Answer:
[tex]\boxed{\mathrm{Nucleus}}[/tex]
Explanation:
The nucleus directs all cellular activities. The nucleus also contains genetic material of eukaryotic organisms.
Enzymes are protein molecules which are made up of long chains of
Which of the following properties is the temperature at which a liquid turns to gas? (3 points)
оа
Magnetism
Ob
Thermal conductivity
ос
Melting point
Boiling point
Od
thermal conductivity
Answer:
Boiling point
Explanation:
I did the test
if a short sequence of dna reads 3'TAACGTCCAGGCAAA5', what is the complementary sequence in the other strand of dna g
Answer:
5' ATTGCAGGTCCGTTT 3'
Explanation:
Complimentary strands of DNA run anti-parallel to each other, the ends facing in opposite directions.
The complimentary base pairs for DNA are:
A=T and C=G and when finding the complimentary strand these pairs are only paired with each other.
The levels of nitrogen and ___ in the atmosphere remain fairly constant but the levels of other gases may vary.
Answer:Oxygen
Explanation:
Which is a feature of a tonic receptor? Select one: a. The action potential occurs when there is a change in response to a change in condition. b. For this receptor, the stimulus begins with a burst of action potentials. c. They are normally inactive. d. When the stimulus changes the action potential generation changes. e. Provides information about the rate and change of the stimulus. f. The action potential is generated for a short time period.
Answer:
For this receptor, the stimulus begins with an explosion of action potentials.
that would be the correct option.
Explanation:
A tonic receptor is one that is activated when the action potentials were maintained over time and during the signaling of the receptor.
Tone receptors require continuous stimulation over a period of time to trigger a response and deliver it to the central nervous system.
It keeps the nervous system constantly active in the environment that surrounds it.
They are slowly adaptable, an example of these receptors are the merkel and ruffini receptors.
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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Paleobiologists generally recognize the five most severe mass extinctions during the Phanerozoic Eon. Along with the other students in this class, determine when each of these “Big Five” mass extinctions occurred. Include in the discussion a brief list of which organisms went extinct during each of these events and which organisms radiated afterwards. You should use the information from this unit and from Internet searches to help you in this discussion. NO plagiarism will be permitted.
Extinction of the Ordovician: This extinction was responsible for the disappearance of about 60-70% of the species of oceanic life, This is because in that period most of the life on Earth was in the ocean. This extinction is believed to have been caused by an intense glacial period, although short that froze most of the planet's water. It is believed that the species most affected were sponges, algae, jawless fish, molluscs and cephalopods.
Devonian Extinction: Responsible for the disappearance of about 75% of marine species. This extinction was caused by the depletion of oxygen in the ocean, making life impossible for all aerobic marine organisms. This depletion of oxygen was caused mainly by the variation in the level of the ocean and climatic changes that may have been caused by asteroids.
Permian Extinction: Responsible for the death of 95% of life on earth. This extinction is known as the "mother of all extinctions" due to its destructive character, which devastated life on land and in the ocean. It is believed that this extinction was caused by volcanic activities triggered by the impact of asteroids. Almost all living species on earth were affected, such as insects, trilobites, sharks, boneless fish, reptiles, among others.
Extinction of the Triassic: It is not known exactly what caused this extinction, but it is estimated that it had the power to decimate 70-80% of terrestrial life, mainly arcosaurs and amphibians. The most accepted theory today is that the extinction was caused by recurrent volcanic activities that influenced huge volumes of carbon dioxide in the atmosphere, generating intense heating.
Cretaceous extinction: It was caused by the impact of an asteroid with the earth. The impact had the power to exterminate 75% of terrestrial life. This is the most well-known extinction, since it was responsible for decimating dinosaurs, however, some animals survived this impact, such as most marine species, frogs, birds, reptile and mammals.
Without this, many cycles such as the water cycle and photosynthesis would not exist. What could all these cycles not exist without?
Answer:
This question appears incomplete
Explanation:
This question appears incomplete. However, one similar substance that, if missing, many cycles (particularly the two cycles/processes provided in the question) will not exist/proceed is the sun/sunlight.
In the water cycle for instance, if there is no sun/sunlight, there will be no heat to allow for evaporation of water from the water-body (ocean, sea, stream or lake) hence there will be no cloud of water droplets in the atmosphere. The implication of this is that, the first process of the water cycle will not proceed, hence the cycle will not exist.
During photosynthesis, carbondioxide reacts with water in the presence of sun/sunlight to produce glucose and oxygen. The absence of sun in this reaction will not lead to the production of glucose which is the useful product of photosynthesis for plant.
From the explanation above, it can be deduced that the absence of the sun/sunlight will prevent the two cycles from existing.
whats the biggest mystery about the virus
Explanation:
Hey, there!!
According to what i have learned, the biggest mystery about virus is; It acts like both living and non- living beings.
living characters includes :
It reproduce to make more no. of themselves. It feeds on various substances.non- living characters:
It can be crystallize. It dont respire and feed eith it is out from a host body.Hope it helps...
Evidence without_______
increases its reliability.
a
bias
b
question
с
repetition
d
replication
Answer:
I'll say its B ) Question
Explanation:
because when your proving something with evidence then you start asking questions (or still have questions) makes it seem like you don't really know.
what are biogeochemical ?
What is the wavelength of an earthquake wave if it has a speed of 10 KM/S and a frequency of 5HZ?
Answer:
2nm
Explanation:
(10KM/S)/(5HZ) = 2nm
The most abundant element in the body of living things is: a) Hydrogen b) Nitrogen c) Carbon d) Oxygen
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.
which type of soil is likely to be found in horizon E
Answer:
A layer of pale,Sandy soil lacking clay and iron is likely to be found in horizon E
Answer:
bedrock
Explanation: A layer or bedrock is the type of soil is likely to be found in horizon E. Hence the correct answer is option A among the options. The bed rocks can be regarded very hard and it cannot be breakable.
What cell feature is used by scientists to classify an unknown cell as prokaryotic or eukaryotic?
Answer:
Like a prokaryotic cell, a eukaryotic cell has a plasma membrane, cytoplasm, and ribosomes, but a eukaryotic cell is typically larger than a prokaryotic cell, has a true nucleus (meaning its DNA is surrounded by a membrane), and has other membrane-bound organelles that allow for compartmentalization of functions.
Explanation:
Hope this helps ;)
C. If you combined the elements in the product of this reaction, what type of
reaction would it be? Hint: What is the reverse of the equation you just
balanced? (1 point)
Answer:
Reversible reaction
Explanation:
If the elements of the product combine together in a chemical reaction, this reaction is called reversible reaction because the reaction moves in backward or reverse direction. For example, if acid i. e. HCl combine with base i. e. NaOH which are the reactants, it produces two products salt (NaCl) and water (H2O). If salt (NaCl) and water (H2O) combine again with each other, it produces acid i. e. HCl and base i. e. NaOH, such type of reaction is called reversible reaction.
Following antigenic stimulation, phosphorylation of _________ relieves inhibition of the transcription factor
Answer: zap70, ITAM.
Explanation:
An antigen is any substance that is capable of stimulating an immune response by activating lymphocytes, which are the body’s infection-fighting white blood cells. Examples of antigens could be proteins that are part of bacteria or viruses or components of serum and red blood cells from other individuals, all of them are foreign antigens originated outside the body. However, there can also be autoantigens (which are self-antigens), originated within the body. In normal conditions, the body is able to distinguish self from nonself. And the antigens that represent a danger induces an immune response by stimulating the lymphocytes to produce antibody or to attack the antigen directly. This is called an antigenic stimulation of the immune system.
ZAP-70 (Zeta-chain-associated protein kinase 70) is a protein that is part of the T cell receptor, thereby it plays a critical role in T-cell signaling. When the TCR (receptor of T cells) is activated by the presentation of the specific antigen through the MHC, a protein called Lck acts to phosphorylate the intracellular CD3 chains and the ζ chains of the TCR complex, allowing the binding of the cytoplasmic tyrosine kinase, ZAP-70. Lck then phosphorylates and activates ZAP-70, which in turn phosphorylates another molecule in the signaling cascade called LAT (short for Linker of Activated T cells), a transmembrane protein that serves as an anchor site for several other proteins. The tyrosine phosphorylation cascade initiated by the Lck culminates in the intracellular mobilization of calcium ion (Ca2+) and the activation of important signaling cascades within the lymphocytes. These include the Ras-MEK-ERK pathway, which is based on activating certain transcription factors such as NFAT, NFκB and AP-1. These transcription factors regulate the production of of certain gene products, most notably cytokines such as interleukin-2 that promote the long-term proliferation and differentiation of activated lymphocytes.
The ITAM motifs (immunoreceptor tyrosine-based activation motif) are sequences of four amino acids present in the intracellular tails of certain proteins that serve as receptors within the immune system. Thus, some receptors such as the TCR have ITAM sequences that, when activated, trigger an intracellular reaction based on consecutive phosphorylations. Kinases are recruited for this purpose.
So, ZAP-70 is a protein tyrosine kinase with a role in T-cell receptor signal transduction. During T-cell activation, ZAP-70 binds to ITAM and becomes tyrosine phosphorylated. The binding of ZAP-70 to the phosphorylated ITAM is able to activate its kinase activity, and relieves the inhibition of the transcription factor which regulates genes that are involved in the immune reaction.
Under normal conditions glomerular filtration depends on three main pressures. What pressures is a pressure that favors the process of filtration?
Answer:
Glomerular Hydrostatic pressure .
Explanation:
The basic function of the kidney is the formation of urine for elimination through the urinary excretory system. Two different processes determine this formation: the filtration of fluid through the glomerular capillaries into Bowman's space and the modification of the volume and composition of the glomerular filtrate in the renal tubules. The fluid passes from the glomerular capillaries to Bowman's capsule due to the existence of a pressure gradient between these two areas. This process is favored by two structural characteristics that make renal corpuscles particularly effective filtration membranes: glomerular capillaries have a much higher number of pores than other capillaries, and the efferent arteriole has a smaller diameter than the afferent arteriole, causing greater resistance to outflow of blood flow from the glomerulus and increasing glomerular hydrostatic pressure. Increased glomerular hydrostatic pressure (due to increased blood flow through the glomerulus) increases filtration, while increases in Bowman's hydrostatic pressure or urinary space (which remains constant, unless there is disease at that level, usually due to fibrosis) and plasma P. oncotic (determined by proteins, which tend to "drag" plasma into the glomerulus) decrease filtering. Resulting in a filtering pressure of 10 mmHg.
Assume that you have an antimicrobial agent specific for each of the targets listed below. Indicate which type of microbe would be most susceptible to the agent.
A. All bacteria
B. Gram-positive bact.
C. Gram-negative bact.
D. Viruses
E. All bact. and viruses
Answer is given below
Explanation:
If you have an antimicrobial agent specific for each of the targets listed below
all bacteria - glycolytic enzymes and membrane proteinsgram-positive bacteria - peptidoglycanGram-negative bacteria ceftiaxon, cefpodoximeviruses - envelope proteinsall bacteria and viruses - nucleic acidsWhen brown iodine is exposed to starch it turns dark purple. In an experiment, you placed a cornstarch solution in a small plastic bag. Next, you placed the bag in a beaker of water containing 10 drops of iodine. If the solution in the plastic bag turned dark purple (Select all that apply)
a) the plastic bag was permeable to cornstarch.
b) the plastic bag was permeable to iodine.
c) the iodine moved into the plastic bag.
d) the cornstarch moved into the beaker.
e) the plastic bag was selectively permeable.
Answer:
b) the plastic bag was permeable to iodine
the plastic bag was selectively permeable
the iodine moved into the plastic bag.
the plastic bag was permeable to cornstarch.
Explanation:
The experimental set up represented the concept of osmosis.That is the movement of water molecules from the region of higher water potential to region of lower water potential through a semipermeable membrane(a cell membrane which only allows water and certain molecules to pass through,but restrain other molecules through its pores.
Generally potassium iodine test is the standard test for starch.
Therefore, in the question,water moves iodine molecules as iodine solution(from high water potential) to move across the paper bag(semi permeable membrane) to the reach the corn solution.(low water potential,high solute potential).This turns the starch in the corn solution blue-black/purple.
If the solution in the plastic bag turned dark purple:
the plastic bag was permeable to iodine. That is option (B)the iodine moved into the plastic bag. That is option (C)the plastic bag was selectively permeable. That is option (E)To test for the presence of starch in a solution, iodine solution is used. This gives a blue-black coloration which occurred due to the formation of a starch-iodine complex.
From the above experiment, the plastic bag which contains the corn starch solution, separates the solution from water containing 10 drops of iodine.
The turning of the solution, inside the plastic bag, into blue black signifies the formation of starch-iodine complex between the cornstarch solution the iodine water.
The reaction was able to occur because the plastic bag, which is selectively permeable to iodine, allowed the movement of iodine from the beaker of water into it. This is called diffusion.
Therefore, if the solution in the plastic bag turned dark purple it means that the plastic bag which is semi permeable allowed the movement of iodine into it.
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Explain in detail if the water molecule (top) can interact with the glucose molecule (bottom)
Answer:
Yes, glucose dissolves in water.
Explanation:
Hello,
In this case, we can state that glucose is largely soluble in water due to the fact that it has five of polar hydroxyl groups which can interact with the hydrogen atoms in the water throughout the hydrogen bonds. Such hydrogen bond is known as an intermolecular force that forms a special type of dipole-dipole attraction when a hydrogen atom bonded to a strongly electronegative atom (like the oxygen in the hydroxyl groups) exists in the vicinity of another electronegative atom with a lone pair of electrons, therefore they occur among molecules.
1. A star is 520 light years from Earth. During what event in history did the
light now arriving at Earth leave the star?
Answer:
A light year is the distance which is equal to 9,460,730,472,580.8 km, so:
= 4.91957985 X [tex]10^{15}[/tex]km
which is distance travels by the light. Now what time it takes light to travel distance we found.
A year has 365.25 days, so,
[tex]1 (\frac{365.25)}{1 year}) (\frac{24}{1 day}) (\frac{3600 s}{1 hr} )[/tex] = 31557600 seg/year
The light speed in the space is equal to 299,792.458 km/s, so:
4.91957985 x [tex]10^{15} (\frac{1 seg}{29792.458}) \frac{1 year}{31557600}[/tex] = 520 years
if today, August, 2020, then
2020 - 520 = 1500
Spanish and Portuguese spread out over the southern part of the Western Hemisphere and bring in America brought to Spanish colony of Santo Domingo in year 1500.
[tex]t=2019-520\\ t=1499 AD[/tex]
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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:
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.
4
THE EAR
4.1
4.1.1 Which ONE of the following is a part of the ear where grommets
are inserted?
A Oval window
B Semi-circular canal
C Tympanic membrane
Pinna
noreen with middle ear infection is usually advised not to fly in an
Answer:
C
Explanation:
Zlrozdzidkzzkxhkcgkxtkckgcgkckzxtixkgx
Is it normal to have two baby fish that are stuck together?
Answer:
No its not
Explanation:
Hey there!
Conjoined fish is a very rare occurrence that does not happen often.
Hope this helps :)
If the gametes from a diploid individual only contain one of the two possible alleles, it is not difficult to determine the genotypes of the gametes from that individual. For the diploid genotypes indicated below, write all of the possible genotypes of the gametes. If more than one type of gamete is possible, separate them with a comma. Diploid Genotype Genotype of Gametes RR rr Rr
Answer:
Diploid Genotype - Genotype of Gametes
RR - R
rr - r
Rr - R, r
Explanation:
According to Gregor Mendel in his law of Segregation, he stated that the alleles of a gene gets separated into gametes during gamete formation (meiosis). This occurs in such a way that only one allele of the two possible alleles that make up the genotype of a diploid organism is found in the gamete.
Meiosis is a cell division that reduces the chromosomal number of the daughter cells. Hence, a diploid organism (2n) will undergo meiosis to produce haploid (n) daughter cells (gametes). An organism with a diploid genotype RR will produce gametes which have only genotype R. The same applies to organism with genotype rr. Only gametes with genotype, r, will be produced.
However, in a heterozygous genotype like Rr, the different alleles will separate into gametes in an equal proportion i.e. gametes with genotypes R and r will be produced in a ratio 1:1.
Are bacteria eukaryotes
Answer:
yes they very much are!!!!!!!!!
A red flower producing snapdragon plant is crossed with a white flower producing snapgragon plant. Resulting F1 generation is crossed with one another to produce F2 generation.
This is incomplete dominance of genes.
Q1 What type of a breeding is this? (monohybrid, dihybrid or interspecific)
Q2 Draw a genetic chart to show P, F1 and F2 generations clearly indicating genotypes, phenotypes and generations.
Answer:
See the answer below
Explanation:
1. This is an example of monohybrid breeding. A monohybrid breeding is the type of breeding that involves parents with a pair of contrasting characters. On the other hand, a type of breeding involving a single gene is what is known as monohybrid breeding.
2. When a red flower snapdragon is crossed with a white flower snapdragon, the resulting offspring are usually pink - an indication of incomplete dominance of the gene responsible for flower color. Assuming the red flower's genotype is AA and that of the white flower is aa:
AA x aa
Aa Aa Aa Aa
F1 genotype = all Aa
F1 phenotype = pink flower
At F2:
Aa x Aa
AA 2Aa aa
F2 Genotype/phenotype:
1AA - red color
2Aa - pink flower color
1 aa - white flower color.
"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
Question 3 (1 point)
During DNA replication, one of the new strands of DNA is synthesized continuously.
The other strand is synthesized as a number of separate fragments of DNA that are
subsequently linked by DNA ligase. Why does this occur?
- RNA primers only anneal to one of the parental strands of DNA.
- DNA polymerase III only synthesizes DNA in the 3' - 5' direction.
- DNA polymerase III only synthesizes DNA in the 5'-3' direction.
- One of the parental strands is unwound slower than the other by helicase.
Answer:
DNA polymerase III only synthesizes DNA in the 5'-3' direction.
Explanation:
DNA replication is an important phenomenon for every living cell. It is the process whereby the double-stranded DNA is doubled to form two new separate double strands. In order for DNA replication to occur, the double strand of the DNA molecule must first be unwound by an enzyme called DNA HELICASE. This gives two separate single strands, which individually acts as a template for the newly synthesized strands.
DNA polymerase III is the enzyme responsible for synthesizing new DNA strand by pairing complementary nucleotides to the old strands it attaches to. However, one of the old strands called LEADING STRAND runs in the 3'-5' direction while the other strand called LAGGING STRAND runs in the 5'-3' direction.
DNA polymerase III only attaches to the 3' hydroxyll end of the DNA and synthesizes new strand of DNA in the 5'-3' direction. Since the lagging strand runs in 5'-3', it is synthesized in small separate fragments called OKAZAKI FRAGMENTS which are later joined together by an enzyme called LIGASE.
N.B: As DNA polymerase synthesizes DNA strand on the leading strand(5'-3'), it is moving in an opposite direction of the lagging strand. Hence, it has to detach and come back to synthesize on the lagging strand. This causes the lagging strand to be synthesized discontinuously.
