Answer:
D.
Explanation:
Totipotent cells can make any of the 220 cell types in an embryo, as well as the cells found in a placenta!
Any cells of the human body or the placenta can develop from totipotent stem cells. Therefore, the correct option is D.
Totipotent stem cells have the unusual ability to differentiate into any type of human body cell, including placental cells. They have the ability to generate all types of cells, including endoderm, mesoderm, and ectoderm, the three basic germ layers. Because of their potential for development, totipotent stem cells can produce both extra-embryonic and embryonic tissues. Although cells produced from totipotent stem cells can differentiate into any cell type, due to maturation and specialization processes, they are not always identical to the original stem cell.
Therefore, the correct option is D.
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difference between plant and animal cyanide resistance respiration.
Answer:
The cyanide resistant respiration differs in plants as oxidation of reduced coenzyme continues even in the presence of cyanides.
Explanation:
The mitochondrial is the electrons through which the animals and plants perform the aerobic exercise and is blocked by the release of the cyanides that stops the activity and is known as cyanide sensitive respiration Plant mitochondria is different from the animal as they both have different oxidase system pathways. Cyanide resistant is responsible for the climacteric in fruits and for the generate heat in thermogenic tissues.You are doing research on nervous system diseases; particularly, one that dissolves lipids. What part of the neuron would you expect to be affected most by this disease? A. Myelin sheath B. Axon C. Synaptic Cleft D. Dendrites
The answer you are looking for is A
Which activity involves a chemical change?
A. Ice melting into liquid water
B. A metal rod being bent into a ring
C. Colorless egg becoming a white solid
D. Salt dissolving in water
SUBMIT
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option C is correct........
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which type of research study compares disease rates among population groups?
Answer:
Epidemiological
Explanation:
Epidemiology compares disease rates among population groups, and attempts to identify related conditions or behaviors.
Answer: Epidemiological studies
A friend comes to you extremely agitated and says her husband beat her last night.She says she doesn't know what to do.You Can See She's Very panicky.Everything you suggest just seems to provoke more anxiety and fears.When you try to respond to one aspect of the problem.Your friend interrupt's frantically and jumps to another.Explain what can you do to help.
Use the completed Punnett square in Part B to answer the questions below about the F2 generation.
Note that questions 3 and 4 require a more advanced evaluation of probabilities than do questions 1 and 2 because you have to consider all possible groupings and orders of three F2 seeds. If you need help with these calculations, use Hint 3.
Complete question:
Part C: Using the Punnett square to make predictions
Use the completed Punnett square in Part B to answer the questions below about the F2 generation.
Note: You will find the Punnet square in the attached files.
Note that questions 3 and 4 require a more advanced evaluation of probabilities than do questions 1 and 2 because you have to consider all possible groupings and orders of three F2 seeds. If you need help with these calculations, use Hint 3.
1) What is the probability that an F2 seed chosen at random will be yellow?
2) What is the probability that an F2 seed chosen at random from along the yellow seeds will breed true when selfed?
3) What is the probability that three F2 seeds chosen at random will include at least one yellow seed?
4) What is the probability that three F2 seeds chosen at random will include one green seed and two yellow seeds?
Answer:
1) 3/4
2) 1/3
3) 63/64
4) 27/64
Explanation:
Y is the dominant allele and expresses yellow color y is the recessive allele and expresses green colorSeeds with YY and Yy genotype are yellowSeeds with yy genotype are greenParental) Yy x Yy
Gametes) Y y Y y
Punnet Square)
(1/2) Y (1/2) y
(1/2) Y (1/4) YY (1/4) Yy
(1/2) y (1/4) yY (1/4) yy
F2) 1/4 YY
2/4 = 1/2 Yy
1/4 yy
1) There will be 3/4 of probabilities of getting a yellow F2 seed
1/4 YY + 2/4 Yy = 3/4 Y-
The symbol "-" means that in its position there might be either a Y or y allele.
2) As there are only three yellow possible genotypes for yellow seeds, the fourth genotype for green seeds is not considered. Our pool now is only yellow seeds. So the probability of getting a pure breeding yellow seed among all yellow seeds is 1/3.
1/3 YY
1/3 Yy
1/3 yY
In the pool of yellow seeds, there are three genotypes, and only one of them corresponds to pure breeding yellow seed.
3) To calculate the probability of getting at least one yellow seed among the three seeds randomly chosen, we need to multiply and then perform addition. This is, when we take three seeds from the pool, we have different possibilities of getting seeds with different genotypes. So:
the three chosen seeds could be yellow. In this case, we multiply their probabilities. The probability of getting one yellow seed is 3/4, so, the probabilities of getting three yellow seeds are: 3/4 x 3/4 x 3/4 = 27/64There might also be one green sees and two yellow seed, so we multiply their probabilities in the order in which we might get them. This is:
Two yellow seeds and one green seed: 3/4 x 3/4 x 1/4 = 9/64One yellow seed, one green seed, and one yellow seed: 3/4 x 1/4 x 3/4 = 9/64one green seed, and two yellow seeds: 1/4 x 3/4 x 3/4 = 9/64And we could also get two green seeds and only one yellow seed:
Two green and one yellow seed: 1/4 x 1/4 x 3/4 = 3/64one green, one yellow and one green: 1/4 x 3/4x 1/4 = 3/64one yellow seed and two green seeds: 3/4 x 1/4 x 1/4 = 3/64Now, having all the possibilities, we just need to add all the possibilities:
27/64 + 9/64 + 9/64 + 9/64 + 3/64 + 3/64 + 3/64= 63/64.
Note that we did not include the possibility of getting three green seeds (1/4 x 1/4 x 1/4). This is because we were asked to calculate the probabilities of getting AT LEAST one yellow seed.
4) To calculate the probability of getting two yellow seeds and one green seed among the three seeds randomly chosen, we should multiply probabilities and then sum them up.
The orders in which we could get the seeds are:
Two yellows and one green: 3/4 x 3/4 x 1/4 = 9/64One yellow, one green, and one yellow: 3/4 x 1/4 x 3/4 = 9/64one green seed and two yellow seeds: 1/4 x 3/4 x 3/4 = 9/64Now, having all the possibilities, we just need to add all the possibilities:
6/64 + 9/64 + 9/64 = 27/64.
Answer:
1) 3/4
2) 1/3
3) 63/64
4) 27/64
Explanation:
1) yellow seed could either be GG, Gg or Gg from Punnett square, therefore, there is a 3/4 probability
2) true-breeding = either GG or gg. Therefore, out of the 3 yellow seed options, only 1 is true-breeding, therefore, it's a 1/3 probability
3) "there are 7 possible groupings that have at least 1 yellow seed; the sum of their individual probabilities yields the final answer of 63/64. Note that you could have also calculated this answer by subtracting the probability of the one category that doesn’t fit the criterion (Green, Green, Green) from the total probability (1 – 1/64 = 63/64)"
4) "there are 3 possible groupings that have 1 green and 2 yellow seeds; the sum of their individual probabilities yields the final answer of 27/64"
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If the scientist had collected 20 dung samples, would you expect more bands, fewer bands, or the same number of bands on the gel?
Answer:
The expectations of the bands will be more on the gels.
Explanation:
One sample provides several sizes and thickness bands in the gel electrophoresis according to the size of fragments that run in it. An individual would expect more bands due to the increase in the number of samples which is 20 samples now.
An increase in the number of samples produces more bands. So one can easily expect more bands from the 20 samples. Scientists would get more bands on the gels.
Thus, the expectations of the bands will be more on the gels.
If the scientist collects 20 dung samples ; we should expect more band on the gel
Given that each dung sample has varying sizes and varying thickness of bands in the gel electrophoresis which depends on the size of fragments contained in it. Therefore when a number of dung samples ( 20 ) are collected it is only logical that the number of bands on the gel will increase ( i.e. you should expect more band on the gel ).
Hence we can conclude that If the scientist collects 20 dung samples we should expect more band on the gel .
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Although your question lacks some data a general answer within the scope of your question is provided
Which type of mutation occurs only in gametes?
germline mutation
somatic mutation
point mutation
frameshift mutation
Answer:
Germline mutation
Explanation:
A germline mutation is a genetic change produced in the gametic-germline cells (i.e., male and female gametocytes). The mutations in the germline are inherited in the next generation, while changes in somatic cells are not inherited. Moreover, the germline mutations may occur either before the fertilization or during embryonic development (zygote morula, blastula and gastrula). This type of mutation may be induced by both endogenous (e.g., oxidative DNA damage) and exogenous (e.g., exposure to UV light) factors.
The type of mutation that occurs in gametes only is : ( A ) Germline mutation
Germline mutation is a type of mutation that occurs in the gametocytes ( male and female ) and is passed( inherited ) on to the offspring ( next generation ). while mutations occurring in somatic cells are not passed on to the next generation .
Germline mutation is induced by either endogenous factors or exogenous factors and this is because Germline mutations can occur before fertilization or after fertilization.
Hence we can conclude that the type of mutation that occurs in gametes only is Germline mutation.
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List the functions of proteins
Answer:
Growth and Maintenance.
Causes Biochemical Reactions.
Acts as a Messenger.
Provides Structure.
Maintains Proper pH.
Balances Fluids.
Bolsters Immune Health.
Transports and Stores Nutrients.
Humans breathe in oxygen and release ____ gas.
Answer:
Carbon
Explanation:
Reverse photosynthesis
Answer:
carbon dioxide
Explanation:
our body changes it
If there are 12 chromosomes in an animal cell in the G 1 stage of the cell cycle, what is the diploid number of chromosomes for this organism?
Answer:
The gametes are haploid, having half the number of cells as compared to somatic cells. The gametes or sex cells produced by the organism will have 6 chromosomes if the organism has 12 chromosomes
Explanation:
To which domain does the animal kingdom belong? Bacteria Archaea Eukarya
Answer:
The correct answer is Eukarya.
Explanation:
The kingdom Animalia is one of four Kingdoms in the Domain Eukarya. This domain is further subdivided into the Kingdoms Protista, Plantae, Fungi, and Animalia.
Answer:
Eukarya is it
Explanation:
Why does the pattern recognition strategy affect prokaryotic and eukaryotic pathogens as well as hosts
Answer:
A Because all are complex cells with certain functions that cannot be greatly altered without affecting fitness.
B Because VSG switching may not be restricted to trypanosomes.
C Because all have proteins in their cell walls or cell membranes.
D None of these answers are correct.
The answer is C .Because all have proteins in their cell walls or cell membranes.
Explanation:
Pattern recognition strategy employs the use of pattern recognition receptors which are proteins for the recognition of molecules frequently found in pathogens. When they are recognized, the body then marks the cells for destruction.
Pattern recognition strategy affect prokaryotic and eukaryotic pathogens as well as hosts because all have proteins in their cell walls or cell membranes and they help in the recognition of pathogens or host for further action on them.
How is the top layer of the ocean different than the deepest layer
A the top layer is colder and doesn’t receive sunlight
B the top layer receives sunlight and is warmer
C the top layer is warmer and doesn’t receive sunlight
D the top layer is colder and receives sunlight
Answer:
B the top layer receives sunlight and is warmer
Explanation:
Answer:
B. the top layer receives sunlight and is warmer.
Huntington's disease, which causes degeneration of neurons, runs in Mark's family. Huntington's is a heritable autosomal dominant genetic disease, in an autosomal dominant disease, only one abnormal allele is required to cause the disease in offspring. However, it is possible not all offspring will receive the abnormal allele from an affected heterozygous parent.
Family members that have the disease include Mark's maternal grandfather and Mark's maternal aunts. Mark's maternal grandmother does not have the disease.
Because symptoms of Huntington's disease do not appear until later in life, genetic testing is necessary to determine whether a younger individual possesses the allele for the disease. These tests determined that Mark's mother does not carry the allele for Huntington's disease.
Mark's paternal aunt suggested that Mark create a pedigree to track the disease in his family. Use the symbols provided to complete the pedigree. Each symbol may be used more than once.
Answer:
As it is given that Mark's paternal grandmother and grandfather does not carry the disease. The symbols on the right side of the Punnett represent Mark's paternal family. If both his paternal grandparents are unaffected then his father and paternal aunt are also not affected by the disease.
His maternal grandfather is affected as mentioned and also his maternal aunts are affected. Mark's mother is unaffected as she does not have the disease-causing alleles determined by genetic testing.
Both of the marks parents are unaffected which indicates that marks also do not carry the allele of the disease.
Huntington's disease is an autosomal dominant disorder in which the presence of even one dominant allele can cause the disorder.
The autosomal disorder can not skip a generation, have to be present in every generation.
Condition Given:
Affected Member: Mark's Maternal grandfather and Maternal aunts.
Normal: Mark's Maternal grandmother and Mark's Mother.
As MArk's Maternal Grandfather is affected then his genotype must be Aa because Mark's Maternal grandmother is unaffected (aa), this is the only condition where Mark's mother will be normal.
A- Dominant (Huntington)
a- Recessive
Maternal means on the mother's side and paternal means on the father's side.
The diagram is attached below.
I-1: Mark's maternal Grandmother
I-2: Mark's maternal Grandfather
I-3&4: Mark's Paternal Grandparents
II-1&2: Mark's Maternal aunts
II-5: Mark's paternal aunt
II-3&4: Mark's mother and father respectively will not have the disease as both father and mother of Mark are normal.
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What best describes the behavior of nonconservative elements in seawater?
The given question is incomplete due to missing options, however, the options for the question as follows:
A. Nonconservative elements are reactive in seawater and have a long residence time.
B. Nonconservative elements are reactive in seawater and have a short residence time.
C. Nonconservative elements are non-reactive in seawater and have a short residence time.
D. Nonconservative elements are non-reactive in seawater and have a long residence time.
Answer:
The correct answer is option C.
Explanation:
Non conservative components are components that enter the ocean water and have little concentration in the sea and show spatial variations. These components have short home time that is short replacements time and they are non responsive.
Constituents, for example, phosphate, nitrate, and different supplements, and broke up oxygen, carbon dioxide, and so on are non-conservative on the grounds that their concentrations are later altered by chemical reactions in the ocean.
Thus, The correct answer is option C.
A microbiologist would like to use a noncompetent genus of streptococcal bacteria, Enterococcus faecalis, as a cloning host to express genes from Streptococcus pneumoniae, which is naturally competent. Is this possible? Group of answer choices
Answer:
Yes, with the assistance of electroporation the bacteria Enterococcus can be made competent, and afterward, the Streptococcus bacterial genes can be presented within the Enterococcus with the help of a cloning vector.
The electropermeabilization or electroporation refers to a microbiology technique in which the permeability of the cell membrane is enhanced with the application of the electric field. This permits drugs, chemicals, or DNA to get presented within the cell.
The procedure of electroporation is generally used in the field of microbiology to transform yeast, bacteria, or protoplasts of plants by presenting novel coding DNA.
The sex of the person that you are attracted to would
determine your
sexual identification.
o gender-role attractions
sexual orientation.
sexuality
Answer:
Gender role attraction determine your sexual identificaton.
Explanation:
Uncellular eukaryotic organisms belong to?
All eukaryotic unicellular organisms belong to Protista.
Answer:
Unicellular eukaryotic organisms belong to kingdom protista.
Explanation:
Unicellular eukaryotic organisms belong to kingdom protista. They depend on other organisms for their food since they're not able to produce their own.
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The mechanism by which proteins synthesized in the cytosol are imported into the mitochondrial matrix is different from the mechanism by which proteins enter the nucleus, yet the two mechanisms do share some features. Indicate whether each of the following statements applies to nuclear import, mitochondrial import, both, or neither. Sort each statement into the appropriate bin.
1. The polypeptide to be transported into the organelle has a specific short stretch of amino acids that targets the polypeptide to the organelle.
2. The imported protein enters the organelle through some sort of protein pore
3. The pore complex consists of more than two dozen proteins and is large enough to be readily seen with the electron microscope.
4. ATP hydrolysis is known to be required for the translocation process
5. The signal sequence is always at the olypeptide's N-terminus and is cut off by a peptidase within the organelle
6. The signal sequence is recognized
GTP hydrolysis is known to be required for the translocation process
7. There is strong evidence for the involvement of chaperone proteins polypeptide's N-terminus and isctand bound by a receptor protein inrequired for the translocation during translocation of the protein the organelle's outer membrane process.
Nuclear import Mitochondrial import Both Neither
Answer:
Nuclear import:
1. GTP hydrolysis is known to be required for the translocation process
2. The pore complex consists of more than two dozen proteins and is large enough to be readily seen through electron microscopy
Mitochondrial import:
1. ATP hydrolysis is known to be required for the translocation process
2. The signal sequence is always at the polypeptide's N-terminus and is cut off by a peptidase inside the organelle
3. There is strong evidence for the involvement of chaperones during the translocation process
4. The signal sequence is recognized and bound by a receptor protein in the organelles outer membrane
Both:
1. The polypeptide to be transferred into the organelle has a specific short stretch of amino acids that targets the polypeptide to the organelle
2. The imported protein enters the organelle through some sort of protein pore
Explanation:
why was an Archea considered as a prokaryote
Answer:
Archaea is prokaryotic because it doesn't have a nucleus, nor any membrane-bound organelles. The reason why Archaea is different from Bacteria is it is more prominent, most of them are extremophiles.
because they are composed of cell which doesn't have nucleus and other some cell organelles and also considered as incomplte cell.
what is a part of interphase A. cytokinesis B. telophase C. anaphase D. s phase
Answer:
D
Explanation:
I did it on my own soci know it's the answer
In the context of interphase, the S phase is the specific part where DNA replication occurs, contributing to the overall process of cell division and ensuring the transmission of genetic material to new cells and the correct option is option D.
Interphase is the longest phase of the cell cycle, and it is divided into three main stages: G1 phase, S phase, and G2 phase. Among these stages, the S phase is specifically responsible for DNA replication. It is during the S phase that the cell's DNA is duplicated in order to prepare for cell division.
During the S phase, the DNA helix unwinds, and the two strands separate. Enzymes and proteins then work together to synthesize new complementary strands for each of the separated DNA strands. As a result, two identical copies of the DNA molecule are produced.
The S phase plays a crucial role in maintaining the genetic integrity of the cell. By accurately duplicating the DNA, the cell ensures that each daughter cell formed during cell division receives an identical set of genetic information. This is vital for the proper functioning and development of organisms.
Thus, the ideal selection is option D.
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1.
behavior is inherited and
2. When a deer hears a sound in the forest, such a
then bolts through the bushes. This is an examp
3. Mating rituals or displays can be very complex
4. When animals fight and one wins and one subm
This means that one animal (the winn
behavior is used to intimidate other an
baring teeth or growling.
Answer:
Behavior is inherited in many animals which helps them to survive.
Explanation:
When animals fight with one another and one wins. This means that one animal (the winner) behavior is used to intimidate other. Two deer fight each other in order to show dominance over each other and to impress the female deer. The winner deer will be the leader and also mate with the females of the group. This is a type of behavior which is inherited in animals through genetics is called instincts.
 What is the most likely explanation for how speciation occurs?
A. It occurs by more than one method.
B. It occurs by gradual changes.
C. It occurs only by geographic isolation.
D. It occurs only by natural selection.
Answer:
The answer is (C) geographical isolation
Explanation:
Speciation is most likely to occur when populations become geographically isolated from one another, and come to inhabit environments that differ significantly from one another.
the clean water act ___
set maximum contaminant levels for drinking water
required industries to reduce or eliminate point source pollution in surface waters
established national ambient air quality standards
requires that pollution control devices be installed on cars and trucks and at factories
Answer: The Clean Water Act required industries to reduce or eliminate point source pollution in surface waters
What was the population density of the original sample? What would have happened if you had inoculated an agar plate with 1 mL of the original sample?
Answer:
The sample was at a population density of 8.75x10^4 CFU/mL originally.
Explanation:
Had it been inoculated in the way as described in the question statement, the bacterial growth would have covered all of the agar plate and we would not have been able to distinguish individual colonies of bacteria. 1 mL is too much of a sample volume to use in cases like these.
The graph below shows the temperature of ocean water at different depths. Which part of the graph represents the thermocline? a. AB b. BC c. CD d. DE
Answer:
C
Explanation:
Generally temperature changes rapidly with the dept in aquatic bodies especially the oceans.The thin distinct or transition layer where the variation in temperature with depth is most significant is called Thermocline.
Its represents the gaps which separate the upper mixed layer from the lower deeper water bed below,it therefore represent the layer of mixture of the upper surface warmer water (C) below 200 meters, with the cooler water layer(E) below,from 1000 meters check the attachment.
It is caused by variation in intensity of sunlight.The latter hits the surface water layer and therefore raises the temperature.But,because of decrease in sunlight penetration,with increasing dept, the temperature of the lower water dept decreases.
A circular double-stranded DNA molecule contains 4200 base pairs. In solution, DNA is in a B-form helix, with about 10.5 base pairs/turn. The DNA has 12 superhelical turns. What is its superhelix density
Answer:
The correct answer is - 3 × 10⁻²
Explanation:
first find LKo = length of BDNA/ 10.5 bp/turn
= 4200/10.5
= 400
now calculate Δ LK= LK - LKo
ΔLK = 412-400
ΔLK = 12
now, superhelix density = ΔLK/LKo
⇒ 12/400 = 0.03
= 3×10⁻²
Thus, the correct answer is - 3 × 10⁻²
Jessica weighs 125 pounds. She rode a bike at 17mph for 25 minutes. What is the calorie cost of this activity?
a. 185 calories
b. 178 calories
c. 180 calories
d. 190 calories
e. 188 calories
Answer:
b. 178 calories
ok so 17mph is 0.57 weight/min
so what were going to do is multiply 125 times 0.57 which is 7.125
then multiple that answer by 25 mintues: 7.125 x 25 = 178.12...etc
now if we round this up it stays as 178!
Which structure of a paramecium is analogous to a “reserve copy” of all the cell’s genes
Answer:
micro nucleus
Explanation: