Answer:
Sister chromatid Cohesion is essential for the bi orientation of chromosomes on the mitotic or meiotic spindle, and is thus an essential condition for chromosome segregation.
Explanation:
Sister chromatid Cohesion:- it is the process by which sister chromatids are combined and held together during specific periods of the cell cycle.
Why Sister chromatid Cohesion is important?
This is essential for the bi orientation of chromosomes on the mitotic or meiotic spindle, and is thus an essential condition for chromosome segregation.
Protein Cohesion:- The proteins that bind the two sister chromatids, denying any premature sister chromatid partition, are a piece of the cohesion protein family. They regulates the separation of sister chromatids during cell division, either mitosis or meiosis. It hold sister chromatids together after DNA replication until anaphase stage is complete
Separase:- Separase is a protease enzyme which can cleave the Ssc1 subunit of the cohesin ring complex releasing the tension or strechness in the spindle and allowing segregation of sister chromatids.
Mammals feed their young with insects plants and roots. True or False
Answer: False
Explanation:
One of the striking characteristics of mammals that distinguishes them from other vertebrates is that they feed their young ones with breast milk. Therefore, mammals are animals that feed their young ones with breast milk.
Mammals, like bears, platypuses, and opossums, may supplement their diets with insects, plants, and roots, but they do not feed their young directly with such. Instead, they produce milk, which contains all the necessary nutrients for their young's growth and survival.
Explanation:The statement that mammals feed their young with insects, plants, and roots is false. One distinctive attribute of mammals that separates them from other animals is their method of nourishment for the young. Mammals produce milk from their mammary glands to feed their offspring. Some mammals like bears, platypuses, or opossums may supplement their diet with insects, plants, and roots but they do not feed their young directly with such. Rather, the milk they produce is packed with all the necessary nutrients the young need for their growth and survival.
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Which of the following is NOTa sign that a survivor may need stabilization?Select one:a. Excessive talkingb. Glassy and vacant eyesc. Strong emotional responsesd. Uncontrollable physical reactionse. Frantic searching behavior
Answer:
Excessive talking is not a sign that a survivor may need stabilization.
Explanation:
What symbiotic partnerships form between plant roots and fungi, and increase water and mineral absorption by the plant?
Answer: It is a Mutualistic partnership between fungi and roots of plant
Explanation:
Mutualism is a relationship between two organisms in which both of the benefits from the relationship. Fungi live in the roots of plant thereby getting carbohydrates from plants made by photosynthesis and the fungi in turn put mycelia that help plant to increase absorption of minerals and water.
Answer: Symbiosis
Explanation:
The root nodules of legumes contains organisms of the kingdom, fungi that helps the legumes fix atmospheric nitrogen thus meeting it's need for mineral Nitrogen. In turn, the legume plants provides shelter to the fungi.
This codependent partnership is known as Symbiosis
The wild-type (normal) fruit fly, Drosophila melanogaster, has straight wings and long bristles. Mutant strains have been isolated that have either curled wings or short bristles. The genes representing these two mutant traits are located on separate chromosomes.
This question is incomplete. However, I understand that it is mainly dealing with "Scale of gene alterations in Mutant strains of Fruitfly"
Answer: Mutant strains evolved from large-scale mutation that occured in the chromosomes of Wild-type Fruitfly.
Explanation:
Large scale mutations involves massive chromosomal alterations such as change in base sequence of multiple genes.
Note that the change in the base sequence of genes determining Wing shape and Hair length in Wild-type fruitfly, located on different chromosomes is said to be LARGE SCALE mutation; for genes controlling different characters would occupy different locations on different chromosomes.
Thus, the genes representing the two mutant traits are located on separate chromosomes are said to have undergone LARGE SCALE MUTATIONS
The wild-type Drosophila melanogaster has straight wings and long bristles. Mutant strains show curled wings or short bristles, with genes on separate chromosomes.
Curled Wings:
The mutant strain with curled wings displays a phenotype where the wings of the fruit fly are not straight as in the wild type, but instead, they exhibit a curled or abnormal wing structure.
This trait is a result of a genetic mutation affecting the development of wing morphology.
Short Bristles:
The other mutant strain with short bristles demonstrates a phenotype characterized by a reduction in the length of bristles compared to the long bristles of the wild type.
This mutation affects the growth or development of bristles on the body of the fruit fly.
Interestingly, the genes responsible for these two mutant traits are located on separate chromosomes.
This implies that the genetic mutations leading to curled wings and short bristles are governed by genes residing on different chromosomes.
The fact that these genes are on separate chromosomes can have implications for genetic inheritance patterns and the assortment of traits in the offspring.
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Even though RSV infection in infants is common, a vaccine does not currently exist. Imagine you are designing a recombinant vaccine for RSV--what viral components would you use in your vaccine? Justify your choice.
Explanation:
RSV is caused by a group of paramyxoviruses and belong to the group of pneumoviruses. It has two prevalent strains RSV-A and RSV-B.
To test RSV antigen in the sample we can use immunofluorescence test (IFT). IFT is a method used to identify pathogen specific antigen or antibody in sample collected from patient using fluorescence. A fluorochrome protein molecule is used to do this. Depending upon the material used and dye fluorochrome releases energy which results in fluorescence.
In this method sample collected from patient is applied on the plate and antibodies specific to the antigen are added into the plates with antigen. The antibodies used are labeled with fluorochrome to give fluorescence.
2. As viruses genetic material is prone to errors it is difficult to create a vaccine against viruses. RSV has RNA as genome and it has highly conserved lipoprotein which functions as antigen. This lipoprotein sequence is highly conserved in all the strains of the viruses. I would prefer to use this lipoprotein present in the membrane of virus for vaccine research
Contrast the genetic content and the origin of sister versus nonsister chromatids during their earliest appearance in prophase I of meiosis. How might the genetic content of these change by the time tetrads have aligned at the equatorial plate during metaphase I?
Answer:
Sister chromatids are identical forms of chromatids of a chromosomes. They are mostly formed by semi-conservative replication of DNA molecule of a single chromosome.Thus they are like 'photocopies' of original parent chromosomes; joined together at the Centromere.
They are exactly similar in all ramification; with the same gene and allele compositions..
However; slight differences arise between the two identical sisters due to mutation from errors at replication;and also in the length of telomere repeats.
Non-sister chromatids are dissimilar forms of chromatids of a chromosomes formed when each half of a chromosome at fertilisation from separate haploid sex-cells, of each parent. fused.They contain different genetic composition;because they are not on the same homologous chromosomes.Therefore crossing -over ensure variation.
However, they are genetically similar in composition; if they are contained in homologous chromosomes. This is because Synapsis of bivalent of these chromosomes allow genetic material to be shared by chromosomal crossing-over between the non-sister chromatids on the chromosomes ; therefore identical genetic characteristics are shared .
Explanation:
You decide to try Alex’s response to glucagon. This test consists of injecting a high dose of glucagonintravenously and then drawing samples of blood periodically and measuring the glucose content ofthe samples. After the glucagon injection, Alex’s blood sugar rises dramatically. Is this the responseyou would expect in a normal person? Explain
Answer:
YES!
Explanation:
Yes, because the liver is responsible for regulating blood sugar.
Inspired by Louis Pasteur's swan-neck flask experiment, a microbiology student was interested in repeating this experiment using hay (dried grass) infusion instead of nutrient broth. He placed a few strands of dried grass into a sugar solution in an open neck flask, drew the neck of the flask into a swan-neck shape using a flame and then boiled the contents of the flask for 30 minutes. After 3 days of incubation at room temperature he was surprised to see bacterial growth in the flask.Does this experiment support spontaneous generation? If not, how would you explain the growth bacteria in the flask?
Answer:
Bacterial endospores. Some bacteria, specially from the phylum Firmicutes, produce endospores. An endospore is a resistance form of the bacteria, which is not reproductive. This form of the bacteria is resistant to heat, UV radiation, drought, cold, and might remain dormant for long periods. The formation of the endospores is triggered by starvation. In the experiment mentioned, instead of nutrient broth, which production is controled, a hay infusion is used. Since this is a material picked up from the soil, there is a high possibility of the presence of the endospores, could be from Bacillus, since this genera of bacteria is widely present in the soil. Once the endospores are in a suitable envoronment (culture media) they will stop dormancy and became metabolically active. It is mentioned that the culture media was boiled for 30 minutes, and endospores are resistant at 100 °C for several hours.
Explanation:
What type of protist is heterotrophic and includes species whose cells can come together to form a slug that moves to a new habitat?
Answer: Fungus like protists are heterotrophic and feed on organic matter.
Cellular slime mould is a species of fungus like protists that form slug.
Explanation:
Fungus like protists are heterotrophic and the feed on organic matter and mostly unicellular.
Cellular slime mould are protists belonging to class Dictyostelia. They are heterotrophic and decomposers that live on organic matter. When there is deterioration condition, the cells migrates together to form slugs and move to form new habitat. Some of the cells form stalk and others form spores.
The PCR reaction for Lab 2 is listed below. For each component besides water, briefly describe the component’s role in the PCR reaction. PCR amplification recipe (stock concentrations):
14.4 μL high quality nuclease‐free water
6.0 μL 5x Phusion HF buffer
3.0 μL dNTPs (2.5 mM each of four dNTPs – dATP, dCTP, dGTP, dTTP)
0.3 μL Phusion DNA polymerase
0.3 μL pCD122 (a plasmid that serves as template DNA)
3.0 μL sense primer CD474 (5 μM)
3.0 μL antisense primer CD475 (5 μM)
30.0 μL total reaction volume
Answer:
PCR is known as polymerase chain reaction used to amplify DNA sequence of our interest into multiple copies. This technique is been commonly used by researchers to study in depth about the gene of interest during their research work.
Explanation:
PCR:- It is known as polymerase chain reaction, used to amplify DNA sequence of our interest into multiple copies mainlty to millions or trillions
Components
Phusion HF buffer:- This buffer Create optimal reaction conditions for high fidelity amplification of DNA
dNTPs:- They are the building blocks in the synthesis of new copies of DNA.There are four dNTPs used in the amplification process that is dATP, dGTP, dCTP, dTTP. these building blocks are added in equal proportion during the PCR reaction
Phusion DNA polymerase:- This is the enzyme used in DNA amplification, it is generally a fusion of DNA binding domain to a portion of pyrococcus like proof reading polymerase. This enzyme is tolerant to various inhibitors, allowing strong amplification of DNA of interest with minimal optimization
pCD122:- Plasmid that serves as a template DNA for the amplification of desired DNA sequence of our interest.
sense primer CD474:- Sense primer is also known a reverse primer, it attaches to the stop codon of the complementary strand of DNA
antisense primer CD475:- This primer is also known as forward primer, it attaches to the start codon of the template DNA
All these components is added in such a way that the total mixture should have a reaction volume of 30.0μl
Final answer:
In the described PCR setup, specific roles include the buffer for maintaining an optimal enzyme environment, dNTPs as building blocks for new DNA, Phusion DNA polymerase for synthesizing DNA, plasmid template DNA for target amplification, and primers for initiating synthesis. The final concentration of each primer in the PCR reaction will be 0.5 µM.
Explanation:
In the PCR (Polymerase Chain Reaction) described, each component serves a crucial role:
5x Phusion HF buffer provides the optimal pH and ionic environment for the activity of the DNA polymerase.
dNTPs (deoxyribonucleotide triphosphates - dATP, dCTP, dGTP, dTTP) are the building blocks needed for the synthesis of new DNA strands.
Phusion DNA polymerase is the enzyme that synthesizes the new DNA strands by adding dNTPs to the primed DNA template.
pCD122 (plasmid template DNA) contains the specific region of DNA that we aim to amplify.
Sense and antisense primers (CD474 and CD475) are short pieces of single-stranded DNA that mark the starting point of DNA synthesis on each strand of the DNA template.
Regarding your specific query, mixing 5.0 µL of each 2.0 µM primer into a 20 µL PCR reaction will result in 0.5 µM final concentration of each primer in the reaction. This is because the primers are diluted by a factor of four (10 µL combined primers in a total volume of 40 µL).
If thymine makes up 19% of the DNA nucleotides in the genome of a plant species, what are the percentages of the other nucleotides in the genome?
Enter your answer in the following order:
a. % of adenine,
b. % of cytosine,
c. % of guanine.
Answer:
a. 19% of adenine
b. 31% of cytosine
c. 31% of guanine
Explanation:
Chargaff's rule states that in a DNa molecule, adenine pairs with thymine and guanine pairs with cytosine.
For that reason, if 19% of the DNA is made up of T, then 19% as well will be made of A.
Then, 19% + 19% = 38% of the genome is comprised of A+T and the other 62% is comprised of G+C.
Since G and C pair with each other, 31% will correspond to G and 31% will correspond to C.
Final answer:
To determine the percentages of adenine, cytosine, and guanine in the DNA genome of a plant species when thymine makes up 19%, utilize Chargaff's rules. Adenine and guanine percentages can be calculated using the complementary base pairing concept.
Explanation:
a. Percentage of Adenine: Since in DNA, adenine pairs with thymine, if thymine makes up 19%, then adenine also makes up 19% to maintain Chargaff's rules.
b. Percentage of Cytosine: The percentages of adenine and thymine together add up to 38%. As adenine and thymine are complementary base pairs, cytosine pairs with guanine. Therefore, cytosine would make up the remaining 62%.
c. Percentage of Guanine: Following Chargaff's rules and considering the percentages of thymine and cytosine, guanine would also be 31%.
Each of the following statements describes a type of column chromatography. Sort the statements to the type of chromatography they describe. If a statement can describe all of the types, place that statement in the "All" category. (Note that size-exclusion chromatography may also be called gel-filtration or molecular exclusion chromatography.)
Categories: Size-exclusion Chromatography, Affinity Chromatography, Ion-exchange Chromatogrpahy, and All
Statements:
Separate molecules by size.
Separate molecules by charge.
The stationary phase has a covalently bound group to which a protein in the mobile phase can bind.
Uses a mobile phase and a stationary phase to separate proteins.
The stationary phase contains cross-linked polymers with different pore sizes.
Can separate molecules based on protein-lignand binding.
The stationary phase may contain negatively or positively charged groups.
Answer:
Explanation:
Separate molecules by size. > AllSeparate molecules by charge. > Ion-exchange Chromatography, Affinity ChromatographyThe stationary phase has a covalently bound group to which a protein in the mobile phase can bind. > AllUses a mobile phase and a stationary phase to separate proteins. > AllThe stationary phase contains cross-linked polymers with different pore sizes. > Size-exclusion Chromatography Can separate molecules based on protein-ligand binding. > Affinity ChromatographyThe stationary phase may contain negatively or positively charged groups. > Ion-exchange Chromatography, Affinity Chromatography
Cholesterol is an important component of animal cell membranes. Cholesterol molecules are often delivered to body cells by the blood. which transports the molecules in the form of cholesterol- protein complexes. The complexes must be moved into the body cells before the cholesterol molecules can be incorporated into the phospholipid bilayers of cell membranes. Based on the information presented, which of the following is the most likely explanation for a buildup of cholesterol molecules in the blood of an animal? The animal's body cells are defective in exocytosis. The animal's body cells are defective in endocytosisy c) The animal's body cells are defective in cholesterol synthesis. (D) The animal's body cells are defective in phospholipid synthesis.
Answer:
The correct answer is - option B. The animal's body cells are defective in endocytosis.
Explanation:
Endocytosis is the process that internalized the substances inside the cells by the area surrounded by the membrane. This is the process that helps in the internalize cholesterol-protein complexes.
If the cells are not able to perform this then cholesterol-protein would not be internalized by the cell and it will remain in the blood and buildup cholesterol molecule.
Thus, the correct answer is - option B.
The most likely explanation for a buildup of cholesterol molecules in the blood of an animal is that 'the animal's body cells are DEFECTIVE in endocytosis'.
Endocytosis refers to the movement of molecules from the surrounding cellular environment to the interior of the cell by transport vesicles.
This process (endocytosis) is caused by the invagination of the cell membrane to form an intracellular vesicle that contains extracellular fluid.
Cells absorb cholesterol from their surrounding extracellular environment by receptor-mediated endocytosis.
In conclusion, the most likely explanation for a buildup of cholesterol molecules in the blood of an animal is that 'the animal's body cells are DEFECTIVE in endocytosis'.
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What is the wave speed of a wave that has a frequency of 5 hz a wavelength of 10 meters
Answer:
Frequency × wavelength
5 × 10 = 50m/s
Explanation:
Wave speed is the distance a wave travels in a given amount of time. Wave speed is related to wavelength and wave frequency by the equation:
Speed = Wavelength x Frequency.
Speed = 5 × 10
= 50m/s
Answer:
Wave speed= wavelength × frequency . it is measure in meters per seconds.
5 ×10=50 m/s
Explanation:
Wave speed is the distance travel by waves in a given time. It is measured in meter per second.
Wave speed= frequency (Hz) × wavelength.
Wavelength is the distance between two successive crest or trough ,it is measured in meter.
Frequency is the successive occurrence of waves within a fixed place at a fixed time. It is measured in Hertz.
In hybrid zones where reinforcement is occurring, you should see a decline in ________.
Answer:
gene flow between distinct gene pools
Explanation:
In ecological studies, a hybrid zone can be described as an area in which mating occurs between organisms of two closely related species. As a result, viable offspring are produced. When reinforcement occurs, we will see that a decline in gene flow between organisms of these species will be seen. The organisms which are only closely related will be able to reproduce to produce viable offsprings. In a stable hybrid zone, continued production of hybrids will take place.
Final answer:
Reinforcement in hybrid zones leads to a decline in hybridization due to continued speciation divergence. Less fit hybrids result in a reduced gene flow between species, enhancing reproductive isolation.
Explanation:
In hybrid zones where reinforcement is occurring, you should see a decline in hybridization. Reinforcement is a process in evolutionary biology that leads to the continued speciation divergence between two related species due to the low fitness of hybrids between them. As the two species continue to diverge, hybrids become less common because they are less fit compared to the purebred species, ultimately leading to a reduction in gene flow between the species.
What phenotypic ratio is expected in the offspring? Write your answer as numeric values and use a colon to separate numbers (e.g., 1:1)
Answer:
According to the statement of the question, this problem is solved by analyzing the sex chromosomes for the determination of sex, that is, as it does not present the problem (genotype and phenotype) of the parents, the indication that refers to the determination of sex, in women and his sex chromosomes are "XX" and the man who determines sex is "XY".
The result of sex in the offspring corresponding to the man, if the chromosome in the sperm fertilizes the ovule is "X", then the sex or phenotype of the individual will be "XX" (female) and if the fertilizing sperm is "Y" then the sex will be "XY" of male, the probability of being born female or male will be 50% and 50% and the ratio is 1: 1
A tall (dominant trait) pea plant is crossed with a short (recessive trait) pea plant to determine if the tall pea plant carries one or two factors for tallness. This is called ________.
a.a Punnett square test
b.a Punnett cross
c.a test cross
d.a parental cross
Answer:
c. A testcross
Explanation:
A dominant organism can be heterozygous or homozygous for the gene. To determine the genotype of a dominant organism, it is crossed with a homozygous recessive organism for the same gene. In the given cross, a tall pea plant with an unknown genotype is crossed with a pure breeding short pea plant. This is called a testcross.
If the obtained progeny express both tall and short phenotypes, the tall parent plant was heterozygous (Tt) for the trait. If the cross gives only tall progeny, the genotype of the tall plant is homozygous dominant (TT). Here, T represents the allele for tallness while the allele "t" gives short phenotype.
Which of the following model organisms is a multicellular worm usedto study the process of development? a)Saccharomyces cerevisiaeb) Caenorhabditis
Answer:b) Caenorhabditis
Explanation:
The C.elgans is a species of Caenorhabditis. It is a namatode which can be male or hermaphrodite. It lays 1000 of eggs which can be used for the purpose of research studies. These worms share common genes and molecular pathways like in humans.
Many of the genes present in C. elegans have functional counterparts as that of humans which makes them useful for determining the cause of human diseases. The mutant forms of genes of C.elegans can be used for screening the thousands of drugs. The genes have been used to determine the effects of ageing in humans, cancer, diabetes and other diseases.
Due to above reasons the C.elegans are considered as model organisms.
Lauren is a senior at a nearby high school. She is a good student who does her work on weekdays and likes to party on the weekends with her friends. A few months ago, Lauren developed a persistent cough that has worsened over time. She has also had trouble breathing over the past week or so. Her parents decided it was time she see a doctor and made an appointment with the family physician. Lauren arrived at the doctor's office the following day. The doctor asked her a number of questions about her health history and daily habits, including, whether or not she smoked cigarettes. Lauren responded that she started smoking cigarettes (about 2 per day) her freshman year of high school and does not intend to quit anytime soon. The doctor decided it was necessary to take a biopsy of the cells that line the bronchus (passageway to the lungs). After several days, the biopsy report has come back along with a micrograph of the bronchial cells. Lauren's cigarette smoking has badly damaged these organelles which are responsible for mucous and dirt moving up-and-out of her respiratory system. Analyze the micrograph below and determine what cell organelles have been damaged. Cella OA smooth ER B Peroxisome C Mitochondria D Cilia E Golgi Apparatus F Rough ER
Cilia have been damaged
Explanation:
If the cilia are damaged the mucous and dirt do not come out of the lungs leading to clogging of the bronchia. Lauren smoking habit has damaged her cilia and thus had troubled breathing. The bronchi of the lung are lined with cilia.
Cilia are tiny hair-like projections that keep the airways clean by sweeping away mucus and dust particles and keeping the lungs clear. When cilia do not work the air passage gets clogged with dust and dirt causing trouble breathing.
Lauren's smoking has damaged the cilia in her bronchial cells.
Explanation:The cell organelles that have been damaged in Lauren's bronchial cells are the cilia. Cilia are hair-like structures on the surface of cells that help in moving mucus and dirt out of the respiratory system. Smoking cigarettes has a detrimental effect on cilia, causing them to become damaged and less effective in their function.
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Identify the moth variant that was selected for during the Industrial Revolution, when pollution began to kill the lichen that previously covered all of the trees.
Answer:
Black moth
Explanation:
During the time of the industrial revolution, the lichen that covered all the trees previously got killed. Due to the pollution arising from the industries, the colour of the woods of the trees turned into black. As a result, the light- coloured moths were not able to hide from their predators as their colour no longer matched with the colour of the trees. The black coloured moths were more adapted to survive in this environment because their colour matched with those of the woods of the trees. They were better adaptable to live in the Industrial Revolution.
You are studying a new variant of a eukaryotic cell. The variant cell has mutated so that it no longer attaches well to surfaces or initiates the formation of a biofilm. The mutation in this cell has most likely affected the _____.
You are studying a new variant of a eukaryotic cell. The variant cell has mutated so that it no longer attaches well to surfaces or initiates the formation of a biofilm. The mutation in this cell has most likely affected the _____.
Mitochondria
Flagellum
Cell membrane
Glycocalyx
Answer:
Glycocalyx
Explanation:
Biofilms refer to the irregular layers of cells. These are formed when a group of cells arranges themselves at a solid or liquid surface. The glycocalyx forms a layer around some animal cells. It has glycolipids and glycoproteins. These substances allow the cells to recognize one another and to make contact with one another. The formation of adhesive communications that hold the cells in the biofilms is mediated by the components of the glycocalyx. Therefore, any mutation in the genes coding for components of glycocalyx would affect the ability of the cells to form biofilms.
In a eukaryotic cell, mutations hindering attachment and biofilm formation likely affect the cell adhesion mechanisms, particularly the functioning of adhesion molecules such as cadherins, selectins, and integrins.
Explanation:The mutation in the eukaryotic cell described has most likely affected the cell adhesion mechanisms, specifically the functioning of proteins known as adhesion molecules. These molecules help cells stick to each other and to their surroundings, a crucial process for the formation of biofilms.
Depending on the organism and environment, various types of adhesion molecules might be present. Nonetheless, some universally key ones would include cadherins, selectins, and integrins. If a mutation affects these proteins, the cell may lose its ability to properly adhere to its environment or form biofilms.
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The GART, a gene which is involved with Down syndrome, is found toward the bottom of chromosome 21. This is an example of a ________.
Answer: It is an example of TRISOMY 21.
Explanation:
Trisomy 21 is a genetic condition in which 21 chromosome fail to separate during the growth of egg cell and sperm which make either of the two to an extra copy of chromosome 21. This is also known as down syndrome. This lead to slow and in appropriate mental development in affected individual.
Phosphoriboglycinamide transformylase(GART) is agene found on chromosome 21, which lead to down syndrome is an example of trisomy 21.
Insulin is taken up, via endocytosis, by the endothelial cells that line blood capillaries. Then, it's transported across the cell to the other side, where it is released. This transport is called __________.
Answer: Insulin is transported in the cell by TRANCYTOSIS transport.
Explanation:
Insulin is a hormone that is secreted by pancreatic islet that allow glucose from the blood to enter cell. It is transported by trancytosis after been taken up by endocytosis.
Trancytosis is a process that transport material from the cells and move it to the other side where it is released. In the case of insulin, after the insulin is taken up through endocytosis at the vascular lumen ,there is traffic of the hormone bearing vessicles and exocytosis of of the insulin vesicles present at the basal membrane which lead to insulin been released in the cell.
The transport of insulin across the endothelial cells lining blood capillaries is called transcytosis. This involves endocytosis, in which insulin is enveloped into the cell, and exocytosis, in which insulin is released to the other side.
Explanation:The transport of insulin across the endothelial cells that line blood capillaries is called transcytosis. This happens through a process of endocytosis, where the insulin molecules are enveloped by the cell membrane and brought into the cell. The insulin-filled vesicle then travels across the cell and merges with the cell membrane on the opposite side, releasing the insulin to the other side. This is the final step in the transport process known as exocytosis.
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Which of the following statements is FALSE? A. Biological systems are highly ordered so entropy changes are not relevant.B. The entropy of a biological system can decrease. C. Entropy is a measure of disorder. D. The entropy of an isolated system will tend to increase to a maximum value
Answer: Option A is false.
Biological system is highly ordered, entropy changes is irrelevant.
Explanation:
Biological systems is the network of complex important biological individual like cells, organelles, ordans, macromolecules. Biological systems obey the second law of thermodynamics in that entropy changes with gain or loss of energy and for this to happen, it must increase the entropy of the universe. Living organisms taking in food to decrease their entropy yet the overall entropy is increased when entropy within the organism decreased.
The false statement is that entropy changes are not relevant in highly ordered biological systems. Entropy is certainly relevant as it can decrease with an input of energy to the system, and it is a measure of disorder, with the entropy of an isolated system tending to increase over time.
Explanation:The statement that is FALSE among the provided options is: "Biological systems are highly ordered so entropy changes are not relevant." This statement contradicts the principles of thermodynamics which apply to all systems, including biological ones. Biological systems, while highly ordered, still adhere to the laws of thermodynamics, and thus changes in entropy are indeed relevant.
Entropy can decrease in a biological system but only if there is an input of energy. This is because biological systems are not closed systems; they exchange energy with their surroundings. The second statement that says the entropy of a biological system can decrease is true given that energy is constantly being input into these systems to maintain their order and function. Thus, entropy in a local system can decrease if work is done on it, although the overall entropy of the universe still increases.
Entropy is a measure of disorder; a higher level of entropy corresponds to a higher state of disorder in the system. Finally, the entropy of an isolated system will tend to increase to a maximum value, which aligns with the second law of thermodynamics.
Cindy is 63 years old and at risk for osteoporosis. Which nutrients would be the most important to consume in adequate amounts to preserve bone mass?
Answer:
Consumption of adequate calcium and vitamin D
Explanation:
Osteoporosis is a condition whereby there is significant amount of porous partition on the surface of the bone,this disease reduces the density and quality of bone. As bones become more porous and fragile, the risk of getting fractured is greatly increased.
In a bid to overcome or to be resistant to Osteoporosis, Consumption of adequate calcium and vitamin D is required by the body. This because Calcium performs various function which include;building strong bones, regulating heart beat and fluid balance within cells.
Also Vitamin D works in conjunction with calcium to slow down or even reverse osteoporosis. That is to say, the body cannot absorb calcium at all without some vitamin D. Hence, Vitamin D is vital in assisting the body absorb and use calcium.
The most important nutrients to preserve bone mass and prevent osteoporosis are calcium and vitamin D. Regular exercise, particularly resistance training, is also crucial for maintaining bone density. Other nutrients like vitamin K, magnesium, and omega-3 fatty acids can also support bone health.
Explanation:The most important nutrients to consume in adequate amounts to preserve bone mass for someone at risk for osteoporosis are calcium and vitamin D. Calcium is a critical component of bone and must be obtained from the diet, while vitamin D is necessary for the absorption of calcium. Both nutrients play crucial roles in bone health and can help prevent bone loss.
Regular exercise, especially resistance training, is also important for preserving bone mass and preventing osteoporosis. Exercise stimulates the deposition of bone tissue and helps improve bone density.
Additionally, other nutrients like vitamin K, magnesium, and omega-3 fatty acids also support bone health.
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niridia is a type of blindness due to a single dominant gene. Migraine headache is the result of a different dominant gene. A man with aniridia, but normal headaches whose mother was not blind, marries a woman who suffers from migraines and has normal vision but whose father did not have migraine headaches. What is the expected proportion of their children that would have both aniridia and migraines together
Answer: 25% of the children will have both anirida and migraine headache
Explanation: The man has a dominant gene for anirida and normal headache with a mom with no blindness. Therefore, the man gene is (Nn) for heterozygous anirida and (mm) for normal headaches and his wife has heterozygous gene for migraine (Mm) 'cause her dad has normal headache and homozygous gene for normal vision(nn) .
Therefore
Nnmm × Mmnn= NnMm Nnmm Nnmm nnmm.
The crossing would give birth to Offspring with nirida and migraine (NnMm) =1/4 ×100= 25%
Explain the contributions of John Muir and Gifford Pinchot in the history of environmental ethics.
Answer:
Mr Mr. Gifford Pinchot supports wilderness protection of a forest area while Mr. John Muir focused on forest and its wildlife conservation
Explanation:
During the nineteenth century, both the environmentalists Mr. John Muir and Mr. Gifford Pinchot headed the environmental movement. However, both the environmentalists have opposite beliefs - Mr. John Muir believed in preserving the entity of the forest by protecting its wilderness while Mr. Gifford Pinchot believed in conserving the environment. Mr. John wrote several books on protecting wilderness area.
Mr. Gifford Pinchot believed on conserving the forest and its wildlife.
John Muir and Gifford Pinchot were key figures in the conservation movement. Muir advocated for the preservation of wilderness while Pinchot pushed for utilitarian conservation. Their contributions significantly influenced environmental policy and practice.
The environmental ethics landscape was significantly shaped by two pivotal figures in the conservation movement: John Muir and Gifford Pinchot. Muir, the founder of the Sierra Club, was a staunch advocate for the preservation of wilderness areas for their inherent, spiritual value. In contrast, Pinchot, as the first Chief of the US Forest Service under President Teddy Roosevelt, promoted utilitarian conservation, which held that natural resources should be managed sustainably to benefit the public good. Both contributed to a shift from the frontier ethic, which assumed limitless resources, to a modern understanding that emphasizes conservation and sustainable use of the environment. Muir is celebrated for his role in the creation of Yosemite National Park and his eloquent writings that inspired the preservationist movement. Pinchot's impact includes the increase in protected forests, mandatory reforestation policies for lumber companies, and the pioneering of sustainable resource management practices.
What multicellular life forms existed on Earth BEFORE dinosaurs existed on Earth (Mark all that apply) corals some land plants early molluscs, like giant cephalopods (distant relatives of squids and octopi) ■ birds mammals fishes early arthropods, like trilobites
Some land plants and arthropods like trilobites existed before the dianosaurs existed on Earth.
Explanation:
The first reptiles are reported to have appeared during the Carboniferous period. The Dinosaurs were large reptiles that became dominant during the triassic and jurassic. By the end of cretaceous the dinosaurs became extinct.
In the given options we find that some land plants and trilobites are suitable choice because, firstly, the fossil plants are recorded during carboniferous period and some of those species became extinct before the appearance of dinosaurs. Secondly, trilobites became extinct by the end of permian period.
Rest of the options like birds, mammals, etc came into existence before dinosaurs but they continued to exist along with dinosaurs.
How is DNA technology used in science, society, and in medicine? Which of the following would be considered an example of biotechnology?
A. Growing cancer cells in a petri dish
B. Developing a rice plant that produces a precursor to Vitamin A
C. Brewing beer
D. Crossing a wild corn plant with a domesticated corn plant
Answer: Option B
Explanation:
The DNA technology can be defined as the process by which a small manipulation in the genetic material can lead to the production of the products that is very important for the welfare of the society, science and medicine.
The golden rice is an example of genetic engineering which produces the precursor for the synthesis of vitamin A in the body.
This crop is produced by genetic engineering and is very high economic and nutritive value.
Which lymphatic tissue is associated with mucous membranes and is called mucosa-associated lymphatic tissue, or MALT?
Answer:
MALT or mucosa-associated lymphoid tissue refers to a bundle of lymphatic cells, known as lymphatic nodules, situated inside the mucous membranes, which envelopes the respiratory, gastrointestinal, urinary, and reproductive tracts. These nodules comprise macrophages and lymphocytes that fight against the entering bacteria and other pathogens, which moves into these pathways along with air, food, or urine. These nodules can be grouped together in clusters or can be present solitary.
The major clusters of lymphatic nodules comprise adenoids, tonsils, and Peyer's patches.