Answer:
Tight junctions - prevent liquid from seeping between cells
Desmosomes - act as rivets to hold adjacent cells together when epithelial tissue moves
Gap junctions - allow movement of cytosol, ions, and small molecules between animal cells
Plasmodesmata - openings through the cell walls of plant cells that allow adjacent cells to share materials
Explanation:
1) Tight junctions are one of the cell junctions found in animal cells. Tight junctions function to prevent the flow of liquid materials between cells.
2) Desmosomes are another type of cell junctions whose function is to form a connection between two adjacent cells. The structure formed by this connection confers strength upon the tissues involved.
3) Gap junctions are the most commonly found cell junctions found in animal cells that connects adjacent cells allowing the passage of cytosol, ions and other small molecules in them from one adjacent cell to another.
4) Plasmodesmata is a cell junction found in plant cells. They are small openings lying across the cell wall of plant cells whose function is to connect the cells and facilitate the movement of materials from one cell to another.
Tight junctions prevent liquid seepage between cells, desmosomes hold cells together, and gap junctions allow movement between animal cells.
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When a cell with 18 chromosomes divides by meiosis, the result would be four daughter cells, each with 9 chromosomes, option(c). This is because meiosis reduces the number of chromosomes in each daughter cell by half.
The best description of the expected daughtercells when a cell with 18 chromosomes divides by meiosis is: The result would be four daughter cells, each with 9 chromosomes. This is because meiosis is a type of cell division that results in four daughter cells each with half the number of chromosomes of the parent cell. The process of meiosis begins with a parent cell that is diploid, meaning it has two complete sets of chromosomes. During meiosis, this cell undergoes two rounds of division, resulting in four haploid daughter cells, with each containing half the number of chromosomes as the original parent cell. In this case, the original cell has 18 chromosomes, so each of the four daughter cells will end up with 9 chromosomes, option(c).
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Question: A cell with 18 chromosomes divides by meiosis. Which of the following statements best describes the expected daughter cells?
a. The result would be two daughter cells, each with 18 chromosomes.
b. The result would be four daughter cells, each with 18 chromosomes.
c. The result would be four daughter cells, each with 9 chromosomes
d. The result would be two daughter cells, each with 9 chromosomes.
Answer: there are four daughter cells. Each daughter cell has different characteristics from the parent
Explanation: did it
Place a sterile bandage.
Clean the wound
Apply pressure
the heart
Answer:
clean the wound
Place a sterile bandage
Apply pressure
Raise the wounded area above the heart
Explanation:
this is the correct answer on Edge
b. False
It is true that in the DNAisolation process, cells are mixed with sodium chloride (i.e. NaCl) because the sodium (Na ) neutralizes the negative charge of DNA.
DNA extraction is a method of separating DNA from cell membranes, proteins, and other cellularcomponents using physical and/or chemical methods from a sample. In 1869, Friedrich Miescher isolated DNA for the first time.
The ability to extract DNA is critical for studying the genetic causes of disease and developing diagnostics and drugs.
It is also required for forensicscience, genome sequencing, detecting bacteria and viruses in the environment, and determining paternity.
Because sodium (Na+) neutralizes the negative charge of DNA, cells are mixed with sodium chloride (i.e. NaCl) during the DNA isolation process. It makes homogenization easier.
Thus, the given statement is true.
For more details regarding DNA isolation, visit:
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Answer:
this is true it neutralise easily
The correct answer is D Vitamin D. The lack of this vitamin alters bone maturation and development.
Answer:
Explanation:
Fungi grows more in damp conditions. There is also more food availibility of food in the forest for fungi. There is more dead matter in forests that fields.
Fungi and other decomposers feasst on dead matter and break them down. This increases soil fertility and allows more plants to grow.
Answer:
The role fungi plays is a decomposer.
Explanation:
Decomposers are plants that don't make their own food. They get their food from breaking down dead plants and animals. Usually you see fungi in forests because its more moist and has more greens. A field is an open space and has too much sunlight, not many animals die in a field as in a forest its more common.
I hope this answers your question. :)) ~ kinzey
Answer:
0.549 is the frequency of the F allele.
0.495 is the frequency of the Ff genotype.
Explanation:
FF or Ff genotypes determine freckles, ff determines lack of freckels.
In this class of 123 students, 98 have freckles (and 123-98= 25 do not).
If the class is in Hardy-Weinberg equilibrium for this trait, then the genotypic frequency of the ff genotype is:
q²= 25/123
q²=0.203
q=
q= 0.451
q is the frequency of the recessive f allele.
Given p the frequency of the dominant F allele, we know that:
p+q=1, therefore p=1-q
p=0.549 is the frequency of the F allele.
The frequency of the Ff genotype is 2pq. Therefore:
2pq=2×0.549×0.451
2pq=0.495 is the frequency of the Ff genotype.
The frequency of the dominant allele, F, in this class is 0.55. The frequency of the heterozygous genotype, Ff, is 0.495. This is calculated using Hardy-Weinberg equilibrium and observed phenotype frequencies.
To start, we need to calculate the frequency of the recessive allele, f, which is easily calculated as those who do not have freckles. From a total of 123 students, 98 have freckles, leaving 25 students with no freckles, which represents individuals who are homozygous for the recessive trait (ff). As these are the only individuals we can be sure of, we take the square root of their frequency to get the frequency of the recessive allele, q. In this case, q = sqrt(25/123) = 0.45. To find the frequency of the dominant allele, p, we subtract q from 1 (because p + q = 1), so p = 1 - q = 0.55.
Next, we'll calculate the frequency of the heterozygous genotype Ff.
Using Hardy-Weinberg equilibrium, we know this is represented by 2pq. Hence, the frequency of genotype Ff would be 2 × 0.55 × 0.45 = 0.495.
This process offers an example of applying the principles of population genetics and Hardy-Weinberg equilibrium to determine the likely genotype frequencies in a given group of individuals based on observed phenotype frequencies.
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