermentation by microorganisms converts organic materials into alcohol
what is haemopoiesis??
Haemopoiesis is from greek meaning “ to make. new blood” •
Explanation:
It refers to the formation of blood cellular. components.
While not an official step in the process, Acetyl CoA formation is a transition between glycolysis and krebs, and occurs only in the presence of oxygen.
A. True
B. False
Answer:
A. True
Explanation:
Acetyl coA is a compound which helps in the transport and availability of the acetyl group to the citric acid cycle (Krebs cycle).
Though it is not an official step in the process, Acetyl CoA formation is a transition between glycolysis and krebs, and occurs only in the presence of oxygen.
The Acetyl group helps in the oxidation for energy production.
What are the four main components of soil?
A. Water
B. Sediment
C. Air
D. Humus
E. Bedrock
Answer:
The four (4) components of soil are;
WaterAirMineralsorganic¿Qué características crees que deben tener los métodos anticonceptivos para ser eficaces en el control de la natalidad?
Answer:
Los métodos anticonceptivos para ser eficaces en el control de la natalidad deben ser seguros para las personas que los utilizan, es decir, no tener efectos adversos, o al menos, la menor cantidad posible. Ser reversibles, es decir, que no generen infertilidad de por vida. Deben ser además fáciles de usar, cómodos y fáciles de conseguir, esto es, económicos. Además deben ser eficientes en su objetivo de evitar la concepción, que es para lo que fueron creados.
Explanation:
La tasa de eficacia de un método anticonceptivo se mide por la cantidad de embarazos que se producen cada 100. Esta parece ser la característica más importante, y si bien es una de las más importantes, hay que tener en cuenta todos los otros factores mencionados, es decir que la gente pueda acceder a ellos, sean cómodos, seguros y a un precio coherente.
Is talc renewable???
Answer: Talc is a nonrenewable resource.
Explanation:
Talc is a mineral, which counts as a nonrenewable resource. Minerals, fossil fuels, and ores are nonrenewable resources because they do not regenerate or renew themselves at a quick enough rate and often take long periods of time to replenish.
Role of mitosis in plant growth and development
Answer:
The mitosis cell division helps in vegetative growth in plants. It also helps in healing of wounds
Nucleus is where dna is stored and where rna transcription occurs
Answer:
Yes that is correct
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
Note the mechanism of absorption (passive or active transport) of the following food breakdown products, and indicate by a check mark whether the absorption would result in their movement into the blood capillaries of the lymphatic capillaries (lacteals).
Substance Mechanism of absorption Blood Lymph
Monosaccharides
Fatty acids and glycerol
Amino acids
Water
Ca2+ , Na+ , Ca2+
Answer:
The absorption of monosaccharides takes place via facilitated and cotransport mechanism and is transported through blood capillaries present in the villi.
The absorption of glycerol and fatty acids takes place through diffusion and the majority of them are transported via lymph capillaries, while some are transported through blood capillaries.
The absorption of amino acids takes place via a cotransport mechanism with sodium ions and is transported through blood capillaries.
The absorption of water takes place by the process of osmosis via diffusion and is transported with the help of blood capillaries.
The absorption of calcium and sodium ions takes place via an active transport mechanism, while the absorption of chlorine takes place via the process of diffusion. The transportation of all these ions takes place with the help of blood capillaries.
PLEASE ANSWER QUICK!The Greek roots of the word prokaryote mean “before nucleus.” Describe the way that DNA is organized in prokaryotic cells without the help of a nucleus. How does this approach differ from the way that eukaryotic cells organizes their DNA
Answer:
Prokaryotic cells' DNA are located in the cytoplasm of the cell rather than in the nucleus, like in eukaryotic cells. DNA aids in protein synthesis and determines functions of the cell in cells, regardless of being within the membrane of a nucleus or not.
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I hope this helps a little.
The word prokaryote in Greek means before kernel (nucleus). Unlike the eukaryotic cells, the nuclear material is located in the cytoplasm of the cell in a nucleoid.
What are the characteristics of prokaryotic cells?The prokaryotic cells are the primitive karyons that are defined by the lack of the true nucleus and organelles. Unlike the eukaryotes, the organelles lack the membrane that covers them but has a tough cell wall.
The prokaryotes include archaea and bacteria which are unicellular and microscopic organisms that are simple and have their genetic material organized into nucleoids in the center of the cell. They have the ability to live in harsh conditions.
Therefore, the eukaryotes and prokaryotes differ in the arrangement of the genetic material.
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Which of the following describes the most likely impact that exposure to pollutants in the atmosphere would have on one’s personal health?
Answer:
It will lead to upper respiratory infections and pneumonia.
Explanation:
Watch the video and answer the question in 2 or 3 paragraphs. What are the main causes of air pollution? Discuss some effects of air pollution and provide possible solutions to this environmental issue. https://youtu.be/_dTtvtlct9k
Answer:
Explanation:
Exhaust from vehicles, factories, volcanoes and wildfires are the major causes of air pollution.
Effects of air pollution include many diseases. For example heart disease, lung disease, respiratory diseases etc... They also damage nerves, brain, kidney, lungs, heart etc...
Solution to air pollution:
We should replace the fossil fuels with natural energies like solar, wind, geothermal, hydro etc...
Waste water treatment
Dust control
Hope this helps
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The valve present between right atrium and ventricle is
explain the scientific theory of evolution and the evidence scientist found that proves it to be valid ?
Answer:
There are many theories
But main evidences are
Fossils of Plants animals and other organisms who lived on this planet at sometime. Oprain And Halden theory which suggests the chemical and biological evolution.Tissue found in leaf stalks below the epidermis is________.
Explaination
Collenchyma is the tissue found below the epidermis of a leaf stalk. These are simple permanent tissue, perform photosynthesis when chloroplast is present.
Which statement best describes a hypothesis?
Answer:
A hypothesis is an idea or explanation you can test through study and experimentation.
Explanation:
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Answer:
D. Their ability to replicate very fast and natural mutations combine to allow them to evolve rapidly
Explanation:
They develop mutations in their DNA that can be resistant to the drugs given.
g why did you digest the plasmid with XhoI? What is your expectation if the plasmid is a closed vector without an insert? What is your expectation if your plasmid has an insert (how many bands, what are their sizes)?
Answer:
I digest the plasmid with Xholin so as to find out the total size of the plasmid after insertion of the gene of interest and whether or not the insert went into the vector. Ao the plasmid is a reclosed vector without an put we expect to see one band. However, if the plasmid has an insert, we expect to see two bands on the gel
A 23 year old female presents with bilateral loss of central vision. You obtain a detailed family history and produced the associated pedigree. Which of the following transmission patterns is most consistent with this patient's family history?
a. Autosomal dominant
b. Autosomal recessive
c. X-linked dominant
d. X-linked recessive
e. Mitochondrial
Answer:
e. Mitochondrial
Explanation:
This is a case of mitochondrial heredity so correct answer is e. Mitochondrial In this case, the mother transfers her mitochondria to her children. So the child is affected. The male does not pass this trait on to his children. We can see from the lineage that the mother has defective mitochondria so all their children will have defective mitochondria. One of his sons has mitochondrial, but is unable to pass it on to the next generation.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.
Breast milk is naturally formulated to meet the nutritional needs of a growing infant.
a. True
b. False
Biochemical and genetic experiments have demonstrated that the _________ of tRNA are important for recognition by its cognate aminotransferase-tRNA synthetase.
Answer: Acceptor stem and anticodon loop.
Explanation:
Transfer RNA (tRNA) is a small RNA nucleic acid involved in protein synthesis (translation). Each tRNA molecule has two important areas:
A region of trinucleotides, called the anticodon A region where a specific amino acid binds.During translation, the ribosome reads the sequence of the mRNA in groups of three bases to assemble the protein. So, in the mRNA chain there are codons, set of three bases, which determine the amino acid to be added to the peptide chain. The tRNA transfers the amino acid to the ribosomes, and then arranges them along the messenger RNA (mRNA) molecule. Then, the tRNA must have an anticodon that is complementary to the codon. Each type of tRNA is specifically combined with 1 of the 20 amino acids to be incorporated into proteins.
This means, during translation, each time an amino acid is added to the growing chain, a tRNA molecule is formed whose base pairs have a complementary sequence with mRNA molecule, ensuring that the appropriate amino acid is inserted into the protein. So, tRNA is a key link between RNA transcription and the translation of that RNA into protein. On the other hand, aminotransferases are enzymes responsible for attaching amino acids to the 3ʹ‐end of cognate tRNAs.
The acceptor stem is the site of attachment of amino acids to tRNA, and anticodon loop is the site of tRNA that is complementary to the codons found in mRNA (that determine the amino acid that will be added) This means, both parts are important for recognition, because the acceptor stem is where the amino acid is, and the anticodon loop ensures that the appropriate amino acid is inserted into the protein.
the oceanic zone is
Answer:
The oceanic zone is the region of open sea beyond Continental shelf.
Which of the following statements is compatible with Darwin’s theory of evolution? much of the variation in individuals of a population is heritable resources are unlimited survival depends mostly on luck members of a population are usually very much alike in their characteristics
Answer:
A. much of the variation in individuals of a population is heritable
Explanation:
Charles Darwin is a scientist well known for his development of the concept of evolution. Darwin was able to deduce that the species of a population have a common ancestor and that as time passes there is a modification in the characteristics of these species that makes it easier for them to adapt to their environments. These traits that were inherited from past descendants, were helpful traits that made it possible for their own descendants to be able to survive better than their peers. He noted that there were limited resources for survival, therefore it was a case of the survival of the fittest.
Natural selection was the term used by him to describe how species of a population survived over time. Members of a population had varied traits that enabled them to survive. For example, the descendants of an ancestor could have slightly different shapes, colors, sizes, and other features but still maintain traceable heritable traits.
Please answer the following Anthropology questions fully, thoroughly, and in depth.
Part i. What adaptations are associated with an insectivory diet? Give an example of a primate that has this diet.
Part ii. What adaptations are associated with a gummivory diet? Give an example of a primate that has this diet.
Part iii. What adaptations are associated with a frugivory diet? Give an example of a primate that has this diet.
Part iv. What adaptations are associated with a folivory diet? Give an example of a primate that has this diet.
Part v. What adaptations are associated with vertical clinging and leaping? Give an example of a primate that practices this form of locomotion.
The examples that you need are 1-Gummivory: marmoset, tamarins; 2-Frugivores: spider monkey, ray-bellied night monkey (owl monkey); 3-Folivory: howler monkey, leaf monkey, colobine; and 4-Vertical clinging and leaping: lemurs and tarsiers
A gummivore is an animal (in this case, a primate) with a feeding strategy that depends on the sap or gum from trees. It is a type of diet that consists primarily of exudates. Some adaptations observed in these species include: 1-specialized anterior teeth in order to stab bark; 2-well-developed claws for clinging to trees
Frugivores primates have a feeding strategy mainly based on raw fruits. Some adaptations observed in these species include 1-low rounded molar cusps and 2-broad incisors.
Folivore primates (also called leaf-eating monkeys) have a feeding strategy mainly based on leaves. Some adaptations observed in these species include 1-narrow incisors, 2-broad molars (high shearing crests), 3-thin enamel molars.
Vertical clinging and leaping is an arboreal locomotor pattern practiced by lemurs and tarsiers. These animals push off from one vertical support with their hindlimbs to land on another vertical support. Some adaptations observed in these species include 1-hips, knees, feet, arms, and tails that facilitate climbing and leaping, 2-large hip extensor muscles and 3-specialized minor muscles.
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