Answer:
d
Explanation:
stomach acid secretion in the digestive tract : epithelial tissue.
Insulation of the body : Nervous tissue relates a function with a type of tissue. So, the correct option is C.
What are Nervous tissues?Nervous tissue is defined as the main tissue of our nervous system that monitors and regulates body functions. Nervous tissue consists of two cells: nerve cells or neurons and glial cells which help in the transmission of nerve impulses and also provide nutrients to the neurons.
The brain, spinal cord, and nerves are composed of nervous tissue, which are specialized to be excitable, to rapidly transmit stimuli from one part of the body to another. Nervous tissue is made up of nerve cells or neurons, all of which have an axon which is a long stem-like projection extending from the cell, which is responsible for communicating with other cells called target cells, thereby conducting impulses.
Thus, Insulation of the body : Nervous tissue relates a function with a type of tissue. So, the correct option is C.
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Which choice has eukaryotic animal cells? Lactobacillus virus liver oak
Answer:
Probably the Liver.
Explanation:
Answer: Its Liver! obviously!
Explanation: well, i dont have an explanation but it should help!
A scientist designed an experiment to test an artificial membrane that mimics the phospholipid bilayer of a
cell.
The scientist built a tube that was divided by an artificial membrane and filled with distilled water. The
scientist put a known amount of a protein into the water on one side of the membrane. After some time, the
scientist measured the concentration of the protein on either side of the membrane but found that there had
been no change.
Which of the following experimental changes would allow the scientist to observe transport of a solute
across the artificial membrane?
A
Increase the solute concentration in the solution
B
Use a small, nonpolar solute instead of a protein
C
Increase the temperature of the solution
D
Add artificial aquaporins to the membrane
Simple diffusion is the free pass of small, non polar molecules through the lipid bilayer.The scientist needs to use small, nonpolar solutes instead of a protein.
-----------------------------------------
Different types of transports occur through the membrane. Among them, we can find simple diffusion, facilitated diffusion, and active transport.
The simple diffusion process allows the passage of only non polar, small molecules. They can freely move through the membrane by momentary openings produced by the lipids' movements.
Simple diffusion is a slow process that requires short distances and pronounced concentration gradients to be efficient. It depends on the molecules' sizes and their concentration on each side of the membrane.
Since proteins (represented as red squares in the attached image) are macromolecules, they can not cross the membrane without carrier or channel proteins.
The scientist should use a small, nonpolar solute (green circles in the attached image) instead of a protein. The diffusion direction will occur from the side where the scientist put the solute to the other side.
-----------------------------------------
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What are the Functions of the skeletal system?
Answer:
The skeletal system works as a support structure for your body. It gives the body its shape, allows movement, makes blood cells, provides protection for organs and stores minerals.
Explanation:
Ms. Brown’s class built series and parallel circuits during science lab. After the results of the experiments are collected, the students should
Answer:
B
Explanation: because the teacher is in charge
Name two pieces of laboratory equipment that can be used
to measure mass.
Answer:
Scientists measure mass with a balance, such as a triple beam balance or electronic balance. In science, the volume of a liquid might be measured with a graduated cylinder.
Explanation:
. Pepper colors range from bright red, orange, and yellow shades, to dark green and brown shades. Coloration in bell peppers is controlled by alleles for two genes: Red (R) is incompletely dominant to yellow (r) and No chlorophyll (G), no green pigment, is dominant to chlorophyll (g), green pigment. A. You cross a red pepper with a green pepper and all the F1 progeny are orange. 1 pt What are the genotypes of the red and green parents? What is the genotype of the F1’s? B. What genotypes and phenotypes, in what proportions, do you expect in the F2 progeny of the orange F1 peppers? 2 extra credit points
Answer:
All crosses and proportions, genotypes and phenotypes are attached.
Explanation:
a. Within living organisms, staining is a characteristic governed by a polygenic inheritance, which means that there is more than one gene involved in the staining process. as you already know, each gene has two alleles, if a trait is controlled by 2 genes, it means that we will have 4 alleles at the crosses.
From the description between dominance and recessivity between the alleles that control the colors of the pepper, shown in the question above, we can see that for the crossing between a red pepper and a green pepper, being able to generate a completely orange offspring, it would be necessary that the genotype of the parent peppers was: Red: RGRG, green: rgrg.
This would generate an orange-colored RrGg offspring, as you can see at the F1 crossing.
b. When individuals of F1 offspring are crossed, the combination of alleles and the determination of genotypes and phenotypes becomes much more complex, because instead of 4 alleles, we will have the combination of 16 alleles among themselves. Once again we will need to rely on the description of dominance and recessivity shown in the question above, so that from the crossing between the alleles of each gene, we can reach a conclusion, as you can see in the F2 crossing.
Answer:
is
Explanation:
vnbmbyi iiigujjg khm,,boul ghjggg
Katherine has recorded the temperature outside each day for one week. The daily temperatures are recorded in the table. Calculate the average temperature for the week. Round to the nearest whole number.
Answer:
71 °F
Explanation:
i had to fail to get the answer lol
what is the average temperature on uranus for day and night?
Answer:
The coldest temp is -353 the warmest it gets is -371
Explanation:
Why is soil important?
What is soil erosion?
What human activities are responsible for soil erosion happening at advanced rates?
What are some of the effects of soil loss?
What can be done to prevent soil erosion?
Answer:
It helps in growing of plants
it is the carrying away of soil by water wind or glaciers
Overstocking the cattle
Infertility
By planting grass
Answer:
Why is soil important?
It is important because it distributes plant species and gives a habitat for a lot of organisms.
What is soil erosion?
Soil erosion is where the topsoil wears off by natural physics of water, wind, and ice. Some nutrients and materials are lost in the process.
What human activities are responsible for soil erosion happening at advanced rates?
Ploughing, unsuitable agricultural practices like using a till to plant things, combined with deforestation and overgrazing.
What are some of the effects of soil loss?
Your dirt might not be fertile anymore, clogged waterways, or an increase in pollution.
What can be done to prevent soil erosion?
To help prevent soil erosion we can use no till or mulch till. We could plant grasses, grains, or legume to help prevent soil erosion.
Explanation:
your welcome.
Which muscle is attached to the gluteus Maximus
Answer:
The Gluteus Medius and Minimus, as well as iliotibial band, and tensor fascia lata.
Explanation:
I hope this helps!
Which property of a rock determines its color?
its grain size
its method of formation
its mineral composition
its coarse-grain content
property of a rock determines its color , is its method of formation
what are the types of rocks ?Metamorphic rocks are always begin as another type of rock which involves the transformation of existing rocks that is igneous or sedimentary rock.
These are either formed by either excessive heat and pressure, or through the chemical activity of fluids.
sedimentary rocks or clastic sedimentary rocks are composed of rock pieces that have been battered from already existing rocks.
It is the grain of sediments which stick and combine to ultimately form the rock and the size of the grain determines the energy of sedimentary transport
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Unlike plant cells, animal cells contain _____. a. cell walls c. nucleoli b. centrioles d. spindles
Answer:
Animal cells contain centrioles. Something plants cells do not have.
Explanation:
Unlike plant cells, animal cells contain centrioles. The correct option is B.
Thus, Animal cells include cylindrical centrioles, which are found specifically in the centrosome area. They are essential for cell division because they organize and direct the formation of the mitotic spindle.
In both mitosis (cell division for growth and tissue repair) and meiosis (cell division for sexual reproduction), the mitotic spindle is in charge of segregating duplicated chromosomes.
On the other hand, centrioles are frequently absent from plant cells. Microtubule organizing centres (MTOCs), which are physically distinct from centrioles but perform comparable tasks, are instead present in plant cells.
Thus, Unlike plant cells, animal cells contain centrioles. The correct option is B.
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Natalie is class reread Shakespeare’s Romeo and Juliet last semester and took a field trip to see it live in community theater. today Natalie or warrant that her class I’ll read another work for Shakespeare this semester
What is the best prediction she can make about this semester‘s text?
It’s a play
It’s an adventure
It is a fairytale
It is a fable
Answer:
itz a play?
Explanation:sory if im wrong TwT
Answer:
its a play
Explanation:
i just took the test
Lipase is an enzyme that breaks down fats
into fatty acids and these fatty acids are
absorbed into the blood stream and act as a
source of energy for cells or are stored as fat.
Absorbed fatty acids support the energy needs
when glucose is unavailable as in the case of
diabetes.
If lipase does not function complex fats
in foods are not broken down and absorbed.
It is this reason why diet pills can often target
this enzyme.
A scientist was searching for a non-
toxic inhibitor of lipase and wanted to test the hypothesis that alginates (found in seaweed) inhibit lipase action. A co-worker alerted her to a diet pill that contain the chemical Orlistat that can also inhibit lipase but has side-effects associated with its use. To test the efficacy of the seaweed inhibitor she tested lipase activity as shown in the table below.
3a. For the reaction below label E, S and P components as explained in the question.
E+S <----> ES<---->E+P
3b. Identify the control group in the experiment.
3c. What is the scientists’ prediction from this experiment?
Answer:
3a. As we know and also mention in the question that lipase is an enzyme that digested and break down the fats into fatty acid and it is also known that the action of an enzyme is as follows:
Enzyme + substrate <===> Enzyme-Substrate <====> Enzyme + Product
and here, E+S <----> ES<---->E+P
So, E is Lipase as the enzyme, S is fats, and P is fatty acids
3b. The control group is in this experiment or research is the group in which activity of the lipase is tested with the help of the chemical Orlistat, present in the diet pill as the control group is used to compare the or test the results of the independent variable's response.
3c. The scientist's prediction would be that in the control group fats will not be digested by the lipase into fatty acids and will not lead to any side effects.
When the human body is subjected to prolonged starvation, cells use up all available glucose first, including glucose bound up in glycogen. When all glucose has been depleted, cells use lipids as a fuel source, and when lipids have been depleted, cells use proteins as a fuel source. Throughout these shifts to new fuel sources, the enzymes of glycolysis and the citric acid cycle remain present and active in the cells. Which statement explains the constancy of these enzymes under conditions of starvation?
The options are missing from the question. They are;
a. These enzymes are inhibited but not destroyed under conditions
of starvation since destroying them would require energy input.
b. These enzymes continue to be needed since catabolism of lipids
and proteins leads to intermediates that are oxidized through
these pathways.
c. Once enzymes are produced in a cell, they continue to function
for as long as the cell lives even if they are not used.
d. All catabolic enzymes are expressed when energy levels are low
in a cell, including enzymes that cannot oxidize the fuel
molecules that are available.
e. The cell continues to produce these enzymes in case glucose or
other carbohydrates become available again.
The correct answer to the question is option B
THESE ENZYMES CONTINUE TO BE NEEDED SINCE CATABOLISM OF LIPIDS AND PROTEINS LEADS TO INTERMEDIATES THAT ARE OXIFIZED THROUGH THESE PATHWAYS.
The enzymes involved in glycolysis are; hexokinase, phosphofructokinase, and pyruvate kinase.
The cell use Amino acids as energy source through the conversion of the amino acids that enter glycolysis or the citric cycle that are eventually catabolized to CO2 at various points through the help of their enzymes (fumarase) or (fumarase hydratase) speeding up the rate of the reaction.(pyrimidines).
Common names given to organisms may be useful locally but may become meaningless on a _________ basis.
Answer:
Global basis
Explanation:
what happens when the birthrate and the death rate are the same?
species with broad niches
Answer:
opossums and bison
Explanation:
Answer:
opossums and bison
Explanation:
Which two characteristics describe all organisms in the domain Eukarya?
A. Their cells are prokaryotic.
B. Their cell walls have peptidoglycan.
c. Their cells are eukaryotic.
D. Their cell walls lack peptidoglycan.
Mycorrhizal fungi are
Answer:
Mycorrhizae are symbiotic relationships that form between fungi and plants. The fungi colonize the root system of a host plant, providing increased water and nutrient absorption capabilities while the plant provides the fungus with carbohydrates formed from photosynthesis.
Explanation:
Mycorrhizal fungi form a mutualistic association with plant roots. They provide benefits to each other.
What are mycorrhizas?A symbiotic relationship between fungi and plant roots is known as mycorrhizae. Their main function is to increase the host plant's intake of nutrients and water by utilizing more soil than roots alone can. Depending on the host plant and fungus' taxonomy, mycorrhizae can take on a variety of shapes.
Mycorrhizal fungi create symbiotic relationships with almost all land plants. These underground fungi are essential to terrestrial ecosystems because they control the carbon and nutrient cycles, affect soil composition, and increase the diversity of ecosystem functions.
Endomycorrhizae and ectomycorrhizae are the two primary classes of mycorrhizae.
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1. Explain why living things need carbon and nitrogen. What are they used for?
Answer:
Explanation:
all living organisms are built of carbon compounds. Nitrogen fixation: special bacteria convert the nitrogen gas into ammonia that plants then use. ... nitrogen is the required nutrient for living things to produce organic molecules. it's the building block of DNA, RNA, proteins, and nucleic acids.
Waves cause erosion through the processes of ______ and abrasion. a. chemical weathering b. impact
Answer:
B
Explanation:
Impact, the strength of the wind carries little pebbles and stone,then it deposits.
Answer:
b
Explanation:
true or false? scientist test explanations with more evidence
Answer:true
Explanation:
Likhon has Parkinson’s disease and is participating in a clinical trial for a new drug that increases the amplitude of EPSPs at neurons only in his basal ganglia. Of the following options, the most pronounced effect of this drug would be: a. Decreased likelihood of action potentials in basal ganglia neurons, which are associated with initiating movements b. Increased likelihood of action potentials in basal ganglia neurons, which are associated with initiating movements c. Increased likelihood of action potentials in basal ganglia neurons, which are associated with emotional processes d. Decreased likelihood of action potentials in basal ganglia neurons, which are associated with emotional processes
Answer:
b. Increased likelihood of action potentials in basal ganglia neurons, which are associated with initiating movements
Explanation:
The substantia nigra is the principal site involved in the physiopathology of Parkinson´s disease.
Cells from this area degenerate and die progressively causing important degeneration in the central nervous system. The cells that die -with no known reason- are those that produce dopamine and proportionate dopaminergic innervation to the encephalon. In fact, almost all the symptoms exhibited by patients with the disease can be explained by the lack of dopamine in the substantia nigra.
This neurotransmitter is used in the area to communicate neurons from the base ganglia. Ganglia are responsible for modulating and modifying movements.
The nigrostriatal system participates in the planning and automatic execution of learned movements. When the ganglia activity decreases, so it does the amount and velocity of movements, something typical of the disease.
If the drug increases the amplitude of EPSPs at neurons in the basal ganglia, the action potentials in basal ganglia neurons would be expected to increase.
Which of the following statements regarding contraction in skeletal and smooth muscle is TRUE? A. In skeletal muscle, calcium initiates contraction by binding to troponin, while in smooth muscle calcium initiates contraction by binding directly to myosin. B. In skeletal muscle, calcium initiates contraction by binding to troponin, while in smooth muscle calcium initiates contraction by binding to calmodulin. C. In skeletal muscle, calcium initiates contraction by binding to myosin light-chain kinase, while in smooth muscle calcium initiates contraction by binding directly to tropomyosin. D. All of the above are TRUE.
Answer:
B. In skeletal muscle, calcium initiates contraction by binding to troponin, while in smooth muscle calcium initiates contraction by binding to calmodulin.
Explanation:
The Calcium ions (Ca2 +) have a key role in skeletal muscle contraction. Moreover, the Troponins are regulatory proteins that also play roles in skeletal contraction. In this muscle, troponin is attached to tropomyosin, when Ca2 + binds to troponin it induces a conformational change, thereby exposing this protein to its binding to myosin filaments on the actin filaments, which initiates muscle contraction.
On the other hand, calmodulin (calcium-modulated protein) is a protein which mediates the contractile state of the smooth muscle by binding Ca2+ ions. The function of this protein is to stimulate the development of force (contraction) in response to Ca2+ ions by activating the phosphorylation of myosin.
Muscle Contraction occurs in both skeletal and smooth type of muscles. Although in smooth muscles, the contraction is slow as compared to skeletal muscles.
Contractions in skeletal and smooth muscles occurs: In skeletal muscle, calcium initiates contraction by binding to troponin, while in smooth muscle calcium initiates contraction by binding to calmodulin. (Option B)
Muscle contraction in both smooth and skeletal muscles is initiated and aided by the influx of calcium ions, the only difference is that in smooth muscle, the influx of calcium ion causes the ions to bind to intracellular calmodulin which causes the activation of an enzyme known as myosin kinase.
Although in skeletal muscles, the calcium ion binds to troponin changing the shape of tropomyosin exposing the myosin binding site
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A scientific theory is NOT an educated guess.
True
False
Answer:
True
Explanation:
A scientific theory is NOT an educated guess as a scientific theory is a well-tested, broad explanation of a natural phenomenon. A theory in science is based on experimentation and research.
Which three structures are found in both prokaryotic
Answer:
The structures that are found in both eukaryotic and prokaryotic cell are-
Explanation:
ribosomes, DNA, cytoplasm.
Ribosomes are biomolecular complex, composed of RNA and protein and act as the site for protein synthesis in all the cells of the living organisms.
DNA (deoxyribonucleic acid) is considered.
write a short note on the process of photosynthesis
Answer:
Explanation:
Plants take in sunlight and water; and then turn them into sugars, which are food for the plant.
Answer: The process by which plants and some microorganisms make substances like carbohydrates. It is an endothermic (takes in heat) chemical process that uses sunlight to turn carbon dioxide into sugars.
Explanation:
everything lol I need help on mi life
Don’t worry, everything is gonna be fine :)
In Drosophila melanogaster, vestigial wings (vg) is recessive to normal wings (vg+), black body (b) is recessive to gray body (b+), and purple eyes (pr) are recessive to red eyes (pr+). All three genes are located on the same chromosome. A female fly with vestigial wings, black homes body, and purple eyes is crossed to a male with normal wings, gray body, and red eyes. The F1 females produced by this cross were mated with vestigial, black, purple males in a test cross. After doing a test cross the following pheonotypes were obtained.
1779 vestigial wings, an ebony body and purple eyes
1665 normal wings, a brown body and red eyes
252 normal wings, an ebony body and purple eyes 241 vestigial wings, a brown body and red eyes
131 normal wings, an ebony body and red eyes
118 vestigial wings, a brown body and purple eyes
13 vestigial wings, an ebony body and red eyes
9 normal wings, a brown body and purple eyes.
1. Determine the order of these genes on the chromosome.
2. Calculate the map distances between the genes. Construct a genetic map for these three loci.
Correct progeny phenotype:
1779 vestigial wings, black body and purple eyes, vg b pr 1665 normal wings, a grey body and red eyes, vg+ b+ pr+252 normal wings, an black body and purple eyes, vg+ b pr241 vestigial wings, a gray body and red eyes, vg b+ pr+131 normal wings, an black body and red eyes, vg +b pr+118 vestigial wings, a gray body and purple eyes, vg b+ pr13 vestigial wings, an black body and red eyes, vg b pr+9 normal wings, a gray body and purple eyes, vg+ b+ prAnswer:
The order of these genes is vg --- pr --- b Map distances between the genes vg/pr = 12.2 MUMap distance between the genes pr/b = 6.4 MUMap distances between the genes vg/b = 18.6 MUExplanation:
We know that
•Normal wings expressed by vg+ is dominant over vestigial wings, vg
•Gray body b+ is dominant over black body
•Red eyes, pr+, is dominant over purple ayes, pr
We have the number of descendants of each phenotype product of the tri-hybrid cross.
•1779 vestigial wings, black body and purple eyes vg b pr
•1665 normal wings, a grey body and red eyes vg+ b+ pr+
•252 normal wings, an black body and purple eyes vg+ b pr
•241 vestigial wings, a gray body and red eyes vg b+ pr+
•131 normal wings, an black body and red eyes vg +b pr+
• 118 vestigial wings, a gray body and purple eyes vg b+ pr
•13 vestigial wings, an black body and red eyes vg b pr+
• 9 normal wings, a gray body and purple eyes vg+ b+ pr
The total number of individuals is 4208.
In a tri-hybrid cross, it can occur that the three genes assort independently or that two of them are linked and the third not, or that the three genes are linked. In this example, in particular, the three genes are linked on the same chromosome.
Knowing that the genes are linked, we can calculate genetic distances between them. First, we need to know their order in the chromosome, and to do so, we need to compare the genotypes of the parental gametes with the ones of the double recombinants. We can recognize the parental gametes in the descendants because their phenotypes are the most frequent, while the double recombinants are the less frequent. So:
Parental)
1779 vestigial wings, black body and purple eyes vg b pr 1665 normal wings, a grey body and red eyes vg+ b+ pr+Double recombinant)
13 vestigial wings, an black body and red eyes vg b pr+9 normal wings, a gray body and purple eyes vg+ b+ prSimple recombinant)
252 normal wings, an black body and purple eyes vg+ b pr241 vestigial wings, a gray body and red eyes vg b+ pr+131 normal wings, an black body and red eyes vg +b pr+118 vestigial wings, a gray body and purple eyes vg b+ prComparing parental with the double recombinants we will realize that between
vg b pr (parental)vg b pr+ (double recombinant)and
vg+ b+ pr+ (Parental)vg+ b+ pr (double recombinant)They only change in the position of the alleles pr/pr+. This suggests that the position of the gene pr is in the middle of the other two genes, vg and b, because in a double recombinant only the central gene changes position in the chromatid.
So, the order of the genes is:
---- vg ---- pr -----b ----
Now we will call Region I to the area between vg and pr and Region II to the area between pr and b.
Once established the order of the genes we can calculate distances between them, and we will do it from the central gene to the genes on each side. First We will calculate the recombination frequencies, and we will do it by region. We will call P1 to the recombination frequency between vg and pr genes, and P2 to the recombination frequency between pr and b.
P1 = (R + DR) / N
P2 = (R + DR)/ N
Where: R is the number of simple recombinants in each region, DR is the number of double recombinants in each region, and N is the total number of individuals.
So:
Parental)
• 1779 vestigial wings, black body and purple eyes vg pr b
• 1665 normal wings, a grey body and red eyes vg+ pr+ b+
Double recombinant)
• 13 vg pr+ b
• 9 vg+ pr b+
Simple recombinant)
• 252 vg+ pr b
• 241 vg pr+ b+
• 131 vg+ pr+ b
• 118 vg pr b+
P1 = (R + DR) / N
P1 = (252+241+13+9)/4208
P1 = 515/4208
P1 = 0.122
P2= (R + DR) / N
P2 = (131+118+13+9)/4208
P2 = 271/4208
P2 = 0.064
Now, to calculate the recombination frequency between the two extreme genes, vg and b, we can just perform addition or a sum:
P1 + P2= Pt
0.122 + 0.064 = Pt
0.186=Pt
The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU). One centiMorgan (cM) equals one map unit (MU).
The map unit is the distance between the pair of genes. Every 100 meiotic products, one of them results in a recombinant product. Now we must multiply each recombination frequency by 100 to get the genetic distance in map units:
GD1= P1 x 100 = 0.122 x 100 = 12.2 MU
GD2= P2 x 100 = 0.064 x 100 = 6.4 MU
GD3=Pt x 100 = 0.186 x 100 = 18.6 MU
---- vg ---------------------- pr ---------------------b ----
R1 R2
-----vg----12.2MU---------pr—
----pr--------6.4 MU----b—
-----vg ----------------18.6 MU--------------------b----
