Answer:
Sickle cell anemia is an autosomal recessive disorder. For the trait to occur, both of the alleles in an individual should be recessive. Parents who both have sickle cell anemia will have 100% results for their children to carry the disease. The results can be depicted from the following punnet square:
s s
s ss ss
s ss ss
The results will be interpreted by the nurse as having 100% chance for the parents to have the child born with sickle cell as both the parents carry the disease which is autosomal recessive.
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.
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).
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.
A cultured cell line appears to be having trouble surviving. You find that the cells do not appear to have normal chromosome separation. You decide to check for the presence of a mutated protein. Which of the following would your best target for analyzing chromosome separation?
a. Actin
b. Melanin
c. Tubulin
d. Shugoshin
Answer:
c. Tubulin
Explanation:
Tubulin protein is polymerized to form the cylindrical structures of microtubules. Microtubules form the spindle apparatus during cell division. The spindle microtubules become attached to the kinetochores of chromosomes and mediate the alignment of chromosomes at the equator of cells during metaphase. The shortening of spindle microtubules is responsible for the movement of sister chromatids during anaphase. The same event also moves the homologous chromosomes during anaphase-I.
Any failure in the formation of the spindle apparatus would not allow the proper separation of chromosomes. Therefore, the cell with abnormal chromosome separation might have a faulty or no tubulin.
Indicate whether below is True/False.
1. Mammary glands are specialized for milk and hormone production.
2. Breasts contain areolar connective tissue but little dispose tissue.
3. Alveolar glands occur in lobes of mammary glands.
Answer:
1. True
2. True
3. False
Explanation:
Mammary glands are made up of glandular tissue. These are meant for the secretion of milk after the newborn birth. The milk ejection is done by different hormones, especially by progesterone and prolactin.
Glandular tissue acts as both exocrine and endocrine glands. The mammary glands are a suitable example as it secretes milk which is an exocrine secretion. The hormone progesterone and prolactin released from the mammary gland are endocrine secretions.
The mammary glands consist of a sac-like structure called alveoli.
These are group alveoli form grape-like appearance called lobules.
The alveolar sacs contain sweat glands. These are the modified sweat glands that secrete milk. It is made up of fibrous connective tissue.
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
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:
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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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.
In humans alkaptonuria is a metabolic disorder in whichaffected persons produce black urine. Alkapotonuria results from anallele(a) that is recessive to the allele for normal metabolism(A). Sally has a normal metabolism, but her brother hasalcaptonuria. Sally's father has alcoptonuria, and her mother hasnormal metabolism.
a) Give the genotype of Sally,her mother,father and herbrother.
b) If Sally's parents have another child what is theprobabilty that this child will have alkaptonuria?
c) If Sally marries a man with alkaptonuria, what is theprobability that their child will have alkaptonuria?
If Sally marries a man with alkaptonuria, there is a 50% chance that their child will have alkaptonuria.
The genotype of an individual refers to the sum total of genes that the individual received from its parents. Since Sally has a normal metabolism, Sally is Aa. Sally's mother is Aa while Sally's father and brother are aa.
If Sally parents have another child, using the Punnet square method, there is a 50% chance that the child will have alkaptonuria. If Sally marries a man with alkaptonuria, there is a 50% chance that their child will have alkaptonuria.
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a) Genotypes: : Aa (carrier of the alkaptonuria allele) ,Mother: Aa (carrier of the alkaptonuria allele)Father: aa (has alkaptonuria)Brother: aa (has alkaptonuria)
b) If Sally's parents have another child:The probability of the child having alkaptonuria (aa genotype) is 25%.The probability of the child being a carrier (Aa genotype) is 50%.The probability of the child having normal metabolism (AA genotype) is 25%.
c) If Sally marries a man with alkaptonuria (aa genotype):The probability of their child having alkaptonuria (aa genotype) is 50% (as Sally is a carrier).The probability of the child being a carrier (Aa genotype) is 50%.
The probability of the child having normal metabolism (AA genotype) is 0% (as the husband has alkaptonuria).
Sally: Aa (normal metabolism carrier)
Mother: Aa (normal metabolism carrier)
Father: aa (alkaptonuria)
Brother: aa (alkaptonuria)
b) If Sally's parents have another child:
Probability of the child having alkaptonuria (aa genotype) is 25%.
Probability of the child being a carrier (Aa genotype) is 50%.
Probability of the child having normal metabolism (AA genotype) is 25%.
c) If Sally marries a man with alkaptonuria (aa genotype):
Probability of their child having alkaptonuria (aa genotype) is 50% (as Sally is a carrier).
Probability of the child being a carrier (Aa genotype) is 50%.
Probability of the child having normal metabolism (AA genotype) is 0% (as the husband has alkaptonuria).
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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 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
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:
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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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%.
Clearly cellulose is very abundant on earth, and it is a long-lasting stable substance. Many animals cannot digest cellulose. Given this, what prevents the bodies of dead plants from filling the earth? Something must decompose cellulose. This is where fungal decomposition comes in--fungi digest cellulose, as do many prokaryotes. Name two organisms that consume cellulose and make an educated guess as to whether each breaks down cellulose or simply excretes it as fiber.
Answer: Termites and herbivores.
Explanation:
The two organism that can digests cellulose are termites and herbivores. The termite contains protists known as mastigophorans carry out digestion of cellulose in the body.
The cellulose in this case is digested and prevents it from getting deposited in the environment.
The other organism are animals like ruminants which can digest cellulose in their gut. They partially digest the cellulose and regurgitate it into the mouth an broken down further.
This process of digestion of cellulose in the gut is anaerobic so methane is released in the environment as a product of digestion.
The human beings cannot digest cellulose and a very less amount of it is considered as fiber and is simply excreted.
Final answer:
Ruminants like cows and fungi have the ability to digest cellulose due to cellulases, allowing them to break down this complex carbohydrate and utilize it as a food source, contributing to nutrient cycles in ecosystems.
Explanation:
Despite the fact that cellulose is a major structural component of plant cell walls and abundantly found on Earth, not all organisms can digest it due to its robust β(1,4) glycosidic bonds. However, certain organisms such as ruminants and termites have symbiotic relationships with microorganisms in their guts that produce enzymes capable of breaking down cellulose. For instance, cows have bacteria in their rumens that secrete cellulases, allowing them to break cellulose down into usable sugars.
Fungi also play a crucial role in cellulose degradation in ecosystems. They secrete exoenzymes into the environment to decompose dead plant material, preventing accumulation of undegraded biomass. These enzymes cleave the cellulose into glucose monomers, which they then utilize for energy and growth, contributing to nutrient cycling in the environment. Therefore, both ruminants and fungi do not simply excrete cellulose as fiber but effectively break it down for consumption.
Describe the biological roots of Behavioral Neuroscience.
Answer:
Explanation:
The biological perspective or roots, a way of looking at neuroscience is by studying the physical basis for animal and human behavior. It involves such things as studying the brain, immune system, nervous system, and genetics.
The biological perspective tends to stress the importance of nature.
The study of physiology and biological processes has played a significant role in neuroscience since its earliest beginnings. Charles Darwin first introduced the idea that evolution and genetics has roles to play in human behavior. Natural selection influences if certain behavior patterns will be passed down to future generations. Behaviors that help survival skills most likely are passed down than those that prove dangerous.
The biological perspective is a way of looking at human problems and actions. For instance in aggression, someone using the psychoanalytic perspective might view aggression as the result of childhood experiences, another might take a behavioral perspective to show how the behavior was shaped by association and punishment. A neuroscientist with a social perspective might look at the group dynamics and pressures that contribute to such behavior. The biological viewpoint, on the other hand would look at the biological roots that lie behind aggressive behaviors by considering how certain types of brain injury might lead to aggressive actions. Or might consider genetic factors that can contribute to such displays of behavior.
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.
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:
Maria is a 121-lb endurance athlete who is planning to employ carbohydrate loading before her next race. Which strategy correctly follows the guidelines for carbohydrate loading?
Choose the statement below that correctly describes a good carbohydrate loading strategy for Maria:
O Four to six days prior to the event, Maria should consume 550 g of carbohydrate daily and decrease to 220 to 275 g of carbohydrate daily 1 to 3 days prior to the event.O One to three days before the event, 550 g of carbohydrate daily is recommended.O One to three days before the event, 1210 g of carbohydrate daily is recommended.O Six days before the event, her carbohydrate intake should be 550 g per day.O Two days before the event, an intake of 33 g of protein daily is adequate.
Final answer:
The correct carbohydrate loading strategy for Maria, a 121-lb endurance athlete, is to consume 550 g of carbohydrate daily for 1 to 3 days before her event to maximize glycogen storage for energy.
Explanation:
The correct carbohydrate loading strategy for Maria—an endurance athlete who weighs 121 lbs—would be to consume 550 g of carbohydrate daily during the 1 to 3 days prior to her event. This is because carbohydrate loading is aimed at maximizing the storage of glycogen in the muscles, which is used for energy during prolonged periods of intense exercise. Ingesting 550 g of carbohydrates daily, following the guidance, should increase her glycogen stores, providing an energy reservoir that can be tapped into during her endurance event.
Concerning the options given, the second option correctly follows the guidelines for carbohydrate loading: "One to three days before the event, 550 g of carbohydrate daily is recommended." This is based on the principle that consuming high amounts of carbohydrates can boost glycogen stores within muscles, effectively increasing the athlete's endurance capacity during the race.
Which of the following answers describes the most direct consequence of tropomyosin binding to F-actin in muscle cells? a. The (-) end of F-actin is stabilizedb. Loss of actin-ADP subunits from the (+) end is preventedc. Myosin binding to F-actin is blockedd. ATP hydrolysis by myosin is blockede. The movement of myosin towards the (+) end is blocked
Answer: A
Explanation:
The "-" end of F actin is stabilized. The myosin head bind to actin and makes the actin filament to slide
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
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.
16.Globular proteins fold up into compact, spherical structures that have uneven surfaces. They tend to form
multisubunit complexes, which also have a rounded shape. Fibrous proteins, in contrast, span relatively large
distances within the cell and in the extracellular space. Which protein is not classified as a fibrous protein?
(a)elastase (b)collagen (c)keratin (d)elastin
Answer:(a)elastase
Explanation:
Elastase is a protease enzyme the function involves the cleavage of peptide bonds after amino acids with small side chains. This is responsible for cleaving the peptide bonds between the elastin fibers and aids in digestion of the elastic protein.
Thus the elastase cleaves the protein fibers this can be said that the elastase is a non-fibrous protein.
Answer:
The answer is A: elastase
Explanation:
Fibrous proteins are insoluble, long, rod-like structured proteins that have high α-helix or β-sheet content, and are made up of a repeating motif. They are mechanically strong, cross-linked and play structural roles in the extracellular matrix. Examples are: collagen, keratin and elastin.
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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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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If the ease of oxygen pickup depends on oxygen concentration, would respiration be easier in air or in water?
Answer:
Air
Explanation:
depending on the concentration of Oxygen in any of the medium- air or water, Oxygen pick up will be easier in air than water. This is because the energy barrier for the exchange of gases is higher in a liquid medium than in a gaseous medium.
Respiration would be easier in air than water
What is Respiration ?Respiration is the process of taking in oxygen and giving out CO₂ through the oxidation of organic substances in the body. The process of respiration leads to the production of energy in the body.
Respiration would be easier in a gaseous medium ( air ) than it would be in water because air molecules are more loosely packed and therefore allows the free movement of oxygen ( lesser energy barrier ) .
Hence we can conclude that Respiration would be easier in air than water.
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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.
Consider a population of 425 diploid giant Sequoia trees. In this population, you observe the following genotypic counts: 100 homozygous dominant genotypes, 250 heterozygous genotypes, and 75 homozygous recessive genotypes. What is the allele frequency of the recessive allele in this population (use two decimal places and the usual rounding conventions)
Answer:
q = 0.42
Explanation:
This question is an example of Hardy-Weinberg question and there are two equations necessary to carry out this question;
p + q = 1
p² + 2pq + q² = 1
where;
p = the frequency of the dominant allele
q = the frequency of the recessive allele
p² = the frequency of individuals with homozygous dominant genotype
2pq = the frequency of individuals with heterozygous genotype
q² = frequency of individuals with the homozygous recessive genotype
Since the total population = 425
q² = [tex]\frac{individuals with recessive genotype}{Total Population}[/tex]
= [tex]\frac{75}{425}[/tex]
q² = 0.1765
To find q; we need to square root both side to eliminate the square from q².
∴ [tex]\sqrt{q^2}=\sqrt{0.1765}[/tex]
q = 0.4201
q = 0.42 (to two decimal places)
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