Answer:
The correct answer is C. be more likely to have cancer at some point in his/her life.
Explanation:
Cell cycle checkpoints are controlled by different proteins which are coded by different genes. They are important in cell cycle regulation and stops cell cycle immediately if they find any problem in the cell like mutation.
The cell cycle will not be continued if the problem does not solve and cell death will occur which prevents cancer by preventing the mutation to spread into new cells.
Therefore if a mutation occurs in certain genes that enforce a checkpoint on the cell cycle then the mutated gene will pass in new cells and the mutated cells will grow uncontrollably due to no checkpoints so the person will probably have cancer at some point in his or her life.
In the Cori cycle, when glucose is degraded by glycolysis to lactate in muscle, the lactate is excreted into the blood and returns to the liver. In the liver, lactate is converted back into glucose by gluconeogenesis. For each given enzyme, identify whether it is involved in the glycolysis pathyway, gluconeogenesis pathway, both pathways, or neither pathway.1. glyceraldehyde 3-phosphate dehydrogenase O glycolysis O gluconeogenesis O both O neither 2. glucose-6-phosphatase O glycolysis O gluconeogenesis both O neither 3. alcohol dehydrogenase O glycolysis O gluconeogenesis both neither 4. phosphoenolpyruvate carboxykinase O glycolysis O gluconeogenesis O both 5. phosphofructokinase-1 O glycolysis O gluconeogenesis both O neither 6. phosphoglycerate mutase O glycolysis gluconeogenesis O both neither 7. hexokinase O glycolysis O gluconeogenesis. O both neither 8. pyruvate dehydrogenase O glycolysis gluconeogenesis
Answer:
Explanation:
1. glyceraldehyde 3-phosphate dehydrogenase: O both
2. glucose-6-phosphatase: O gluconeogenesis
3. alcohol dehydrogenase: O neither
4. phosphoenolpyruvate carboxykinase: O gluconeogenesis
5. phosphofructokinase-1: O glycolysis
6. phosphoglycerate mutase: O both
7. hexokinase: O glycolysis
8. pyruvate dehydrogenase: O neither
Classify the steps involved in the formation of a solution as being endothermic or exothermic. separation of solute molecules separation of solvent molecules formation of solute-solvent interactions
Answer:
Endothermic reactions: separation of solute molecules , separation of solvent molecules
Exothermic: formation of solute-solvent interactions
Explanation:
An endothermic reaction can be described as a reaction which requires energy. Usually, the energy is taken in the form of heat from the environment. For example: the separation of solute and solvent molecules.
An exothermic reaction can be described as a reaction in which energy is released during a reaction. For example, a solute and a solvent reaction. Energy is usually released in the form of heat during an exothermic process.
Classifying the steps involved in the formation of a solution as being endothermic or exothermic:
endothermic
separation of solute moleculesseparation of solvent moleculesexothermic
formation of solute-solvent interactionsEndothermicIn this reaction heat or energy is absorbed in the process of producing the product. Breaking of bonds requires energy, so the separation term is endothermic.
separation of solute molecules and separation of solvent molecules both are endothermic.
ExothermicWith the product formation, energy or heat is also released in such a reaction. The interaction of solute solvent is exothermic as it releases heat in the process.
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What happens to pyruvate in the presence of oxygen? It enters the mitochondrial matrix and is converted to actyl CoA. It remains in the cytosol and is converted to ethanol. It remains in the cytosol and is converted to lactate. It enters the mitochondrial matrix and combines with oxaloacetate to form citrate. g
Answer:In glycolysis , Oxygen + Pyruvate molecules = acetyl-CoA + CO2
Where "=" means in the presence of enzyme complex pyruvate dehydrogenase complex (PDC)
Acetyl-CoA - NADH
Pyruvate oxidation or Pyruvate decarboxylation also known as link reaction, is the conversion of pyruvate into acetyl-CoA by the enzyme complex pyruvate dehydrogenase complex
Explanation:
A carboxyl group is trimmed off of Pyruvate and released as a molecule of carbon dioxide, leaving behind a two carbon molecule.
The two carbon molecule from above is oxidised - Acetyl group, and the electrons lost in the oxidation are picked up by NAD+ to form NADH
An acetyl group is transfered to Coenzyme A, resulting in acetyl CoA
Suppose you are investigating an autosomal recessive disease known as "studius toxicosis" which occurs at a rate in the American college student population of 1 in 16 individuals. Students who inherited two recessive alleles (tt) have the disease. If we assume Hardy-Weinberg equilibrium in a large population of college students, what is the percentage of heterozygous carriers of the studius toxicosis allele in the population? That is, individuals who have the genotype Tt. Refer to the Hardy-Weinberg equation. Be sure to answer in a percentage, not decimal, but do not include the % sign. Example: answer 45, not 0.45.
Answer:
37.5
Explanation:
Acording to the the Hardy-Weinberg equation:
[tex]p^2 +pq+q^2=1[/tex]
[tex]p+q = 1[/tex]
Where p is the frequency of the dominant allele "T", and q is the frequency of the recessive allele "t".
Therefore, p^2 is the frequency of the dominant homozygous genotype "TT", q^2 is the frequency of the recessive homozygous genotype "tt" and
2pq is the frequency of the heterozygous genotypes "Tt" and "tT".
Since 1 in 16 individuals have inherited two recessive alleles (tt) and have the disease:
[tex]q^2=\frac{1}{16} \\q=\sqrt{\frac{1}{16}} \\q= 0.25[/tex]
Thus, the fraction of heterozygous carriers of the studius toxicosis allele in the population is given by:
[tex]p=1-q\\p = 1 - 0.25 = 0.75\\2pq=2*(0.75)*(0.25)\\2pq= 0.375[/tex]
37.5 percent of the population are heterozygous carriers
If skeletal muscles work to the point of fatigue, the muscle cells may not have sufficient oxygen to carry out aerobic respiration. What processes do muscle cells use if oxygen is not available for aerobic respiration?a. lactic acid fermentation b. citric acid cycle c. ethanol fermentation d. glycolysis
Answer:
a. lactic acid fermentation
Explanation:
Under conditions of intense exercise, the oxygen gas obtained by pulmonary respiration may be insufficient to meet the needs of muscle cells in the work of obtaining energy from cellular respiration.
However, even in the absence of oxygen gas, our muscle cells can release the available energy in glucose, leading to even smaller amounts of ATP molecules. Under these conditions, muscle cells perform lactic fermentation, a process that is virtually identical to glycolysis (the first set of cellular respiration reactions), except that pyruvic acid is transformed into lactic acid with the formation of 2 ATPs. Despite the lower energy yield, fermentation ensures the energy supply to the muscle. Lactic acid formed under these conditions has been associated with muscle pain and fatigue characteristic of intense physical exercise. Recent research, however, has shown that pain is caused by muscle fiber micro-injuries rather than lactic acid as it is rapidly metabolized and eliminated.
Which of the following is an example of exploitation?
A. ants eating trash
B. humans that create a beehive
C. flowers blooming
D. rabbits that feed on a homeowner's garden
Answer:
B.
Explanation:
The answer is B, because bee farmers create false bee hives for h.bees to start a hive. They also can purchase a queen, (that reproduce more bees) that is her only job. The worker bees collect pollen and mix it with the bee's saliva to store honey for the hive, which also is food 4 the bees.
the vertices of a triangle are (2,2),(4,4),and (4,2). Reflect the triangle in the y-axis. Give the coordinates of the reflected triangle
Why are non-native tree diseases and insects often so much more virulent than their native counterparts?
Answer:
Non-native plants reduce the diversity of insect populations in gardens, even where the non-native plants are closely related to the native plants, new research shows. ... local insect communities than their non-native counterparts,
Explanation:
Non-native tree diseases and insects can be more virulent in new environments due to factors like the absence of natural predators, a lack of host plant resistance, fewer competing species, and escape from natural constraints.
What happens with the diseasesAdditionally, they may lack genetic diversity, undergo rapid evolution, and be spread by human activities.
These factors, combined with the vulnerability of non-native ecosystems, make them more destructive in their new environments, requiring various strategies to manage their impact.
We can conclude that the reason is because these trees do not competitors as well as the absence of predators that may have destroyed the species.
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Which of the following is NOT part of genetic "linkage mapping"?
A. Calculating recombination frequencies using a testcross
B. Assuming that recombination frequency is proportional to the distance
between two linked genes
C. Finding and studying alternative (e.g., mutant) phenotypes of a character
D. Using information on DNA sequences
E. Studying characters two at a time
Answer:
The correct answer is option D.
Explanation:
Linkage mapping is the mapping process in which the genes present on the chromosomes are mapped on the base of their linkage. The linkage mapping helps in calculating how frequent recombination occurs using testcross.
Linkage mapping does not utilize the information that is on DNA sequences, however, it helps in assuming the distance of two linked genes is proportional of the recombination frequency. By the recombination, frequency mutations can be found and study.
Thus, the correct answer is option D.
The skin color variation of ___________ represents an increase in tissue bili/rubin from RBC destruction or liver cell destruction (hepat/itis). This color is best seen in the sclera of the eye, body membranes, and skin.
Answer:
Jaundice individual.
Explanation:
Jaundice may be defined as the medical condition in which the body of an individual turns into yellow. The different strains of the hepatitis virus is responsible for this infection.
The yellow color is deposited in the eyes, nails and membrane of the skin due to deposition of excess yellow orange pigment known as bilirubin. The alcohol can also cause the jaundice condition in the individual and destroys the proper liver functioning.
Thus, the answer is jaundice.
The law of independent assortment allows for genetic recombination. The following equation can be used to determine the total number of possible genotype combinations for any particular number of genes: 2g = Number of possible genotype combinations (where g is the number of genes)1 gene: 21 = 2 genotypes2 genes: 22 = 4 genotypes3 genes: 23 = 8 genotypesa) Consider the following genotype: YySsTt. How many different gamete combinations can be produced?
Answer:
Gametes that will be produced are:
YST, YSt, YsT, Yst, yST, ySt, ysT, yst
Explanation:
From YySsTt, it is obvious that as there 3 type of genes.
The number of gametes = [tex]2^{3}[/tex] = 8
5 genes = [tex]2^{5}[/tex] = 32 genotypes
10 genes = [tex]2^{10}[/tex] = 1024 genotypes
20 genes = [tex]2^{20}[/tex] = 1048576 genotypes
Below is a hypothetical developmental pathway. This pathway leads to the expression of a black pigment in certain cells of the ectoderm (the spot cells) which results in the ladybug's spotted appearance. Cells of the ectoderm where black pigment is not produced are red.
L ---| D → Y ---| B → black
Based on the pathway above, will the spot cells in ladybug mutant lacking the D gene be able to make black pigment?
Select one:
O not enough information
O yes
O no
No, in a ladybug mutant lacking the D gene, the spot cells would not be able to make black pigment.
Why is No the best choiceThe developmental pathway indicates that the expression of black pigment in spot cells (ectoderm cells) depends on the presence of the D gene. Without the D gene, the pathway is interrupted, and the subsequent steps (Y → B) that lead to the production of black pigment cannot occur.
Therefore, in a ladybug mutant lacking the D gene, the spot cells would not be able to make black pigment.
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Which of the following is an example of homologous structures?
A. Whale hip bones.
B. Embryos of fish and birds both make gill slits.
C. Mice and chicken differ by only 25 amino acids when they make hemoglobin protein.
D. Rabbits and birds have the same bones in the same order in their forelimbs, even though they use them for different purposes.
Final answer:
Homologous structures are those that have a similar structure and origin but may have different functions in various species. The example of rabbits and birds, which have the same bones in the same order in their forelimbs for different purposes, represents homologous structures.(Option D)
Explanation:
The student's question asks which of the following is an example of homologous structures. Homologous structures are those that are found in different species and share a common ancestral origin, despite possibly having different functions in the organisms where they're found.
Considering this, the correct answer is D: Rabbits and birds have the same bones in the same order in their forelimbs, even though they use them for different purposes. This resembles the forelimbs of mammals like humans, cats, and whales, as well as the wing bones of bats, which while serving different purposes, reflect a common structural plan pointing to a shared evolutionary history.
Color-blindness is an X-linked recessive disorder. Mike is color-blind. His wife, Meg, is homozygous for normal color vision allele. Using Punnett squares, derive and compare the genotypic and phenotypic ratios for the offspring of this marriage.
a. What is Mikes's genotype?b. What is Meg's genotype?c. What is the genotypic ratio of offspring?d. What is the phenotypic ratio of offspring?e. If they have eight children, how many of them would you expect to be color-blind?
Mike's genotype is XcY, Meg's is XX. Using Punnett square, the genotypic ratio of their offspring is 1:1 and the phenotypic ratio is 100% normal vision. If they have eight children, none are expected to be color-blind.
Explanation:In terms of genetics, color blindness is an X-linked recessive disorder. Mike, being color-blind, must have the genotype XcY. His wife, Meg, has normal color vision and is homozygous dominant, so her genotype is XX.
When we use a Punnett Square to derive the genotypic ratio of their offspring, we see that all daughters will be heterozygous carriers (XcX) and all sons will have normal vision (XY). Thus, the genotypic ratio is 1:1 (XcX: XcY).
Regarding the phenotypic ratio, all offspring are expected to have normal vision because they will inherit at least one normal vision allele. Therefore, the phenotypic ratio is 100% normal vision.
If Mike and Meg have eight children, because female offspring will be carriers and male offspring will possess the normal vision allele, we would expect none of them to be color-blind.
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During the dissection, you used your hands to follow the digestive tract anteriorly and posteriorly. Outline the path that food takes through the frog’s digestive tract: start at the mouth, indicate the major organs through which food passes, and end at the cloaca
Final answer:
Food in a frog's digestive system flows from the mouth, through the pharynx, esophagus, and stomach, into the small intestine for nutrient absorption, then to the large intestine for water absorption, and finally out through the cloaca as waste.
Explanation:
The path that food takes through a frog's digestive tract begins at the mouth where ingestion occurs. From there, food moves into the pharynx, then down through the esophagus and into the stomach where preliminary digestion takes place. After the stomach, food enters the small intestine where nutrients are absorbed. It then proceeds into the large intestine, where water is absorbed, and finally, the undigested waste products are expelled through the cloaca.
___________ is a measure of the degree to which the phenotypic variation of a given trait is due to genetic factors.
In a long bone, the osteons are:
A. lined up in the same direction as the diaphysis of the bone
B. lined up perpendicular to the long axis of the bone, in the direction of perforating canals.
C. arranged in an irregular pattern
D. are separated by medullary spaces
E. are lacking in the diaphysis of the bone
Answer:
B. lined up perpendicular to the long axis of the bone, in the direction of perforating canals.
Explanation:
Osteons are formations of concentric bone layers (lamellae) that tend to run parallel to the long axis of a bone. They sorround the Haversian canal, a canal that contains blood vessels that supply the osteocytes and are the structural unit of compact bone.
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How often do human skin cells divide? Why might that be? Compare this rate to how frequently human neurons divide. What do you notice?
Answer:
explained
Explanation:
The skin cells divide every day, as the epithelial layer is the outer protective layer, it needs to be replaced from time to time, which helps in healing wear and tear. Our skin cells divide once in each day.
In contrast, the nerve cells do not divide at all. That is their frequency of cell division is zero. They cannot divide because they lack centriole in them.
Human skin cells undergo division approximately every 27 days. This rapid turnover is necessary to replenish and repair the skin, which is exposed to constant wear and tear, environmental factors, and potential damage.
On the other hand, human neurons generally do not undergo division once they are fully developed. Unlike skin cells, neurons are post-mitotic, meaning they have exited the cell cycle and have limited or no ability to divide further.
The notable difference in the division rate between skin cells and neurons highlights the contrasting needs and functions of these cell types.
Thus, skin cells require frequent division to maintain the integrity and functionality of the skin, while neurons focus on specialized functions and long-term stability rather than continuous replication.
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A female rabbit heterozygous at two genes (named A and B) is test-crossed with a male rabbit. Their progeny include 442 A/a;B/b, 458 a/a;b/b, 46 A/a;b/b, and 54 a/a;B/b. Explain these results. Identify which alleles are recessive, whether genes A and B are linked, and, if so, how far apart they are on the chromosome
Answer:
The genes are linked and 10 mu apart.
Explanation:
A female AaBb rabbit is test crossed with a male rabbit (aabb). The male can only produce ab gametes (all the progeny will have ab on one of the homologous chromosomes).
If the genes assorted independently, the female would produce 4 types of gametes with the same frequency: 1/4 AB, 1/4 Ab, 1/4 aB and 1/4 ab.
However, the observed AB and ab gametes were much more frequent than Ab and aB, which means that the genes are linked and alleles on the same chromosome do not assort independently during meiosis.
Recombination is a rare event, so the most abundant gametes are the parentals. That is how we know that the mother had the AB/ab genotype. The recombinant gametes therefore are Ab and aB.
Distance (mu) = # Recombinants × 100/ Total progeny
Distance = (54 + 46) × 100/ 1000
Distance = 100 × 100/1000
Distance = 10 mu
A male Flagus fly with the Barkus phenotype is crossed with a female who has the wild-type phenotype. A second cross is performed in which the female Flagus fly has the Barkus phenotype and the male has the wild-type phenotype. This second cross is known as a:
Answer:
Reciprocal cross
Explanation:
a reciprocal cross is a breeding experiment designed to test the role of parental sex on a given inheritance pattern. In this type of test cross, male of one genotype and female of different genotype are crossed and the opposite is true as well. ( i.e the female of one genotype and male of different genotype are crossed.)
As we can see in the male Flagus fly with the Barkus phenotype is being crossed with a female with the wild-type phenotype.
In the second test, we can see that the opposite is true of our first statement where the female Flagus fly has the Barkus phenotype and the male has the wild-type phenotype after being crossed again thereby reversing the genotype.
Please help me!! I don’t understand
Answer:
The correct option is A. An airplane lifting into the sky during take- off.
Explanation:
The force of the gravitational pull is known to act downwards. The objects which will move along this force i.e will move downwards will have positive action.
A negative action will act against the gravitational pull i.e a force that is applied in the upward direction. As an aeroplane is lifted upwards during taking- off i.e against the gravitational pull of the earth hence it will be considered as a negative action.
Why is a frameshift missense mutation more likely to have a severe effect on phenotype than a nucleotide-pair substitution missense mutation in the same protein? Why is a frameshift missense mutation more likely to have a severe effect on phenotype than a nucleotide-pair substitution missense mutation in the same protein? A frameshift missense will cause the codons to be out of order, but a substitution missense does not change the order of the codons. A substitution missense mutation causes the protein to be shorter and thus non-functional. A substitution missense affects only one codon, but a frameshift missense affects all codons downstream of the frameshift. A frameshift missense mutation will cause an early Stop codon, but a substitution missense might be silent.
A substitution missense affects only one codon, but a frameshift missense affects all codons downstream of the frameshift.
A frameshift mutation occurs during deletion of one of two nucleotides. The arrangement of codons downstream of the mutation will definitely change, hence it being called a frameshift mutation. On the other hand, a substitution mutation will only cause a change in one nucleotide of one codon.
Explanation:
A frameshift mutation causes new codons downstream of the mutation that will code for different amino acids hence changing the properties of the translated proteins. This will have great ramifications for the phenotype of the organism.
On the other hand, substitution mutation will only cause a change in the amino acid at that point of mutation during translation. This will not have a drastic change in the protein hence will not be as lethal as a frameshift mutation.
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A frameshift mutation typically has a more severe effect on protein function than a nucleotide-pair substitution missense mutation because it changes the reading frame, leading to widespread amino acid changes and potentially a nonfunctional protein, while a missense mutation usually alters a single amino acid and may still produce a functional protein.
Explanation:A frameshift mutation is caused by insertions or deletions of nucleotides that are not in multiples of three, leading to a major shift in the genetic reading frame. This results in a different set of amino acids being encoded from the mutation point onward, which often produces a nonfunctional protein. Contrarily, a nucleotide-pair substitution mutation, also known as a missense mutation, typically changes only a single amino acid in the protein. This type of mutation has a less profound effect compared to a frameshift mutation because it does not alter the reading frame for the rest of the coding sequence.
A missense mutation from a nucleotide-pair substitution may sometimes lead to a protein that retains functionality, especially if the new amino acid is chemically similar to the original. However, if that amino acid is in a critical part of the protein, such as the active site, then the consequences can be more severe. In contrast, a frameshift mutation can result in a completely altered protein with many incorrect amino acids, which is more likely to be nonfunctional and have a severe effect on the phenotype.
It is important to note that while a missense mutation affects only the specific codon where the substitution occurs, a frameshift mutation has the potential to affect all downstream codons after the mutation point. Additionally, a frameshift mutation can introduce an early stop codon, prematurely ending the protein, while a missense mutation does not introduce a stop codon and usually only impacts the incorporation of a single amino acid.
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List and describe four stages/processes within a typical animal excretory system.
Answer:
The four stages of the excretory system are:
1. Filtration
2. Reabsoprtion
3. Secretion
4. Excretion
Explanation:
Filtration - Waste is taken out of the bodily fluids by pressure filtration
Reabsorption - Clean bodily fluids are reabsorbed back into the body/blood
Secretion - Adding of toxins and wastes from the body fluids to the filtrate
Excretion - Expulsion/release of the waste products out of the body through urine, sweating, etc
Final answer:
The four stages/processes within a typical animal excretory system are filtration, reabsorption, secretion, and excretion. During filtration, blood is filtered in the kidneys to remove waste products and excess water. Reabsorption involves the reabsorption of useful substances, while secretion eliminates waste materials. The final stage is excretion, where the urine is passed out of the body through the urethra.
Explanation:
The excretory system in animals is responsible for eliminating waste and excess water from the body. The four stages/processes within a typical animal excretory system include filtration, reabsorption, secretion, and excretion.
Filtration: In this stage, blood is filtered in the kidneys to remove waste products and excess water. Small molecules like water, glucose, amino acids, and ions pass through the filtration membrane, while larger molecules like proteins and blood cells are retained.
Reabsorption: During reabsorption, useful substances like glucose, amino acids, and ions are reabsorbed from the filtrate back into the bloodstream. This process helps maintain the body's essential nutrients and ions.
Secretion: Secretion involves the active transport of waste products, toxins, and excess ions from the bloodstream into the filtrate. This further helps in eliminating waste materials that were not filtered during the initial stage.
Excretion: The final stage of the excretory system is excretion, where the filtrate, now known as urine, is passed out of the body through the urethra. Urine is a concentrated solution containing waste products like urea, water, excess ions, and other substances.
Describe the similarities and differences between voltaic and electrolytic cells. Be sure to address the following items in your description: spontaneous reduction or oxidation reactions and which electrode(s) they occur at charges on electrodes .
Answer:
Both cell are opposite to each other in function and make
Explanation:
a) A voltaic cell works opposite to that of electrolytic cell. It produces chemical energy from electrical energy while the later one produces electrical energy from chemical energy.
b) The chemical reactions in voltaic cell are spontaneous while chemical reactions in electrolytic cells are non –spontaneous.
c) The anode and cathode carry opposite charges in the two cells. In case of voltaic cell, anode carries negative charge while cathode carries positive charge.However, in case of electrolytic cell both cathode and anode bear opposite charges wrt voltaic cell
Metabolism can be bisected into two subcategories:
(A) catabolism and anabolism.
(B) takes complex organic molecules and breaks them down into simpler molecules; this is often accompanied by the of energy.
(C) builds up biomolecules from simpler substances; this is often accompanied by the of energy.
Answer: A (catabolism and Anabolism)
Explanation:
Metabolic pathway are basically divided into two categories.
1. Catabolic
2. Anabolic
Catabolic (degradation) pathways, where energy rich complex macramolecules are degraded into smaller molecules. Energy released during this is trapped as chemical energy, usually as ATP.
Anabolic (biosynthesis) pathways. The cells synthesize complex molecules from simple precursors. This needs energy.
Securin is a cytoplasmic protein that binds to separase, an enzyme that degrades cohesin. When separase is bound to securin, separase is inactive. When separase is released, it immediately becomes active. Securin and separase remain attached to each other as long as securin is phosphorylated. Based on the role of securin, it must become dephosphorylated just prior to
metaphase.
anaphase.
prometaphase.
prophase.
telophase.
cytokinesis.
Answer:
Anaphase.
Explanation:
Two main process of cell division are mitosis and meiosis. The regulation of the cell cycle is maintained by the regulation of cyclins, cdks and important protein factors.
Securin and separase are important proteins that allow the transition of one phase of the cell cycle to the another phase. The securin helps in the separation of the chromosome and sister chromatid that occurs in anaphase. Hence, dephosphorylation must occur prior to anaphase.
Thus, the correct answer is option (2).
plasmodesmata are cell junctions that are found between ________. adjacent animal cells in the same tissue type individual cardiac cells in heart muscle tissue the plasma membrane of actively dividing prokaryotes
The question is incomplete. The complete question is:
Plasmodesmata are cell junctions that are found between ________.
a) adjacent animal cells in the same tissue type
b) individual cardiac cells in heart muscle tissue
c) the plasma membrane of actively dividing prokaryotes
d) adjacent plant cells
Answer:
d) adjacent plant cells
Explanation:
Plasmodesmata are the cytoplasmic channels that connect the adjacent plant cells together and maintain the continuity of their cytoplasm. These are the narrow channels with 20-40 nm diameter. Plasmodesmata serve the function analogous to the gap junctions of the animal cells. The presence of plasmodesmata between the neighboring plant cells allows the movement of molecules and ions between them. Plant cells can adjust the diameter of the plasmodesmata according to the requirement to allow the transport of substances across them.
Plasmodesmata are cell junctions found specifically between adjacent plant cells, facilitating transport and communication within plant tissues.
Explanation:The question relates to plasmodesmata, which are cell junctions found between adjacent plant cells. Plasmodesmata are channels that allow the transport of ions, small molecules, and macromolecules between the cytoplasm of adjacent cells, thus facilitating communication and transport of materials within plant tissues. They are found within the cell wall and are sheathed by the plasma membrane, which is an extension of the plasma membrane of adjoining cells. This intercellular connection raises the concept of plant tissue as a potential syncytium—a network of cells sharing a common cytoplasm.
In contrast, animal cells utilize gap junctions for similar cellular communication. Gap junctions are formed by the alignment of pores of connexons—complexes made of six proteins called connexins—between adjacent animal cells, which is particularly important in cardiac muscle tissue. However, these are not found in plants and thus not related to plasmodesmata, which are unique to plant cells.
On the other hand, prokaryotes, such as bacteria, lack specialized intercellular junctions because they do not have multicellular structures like eukaryotes. Instead, they may communicate through direct contact or by releasing and detecting signaling molecules.
Suppose that RNA polymerase was transcribing a eukaryotic gene with several introns. In what order would the RNA polymerase encounter the following elements in the DNA sequence of the gene?A-stop codonB-translation initiation codonC-TATA box elementD-3' UTRE-5' UTRF-splice acceptor site
RNA polymerase, during the transcription of a eukaryotic gene with introns, would encounter DNA elements in the order of TATA box, 5' UTR, translation initiation codon, splice acceptor site, stop codon, and 3' UTR. Transcription occurs in the 5' to 3' direction, beginning with the promoter region and moving downstream.
If RNA polymerase were transcribing a eukaryotic gene with several introns, the order in which it would encounter the elements in the DNA sequence of the gene would be as follows:
C-TATA box element - This is a component of the promoter region where transcription factors assemble and RNA polymerase II binds to start transcription.E-5' UTR (Untranslated Region) - The segment of mRNA that is transcribed first but is not translated into protein.B-translation initiation codon - This is typically the AUG codon, where the ribosome will begin the translation process.F-splice acceptor site - The location within the intron where splicing will occur during mRNA processing.A-stop codon - Signals the end of the protein-coding sequence within the mRNA.D-3' UTR (Untranslated Region) - The segment of mRNA that follows the coding sequence and is involved in regulation and stability of mRNA.It's important to note that RNA polymerase II transcribes the gene from the 5' to 3' direction, starting with the promoter region and moving downstream through the gene.
Endosymbiont theory explains some of the differences and similarities between prokaryotes and eukaryotes. Categorize the following statements as relating to prokaryotes or eukaryotes by dragging and dropping the labels below. Some labels might be used more than once.
a.Prokaryotes
i.Lack membrane-bounded
ii.These were the first organisms on Earth
iii.Includes organisms that are photosynthetic
iv.Aerobic bacteria are these types of cellsv.Includes organisms that carry out aerobic respiration
b.Eukaryotes
i.Have mitochondria ii.Might have chloroplastsiii.Includes organisms that are photosyntheticiv.Includes organisms that carry out aerobic respirationv.Cen be a multicellular organism
Answer:
Endosymbiont theory was proposed by Lynn Margulis in 1967 which suggested the origin of Chloroplast and mitochondria in eukaryotic cells.
The theory suggested that chloroplast and mitochondria organelle originated by engulfing the cyanobacteria and proteobacteria by other prokaryotic bacteria.
This shows that Prokaryotes exhibit the following characters:
1. These were the first organisms on Earth
2. Lack membrane-bounded
3. Aerobic bacteria are these types of cells
Eukaryotes exhibit:
1. Have mitochondria
2. Includes organisms that carry out aerobic respiration
3. Can be a multicellular organism
4. Includes organisms that are photosynthetic
5. Might have chloroplasts
Which characteristic would not be useful in a vector? A) A recognition sequence for a restriction enzyme B) Large size relative to the host chromosomes C) The ability to replicate independently inside the host cell D) A reporter gene E) An origin of replication
Answer:
The correct answer is B) Large size relative to the host chromosomes
Explanation:
A vector for clonation or expression must contain sequence elements which allow it independent replication and identification of host cells which have incorporated the vector. Among these elements are:
- Replication origin (option E): allows it to replicate independently of the host genome. Option B is the same as it.
- Restriction enzyme sites (option A): in order to cut the vector and to inser the desired gene into the vector.
- Reporter gene (option E): it allows to visualize the host cells which are expresing the inserted gene of the vector. For example, a reporter gene can codificate a fluorescent protein, thus the host cells which have expresion of the vector are fluorescent.
The only option which is unuseful is option B because if the vector has too large size, the transformation efficiency will be low. That means that fewer host cells will contain the vector after transformation.