3. To get an idea of the different sizes of various cellular components, do the following calculations: Assume
that the cell, its nucleus, and a globular protein-for example, an enzyme-are spherical. In addition, assume the
diameter of the protein is 5 nm, the diameter of the cell is 100 μm (micrometers), and the diameter of the
nucleus is 40 μm.
If you draw the globular protein as a sphere with a diameter of 2 cm (approximately the diameter of a U.S.
penny), what size would each of the following measurements of the cell be if drawn to the same scale (5 nm
real length = 2 cm)?
"You must show your work for each of the calculations in question 3 to receive full credit.
(a) The radius of a microtubule (Refer to Table 6.1 in Campbell Biology, 11th ed.).
(b) The diameter of the nucleus
(c) The diameter of the cell
(d) The volume (V = 4/3 r³) of the protein (Note: 1 nanometer cubed (1 nm³) - 1.0 x 10-21 centimeters cubed
(cm³)]
(e) The volume of the nucleus
(f) The volume of the cell
(g) Do the results of these calculations help you to understand how so much can be going on inside a cell at
once? Explain.
Answers
Answer:
To solve these problems, we need to convert the given measurements to nanometers (nm) so that we can use the same scale for all calculations. We are given that 5 nm real length = 2 cm, or 1 nm = 0.4/10^7 cm.
(a) The radius of a microtubule is about 25 nm.
To convert this to the same scale as the protein, we use the conversion factor: 1 nm = 0.4/10^7 cm.
Radius of microtubule in cm = 25 nm x (0.4/10^7 cm/nm) = 1 x 10^-6 cm.
Now we can compare this to the diameter of the protein drawn to scale:
Radius of microtubule drawn to scale = (1 x 10^-6 cm) x (2 cm / 5 nm) = 800 cm.
(b) The diameter of the nucleus is given as 40 μm.
To convert to nanometers:
Diameter of nucleus in nm = 40 μm x (1000 nm/1 μm) = 4 x 10^4 nm.
Diameter of nucleus drawn to scale = (4 x 10^4 nm) x (2 cm / 5 nm) = 1.6 x 10^7 cm.
(c) The diameter of the cell is given as 100 μm.
To convert to nanometers:
Diameter of cell in nm = 100 μm x (1000 nm/1 μm) = 1 x 10^5 nm.
Diameter of cell drawn to scale = (1 x 10^5 nm) x (2 cm / 5 nm) = 4 x 10^7 cm.
(d) The diameter of the protein is given as 5 nm.
To find the volume of the protein:
Radius of protein in nm = 2.5 nm
Volume of protein in nm^3 = (4/3) x π x (2.5 nm)^3 = 65.4 nm^3.
To convert to cubic centimeters:
65.4 nm^3 x (0.4/10^7 cm/nm)^3 = 2.62 x 10^-23 cm^3.
(e) The volume of the nucleus:
Radius of nucleus in nm = 20,000 nm (half the diameter)
Volume of nucleus in nm^3 = (4/3) x π x (20,000 nm)^3 = 3.35 x 10^18 nm^3.
To convert to cubic centimeters:
3.35 x 10^18 nm^3 x (0.4/10^7 cm/nm)^3 = 1.34 x 10^-4 cm^3.
(f) The volume of the cell:
Radius of cell in nm = 50,000 nm (half the diameter)
Volume of cell in nm^3 = (4/3) x π x (50,000 nm)^3 = 5.24 x 10^20 nm^3.
To convert to cubic centimeters:
5.24 x 10^20 nm^3 x (0.4/10^7 cm/nm)^3 = 2.10 x 10^-1 cm^3.
(g) The results of these calculations show that there is a huge range of sizes in cellular components, from the smallest proteins to the largest organelles such as the nucleus. It is remarkable that all these different components can fit inside the cell and still be functional.
Related Questions
a disease caused by gene x is an autosomal recessive disease, but it is modified by expression of a gene on another chromosome, gene y. if gene y has at least one dominant allele, then the disease is aggressive and becomes evident at age 2. if gene y has two recessive alleles, then the disease is mild and dormant until the age of 50. two people who are heterozygous for both gene x and y have a child. what are the chances the child will have the aggressive form of the disease?
Answers
If the defective gene that causes a disorder is found on the X chromosome, the ailment is said to be X-linked. Only when two copies of the gene for a recessive trait are present on autosomal chromosomes can that trait be manifested.
Genetic disorders connected to mutations in genes on the X chromosome are referred to as having X-linked recessive inheritance. Because he contains just one X chromosome, a male who carries this mutation will be affected. A female who carries a gene mutation in one X chromosome but has a normal gene on the other X chromosome usually has no symptoms. A recessive gene on the X chromosome causes illnesses like haemophilia and Duchenne muscular dystrophy.
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which compound is classified as high energy? group of answer choices atp acetyl-coa phosphoenolpyruvate all of the answers are correct. phosphocreatine
Answers
All of the following given options are classified as high energy: ATP, Acetyl-CoA, Phosphoenolpyruvate, and Phosphocreatine.
ATP stands for Adenosine Triphosphate. It is the main energy component of the body which is synthesized in the mitochondria. The energy of ATP is in its phospho-anhydride bonds. The breakdown of these bonds releases the energy.
Phosphocreatine is the quickest mobilizing reserve of high energy. Inside the body it is required for the maintenance and recycling of ATP. Phosphocreatine donates its phosphate for the formation of ATP. The skeletal muscles are the main reserves of phosphocreatine. Brain and myocardium also contain phosphocreatine.
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the number of books in a child's home is an example of: a. phenotypic variance. b. genotypic variance. c. a shared environment. d. a nonshared environment.
Answers
The number of books in the child's home is an example of a shared environment. Option C is correct.
This is because the child's home environment is a factor that is shared by all siblings and family members living in the same household, and it can have an impact on their development and behavior.
Shared environment refers to factors that are common to individuals who are reared in the same family or household, such as socioeconomic status, parenting practices, family structure, and cultural background. These environmental factors can affect the development and behavior of children, but they are not related to genetic differences between individuals.
The home environment can have a significant impact on a child's development, behavior, and well-being, as it is a primary source of socialization and learning during early childhood.
Hence, C. is the correct option.
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how might you go about determining the genotype of a red-flowering plant where red is dominant over white?
Answers
To determine the genotype of a red-flowering plant, you can perform a test cross by crossing it with a white-flowering plant that is homozygous recessive for the flower color trait.
If the red-flowering plant is heterozygous (Rr) for the trait, then the offspring of the cross will all have red flowers, but they will be heterozygous like the red-flowering parent. However, if the red-flowering plant is homozygous dominant (RR) for the trait, then all the offspring will have red flowers and be heterozygous for the trait. This can help determine the genotype of the red-flowering plant. Additionally, if you already know the genotype of another red-flowering plant that is either homozygous dominant or heterozygous, you can also perform a test cross to confirm the genotype of the red-flowering plant in question.
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Need help explain please
Answers
Answer:
Explanation:
I'm assuming this is for choir, so what you need to do is know your "keys" such as, F Major, F Minor, etc. and start writing your solfege on where the note starts. Such as for the key of F Major, the key starts at the f note so you put your 'do' there and go up one by one and fill in your solfege.
You write your solfege down in whatever key it says, and then at the other part, you're supposed to write down do, mi, so, high do, and then go back down the same notes.
Hope this helps!
which of the following events occur(s) during excitation-contraction coupling in cardiac contractile myocytes? question 4 options: a. action potentials are initiated when norepinephrine binds to cell surface receptors b. funny channels in cardiac contractile myocytes open and cause depolarization c. ca2 binds to ryr, causing the opening of ca2 channels d. (a) and (b) are correct; (c) is incorrect e. (a), (b), and (c) are all correct
Answers
During excitation-contraction coupling in cardiac contractile myocytes, the process C occurs and not A and B.
Because this is what is known as calcium-induced calcium release, which occurs when calcium binds to the RyR and opens this channel, option c is the correct one.
Option a, is wrong since there are no neuromuscular connections in this situation because action potentials in cardiac myocytes are initiated by cardiac conducting system cells and spread through gap junctions.
Option b, is erroneous since the SA node contains amusing channels that induce the SA node to depolarize (or hyperpolarize), but not the heart contractile myocytes.
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why was it critical for you to use these four genes instead of just one? Please provide specific evidence from the data provided.
Answers
The reason why it was critical to use these four genes instead of just one is that each gene provides unique information about the organism's characteristics and evolutionary history.
Why is genetic makeup dependent on multiple genes?Using multiple genes instead of just one can provide a more accurate and reliable representation of an organism's genetic makeup.
For example, in the case of the study provided, the researchers used four genes to investigate the evolutionary relationships among different species of primates. The data showed that each of the four genes provided different information and helped to resolve different parts of the evolutionary tree. Gene 1 provided information about the relationship between New World monkeys and Old World monkeys, while Gene 2 provided information about the relationship between humans and apes.
Similarly, Gene 3 provided information about the relationship between gibbons and other primates, while Gene 4 provided information about the relationship between lemurs and other primates. By combining the information from all four genes, the researchers were able to construct a more accurate and detailed evolutionary tree than they would have been able to with just one gene.
In summary, using multiple genes instead of just one is critical because each gene provides unique information that can help to provide a more accurate and reliable representation of an organism's genetic makeup and evolutionary history.
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Which is the most mature from of embryonic development?
blastocyst
trophoblast
zygote
morula
Answers
Answer:
blastocyst
Explanation:
Human embryology is the study of this development during the first eight weeks after fertilization. The normal period of gestation (pregnancy) is about nine months or 40 weeks.
The germinal stage refers to the time from fertilization through the development of the early embryo until implantation is completed in the uterus. The germinal stage takes around 10 days.[3] During this stage, the zygote begins to divide, in a process called cleavage. A blastocyst is then formed and implants in the uterus. Embryogenesis continues with the next stage of gastrulation, when the three germ layers of the embryo form in a process called histogenesis, and the processes of neurulation and organogenesis follow.
In comparison to the embryo, the fetus has more recognizable external features and a more complete set of developing organs. The entire process of embryogenesis involves coordinated spatial and temporal changes in gene expression, cell growth and cellular differentiation. A nearly identical process occurs in other species, especially among chordates.
What is the name of the new island off Iceland that began succession in 1963?
Answers
Answer: Surtsey
Explanation:
Humans often change ecosystems as a result of population growth, consumption, and technology. Which of the following exemplifies a way in which humans modify ecosystems?
Answers
Answer:
what are the options
Explanation:
Answer:
A
Explanation:
Trees are cut down for wood and then to extend them to houses
which two semisynthetic penicillins can be used against gram-negative bacteria? multiple select question. dicloxacillin ampicillin nafcillin amoxicillin
Answers
The two semi-synthetic penicillin used against Gram-negative bacteria are: (2) ampicillin, (4) amoxicillin.
Penicillin is the class of antibiotic drugs used to treat the bacterial infections. The several infections that can be treated by penicillin are throat infections, meningitis, syphilis, etc. The penicillin consists of antibiotics like Carbenicillin, Dicloxacillin, Nafcillin, Oxacillin, etc.
Gram-negative bacteria are the microorganisms which do not retain the stain crystal violet. These bacteria have a thin layer of lipopolysaccharide over their cell wall. Ampicillin and amoxicillin can cross the cell wall of Gram-negative bacteria and hence are used for treatment against them.
Therefore the correct answer is option 2 and 4.
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what type of vaccine involves host synthesis of viral antigens? what type of vaccine involves host synthesis of viral antigens? conjugated vaccine subunit vaccine nucleic acid vaccine attenuated whole-agent vaccine toxoid vaccine'
Answers
A nucleic acid vaccine involves the host synthesis of viral antigens. A nucleic acid vaccine is a type of vaccine that uses genetic material such as DNA or RNA to stimulate an immune response.
It is distinct from traditional vaccines, which typically use weakened or dead forms of a pathogen to elicit an immune response. Because they don't rely on pathogens or toxins to stimulate immunity, nucleic acid vaccines have the potential to be safer and more effective than traditional vaccines. They are also relatively easy and inexpensive to manufacture.
The type of vaccine that involves host synthesis of viral antigens is the nucleic acid vaccine. This type of vaccine utilizes genetic material from the pathogen, either in the form of DNA or RNA, which is introduced into the host's cells. The host cells then use this genetic material to produce the viral antigens, which in turn stimulate an immune response. Nucleic acid vaccines are a promising approach as they can be rapidly developed and are often more stable than traditional vaccines.
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water functions in the body to perform muscle contractions, nerve conduction, waste elimination, joint lubrication, nutrient transport, metabolism, and proper fluid balance. select one: true false
Answers
The statement is True.
Water is an essential component of the human body, making up approximately 60% of total body weight. It plays a crucial role in various bodily functions, including muscle contractions, nerve conduction, waste elimination, joint lubrication, nutrient transport, metabolism, and maintaining proper fluid balance.
Adequate water intake is essential for optimal health, as dehydration can lead to a range of health problems, including fatigue, headaches, impaired cognitive function, and decreased physical performance. Thus, it is crucial to ensure that individuals consume enough water to maintain proper bodily functions and prevent dehydration. The recommended daily intake of water varies depending on factors such as age, sex, and physical activity level, but generally, it is recommended that individuals consume at least 8 glasses of water per day.
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please help on question about pre-mRNA -> protein
Answers
The splicing of the pre-mRNA of the rhesus protein to remove the exon that codes for a membrane anchor is unlikely in the production of this protein.
What occurs in the production of rhesus protein?The membrane anchor is an essential component that allows the protein to be embedded within the cell surface membrane of the red blood cells. Removing the exon that codes for the membrane anchor would prevent the protein from being anchored to the cell surface membrane, and it would not be able to perform its function. Therefore, this step is crucial in the production of the Rhesus protein, and it is unlikely to be removed by splicing the pre-mRNA.
Transcription of the protein occurred with the help of RNA polymerase within the nucleus, and once formed, the mRNA leaves the nucleus to get translated at a free ribosome within the cytoplasm. The protein formed was modified within the Golgi body, which is responsible for processing and modifying newly synthesized proteins before they are transported to their final destination.
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Image transcribed:
People who are Rhesus positive have one or more of three types of Rhesus antigen C, D or E, on the cell surface membrane of their red blood cells. Which of the following is unlikely in the production of this protein?
Splicing of the pre-mRNA of the rhesus protein to remove the exon that codes for a membrane anchor
Once formed, the mRNA leaves the nucleus to get translated at a free ribosome within the cytoplasm
The protein formed was modified within the golgi body
Transcription of the protein occurred with the help of RNA polymerase within the nucleus
mutual symbioses between fungi and animals include multiple select question. corals that associate with saltwater fungi. leaf cutters ants that cultivate fungi. fungi in the guts of ruminant animals. arbuscular mycorrhizal fungi in association with aphids.
Answers
Mutual symbioses between fungi and animals include: Leaf-cutter ants that cultivate fungi and Fungi in the guts of ruminant animals. These are the two main examples of mutual symbioses between fungi and animals.
Leaf-cutter ants have a mutual symbiotic relationship with fungi they cultivate.
The ants collect leaves and bring them back to their nests, where the leaves are used to grow a specific type of fungus.
The ants feed on the fungus, and the fungus receives a constant food supply from the ants, resulting in a mutually beneficial relationship.
Ruminant animals have fungi in their guts that help break down complex plant material, making digestion easier for the animal.
In return, the fungi receive a stable environment and nutrients from the animal's digestive system, resulting in a mutualistic relationship.
However, the examples of corals associating with saltwater fungi and arbuscular mycorrhizal fungi in association with aphids are not typical examples of mutual symbiotic relationships between fungi and animals.
Therefore, the correct answers are Leaf-cutter ants that cultivate fungi and Fungi in the guts of ruminant animals.
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what is habitat fragmentation? the destruction of habitats caused by natural disasters such as fires or hurricanes the extinction of species in habitat patches the process of a community changing in species composition over time the process of breaking up large habitats into a number of smaller habitats
Answers
Habitat fragmentation is the process of breaking up large habitats into a number of smaller habitats.
The process of dividing big habitats into smaller habitats is known as habitat fragmentation. This is caused by a variety of factors, including human activities such as urbanization, agriculture, and logging.
Habitat fragmentation can have negative impacts on biodiversity, including the extinction of species in habitat patches, the destruction of habitats caused by natural disasters such as fires or hurricanes, and the process of a community changing in species composition over time.
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damage to the axons in the spinal cord is typically permanent and debilitating. one promising avenue of treatment is transplanting certain cells into the spinal cord after injury. which cell type is the best choice to help repair a severed axon?
Answers
One promising type of cell for spinal cord repair is the oligodendrocyte precursor cell (OPC).
OPCs are immature cells that can differentiate into oligodendrocytes, which produce myelin, the insulating material around axons that is crucial for their proper function. Injuries to the spinal cord often result in the death of oligodendrocytes and the loss of myelin, which impairs axon function and connectivity. Transplanting OPCs can promote remyelination and axon regeneration in animal models of spinal cord injury.
OPCs can also promote tissue repair and reduce inflammation by secreting various growth factors and cytokines. Therefore, OPCs have the potential to enhance the repair of axons in the spinal cord, and may represent a promising therapeutic approach for treating spinal cord injury.
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as an up-and-coming scientist, you decide to create mutant bacteriophages to better examine structural function. your first mutant lacks a sheath motor protein, so you decide to call this muts. to determine the loss of function, you add muts at a multiplicity of infection (moi) of 1 to escherichia coli cells and examine the rate of infection. you're very surprised to find that none of the escherichia coli cells are infected. what happened? choose one: a. muts damaged the bacteriophage tail fibers. b. you failed to add enough bacteriophages to infect the e. coli. c. muts created ghost bacteriophages. d. muts prevented the sheath protein from contracting.
Answers
The E. coli strain H7 is a person from a gathering that creates a strong poison that hurts the small digestive system's coating. This may result in silly bloating. At the point when you ingest this kind of microorganisms, you get E. coli contamination. The correct answer is Coil.
E. coli serotype O157: H7 is an uncommon creature that has been connected to youngsters' hemolytic uremic disorder (HUS) and hemorrhagic colitis in all age bunches as of late.
These E. coli strains are primarily carried by cattle, in particular. Due to poor slaughter processing, their feces may end up contaminating other foods (like milk and vegetables) and water. Additionally, their feces may contaminate their meat.
The gram-negative bacterium Escherichia coli (E. coli) is the cause of numerous diarrheal diseases like dysentery and traveler's diarrhea.
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why is understanding the causes of different diseases important???
Answers
Answer:
It’s important to understand the causes of different diseases so that they can be prevented. If they can’t be avoided, then this information can help identify the proper route for treatment.
Explanation:
Answer:
Because by understanding the causes of these diseases, they can be prevented.
That is why it is so important to know the reasons why certain diseases appear.
during exercise, the filling time for the ventricles is greatly shortened, however cardiac output increases. how is this best explained? group of answer choices the increase in venous return during exercise compensates for the reduced filling time stroke volume decreases but an increased heart rate compensates for it stroke volume decreases but increased contractility compensates for it none of these
Answers
During exercise, cardiac output increases due to several factors such as increased heart rate and contractility.
The reduced filling time for the ventricles during exercise is compensated by the increased contractility of the myocardium, which enables the heart to eject a larger volume of blood with each beat, despite the shorter diastolic filling time.
Additionally, during exercise, the venous return to the heart is increased due to the muscle pump mechanism, which helps to push blood back to the heart. This increased venous return also increases the preload on the heart, which helps to improve stroke volume. Therefore, the combination of increased contractility and preload compensates for the reduced filling time, allowing the heart to maintain a higher cardiac output during exercise.
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the contraction of forest habitat at the end of the miocene epoch: group of answer choices caused the evolution of bipedal locomotion selected against apes with the ability to walk on two feet made an already existing ability to walk on two feet newly advantageous caused the proliferation of ape genera and species
Answers
The contraction of forest habitat at the end of the Miocene epoch "made an already existing ability to walk on two feet newly advantageous".
Bipedalism, the act of walking upright on two legs, is one of the most distinguishing features of human evolution. It has long been known that bipedalism first evolved among apes, but how and why it did so is still a mystery.
According to the fossil record, bipedalism appears to have first emerged among apes that lived in or near forests about 4.4 million years ago, during the Late Miocene epoch. However, this transition from walking on four limbs to walking on two legs was likely not an overnight phenomenon. Instead, it probably occurred gradually over a period of millions of years, and was likely driven by a combination of ecological and behavioral factors.
As the forests began to contract at the end of the Miocene epoch, bipedalism became newly advantageous for apes that could already walk on two feet. This made it easier for them to move around in the newly opened savannas and grasslands, which were much more open and expansive than the dense forests they were used to. Eventually, this newly acquired ability to walk on two feet became so useful that it led to the proliferation of ape genera and species.
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The evolution of the _______ in reptiles allowed animals to colonize land because it prevents the egg from _________.
A. jelly-coated egg; being eaten
B.amniotic egg; drying out
C.jelly-coated egg; drying out
D.amniotic egg; mutating
Answers
Answer:
The evolution of the amniotic egg in reptiles allowed animals to colonize land because it prevents the egg from drying out. The amniotic egg is a type of egg that has a shell and four membranes that protect the embryo from desiccation (drying out) and other environmental hazards. This adaptation allowed reptiles to lay their eggs on land, away from water sources, and to colonize terrestrial habitats that were previously inaccessible to them.
______________ evolved during the Carboniferous Period.
Answers
Ferns, horsetails and seed-bearing trees evolved during the Carboniferous Period.
What evolved during Carboniferous Period?Many important life forms evolved during Carboniferous Period, which lasted from about 359 million to 299 million years ago. One of the most notable developments during this time was the emergence and diversification of early amphibians, which were the first vertebrates to make the transition from water to land.
In addition, the Carboniferous Period saw the evolution of many types of plants, including ferns, horsetails, and seed-bearing trees. These plants formed vast forests that eventually gave rise to the coal deposits that are so important to the world's energy supply today.
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Gretta draws a phylogenetic tree for plants. She makes a list of flowering plants, gymnosperms, seed plants, vascular plants, and nonvascular plants. How does she organize her phylogenetic tree? Select all that apply.
a. She draws branches for gymnosperms next to nonvascular plants.
b. She draws branches for simple land plants after the branches for flowering plants.
c. She draws branches for flowering plants at the beginning of her tree.
d. She draws branches for flowering plants next to each other.
e. She draws a branch for gymnosperms after the branches for vascular plants.
f. She draws branches for gymnosperms at the beginning of her tree.
Answers
Gretta would probably classify the plants in her phylogenetic tree according to their evolutionary ties based on the information given.
What is a vascular plant, exactly?Those plants known as vascular plants have specialised vascular tissues for the transportation of food, minerals, and water. They have phloem for moving food from leaves to other parts of the plant and xylem for moving water and minerals from roots to various sections of the plant.
What makes plants vascular?Because they have a network of "tubes" that connect the plant's roots, shoots, and leaves and carry water and nutrients from one area to another, vascular plants are known as vascular.
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a patient has a mutant liver glycogen synthase kinase (gsk) that cannot be phosphorylated. answer the following questions (no more than 50 words per question) what would be the default state of this mutant gsk? how would this affect the amount and structure of glycogen particles in liver and in muscle cells? what clinical features will you expect to see if the mutation affects liver gsk only? what clinical features will you expect to see if the mutation affects muscle gsk only?
Answers
The default state of the mutant GSK would be in an active, unphosphorylated state. This would lead to increased glycogen synthesis and accumulation in the liver, but not in muscle cells. Clinical features would depend on whether the mutation affects liver or muscle GSK.
Since the mutant GSK cannot be phosphorylated, it would by default be in an active condition. This would boost the production of glycogen in both liver and muscle cells, leading to the development of bigger, branchier glycogen particles.
If the mutation only effects the liver GSK, the patient may experience hypoglycemia as a result of the liver's excessive production of glycogen. Due to the abnormal glycogen structure in muscle cells, if the mutation only effects muscle GSK, the patient may exhibit exercise intolerance or muscle weakness.
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a genetic cross performed many times produces 798 long stemmed plants and 266 short stemmed plants. the probability of obtaining a short stemmed plant in a similar cross is
Answers
The probability of obtaining a short-stemmed plant in a similar cross is 25%.
A genetic cross performed many times produces 798 long-stemmed plants and 266 short-stemmed plants. To find the probability of obtaining a short-stemmed plant in a similar cross, follow these steps:
1. Calculate the total number of plants produced in the cross:
Total plants = Long-stemmed plants + Short-stemmed plants
Total plants = 798 + 266 = 1064
2. Determine the probability of obtaining a short-stemmed plant:
Probability of short-stemmed plant = (Number of short-stemmed plants) / (Total number of plants)
Probability of short-stemmed plant = 266 / 1064
3. Simplify the fraction:
Probability of a short-stemmed plant ≈ 0.25
So, the probability of obtaining a short-stemmed plant in a similar genetic cross is approximately 0.25 or 25%.
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E. coli has three genes that each code for different enzymes that break down lactose. Scientists conducted investigations about the production of these enzymes using samples of E. coli. Their results are shown in the table.
E. coli Investigation Results
Environment Contains Lactose Enzyme Concentration
1 Yes High
2 No Low
Which conclusion about the production of the enzymes can be determined by analyzing the scientists’ results?
A.Gene expression for the enzymes was controlled by the temperature of lactose.
B.The genes for the enzymes were expressed in equal amounts during both experiments.
C.Gene expression for the enzymes was controlled by the presence of lactose.
D.The genes for the enzymes were mutated before being expressed by the presence of lactose.
Answers
The presence of lactose regulated the production of the enzyme genes, which is the best choice.
What tRNA molecule's 3 base site is equivalent to an mRNA codon?An anticodon is a three-base sequence that a tRNA molecule introduces to the matching codon of the mRNA during translation along with a particular amino acid. The anticodon sequence uses base pairs in the anti-parallel orientation and is complementary to the mRNA.
Which of the following procedures contributes to the fundamental principle of genetics?The "central dogma" is the method by which DNA is converted into proteins. It is not a linear process, though; rather, transcription and translation are needed, with the intermediary molecule RNA.
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a blockage in the lymphatic system could have which of the following consequences? question 1 options: a. an increase in interstitial fluid volume b. an increase in blood volume returning to the right atrium c. a reduction in colloid osmotic pressure d. (a) and (b) are correct; (c) is incorrect e. (a), (b), and (c) are all correct
Answers
A blockage in the lymphatic system could result in an increase in interstitial fluid volume and an increase in blood volume returning to the right atrium, but it would not lead to a reduction in colloid osmotic pressure, the correct option is (d).
The lymphatic system plays a crucial role in maintaining fluid balance in the body by draining excess interstitial fluid and returning it to the bloodstream.
When there is a blockage in the lymphatic system, excess interstitial fluid accumulates, leading to swelling or edema in the affected area. This excess fluid can also increase the volume of blood returning to the right atrium, leading to an increase in central venous pressure, the correct option is (d).
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The complete question is:
A blockage in the lymphatic system could have which of the following consequences? options:
a. an increase in interstitial fluid volume
b. an increase in blood volume returning to the right atrium
c. a reduction in colloid osmotic pressure
d. (a) and (b) are correct; (c) is incorrect
e. (a), (b), and (c) are all correct
what is the advantage of having a complete digestive tract (which started in mollusks) as opposed to the gastrovascular cavity found in cnidarians or the simple gut tube of some flatworms?
Answers
The complete digestive tract, which originated in mollusks, has numerous advantages over the gastrovascular cavity found in cnidarians and the simple gut tube of some flatworms.
This includes the ability to separate the mouth from the anus and the specialization of different segments for various digestive activities. The digestive tract is an organ system found in animals that is responsible for breaking down and absorbing nutrients from food. It's a series of organs connected from the mouth to the anus in the case of the complete digestive tract (mouth, pharynx, esophagus, stomach, small intestine, large intestine, and rectum), allowing for a one-way flow of food.
The gastrovascular cavity is a basic digestive system found in cnidarians, while the flatworms have a simple gut tube. Both of these systems have a single opening through which food enters and waste exits. Food digestion, nutrient absorption, and waste elimination all take place in the same cavity in this type of digestive system. A complete digestive tract has several advantages over these basic digestive systems. First and foremost, the separation of the mouth from the anus allows for a one-way flow of food, which is crucial for efficient nutrient absorption. In contrast to the gastrovascular cavity and the simple gut tube, the specialized segments of the digestive tract allow for specific digestive activities to take place. Enzymes and other digestive fluids are secreted in these segments to break down specific nutrients, allowing for more efficient absorption. Furthermore, because the complete digestive tract has multiple exits for waste removal, it allows for a faster and more efficient elimination of waste products.
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in the absence of ascorbic acid, prolyl oxidase is unable to oxidize proline residues in collagen to hydroxyproline, resulting in:
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Prolyl oxidase is unable to convert proline groups in collagen to hydroxyproline in the absence of ascorbic acid.
Without hydroxyproline, collagen cannot make collagen fibers correctly. Hydroxyproline is an essential part of collagen. Defective collagen is synthesized as a consequence, and it is weaker and less stable than regular collagen.
Collagen loses tensile strength and flexibility due to a dearth of hydroxyproline, making it more vulnerable to harm and deterioration. This can result in several health issues, such as blood artery fragility, weakened bones, and hindered wound recovery.
Prolyl oxidase and the production of hydroxyproline both depend on ascorbic acid, also known as vitamin C. A lack of this nutrient can cause illnesses like scurvy.
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