Answer:
RNA was the genetic material
Explanation:
If avary, Macleod, and McCarty would have seen that only RNAse treatment of the heat-killed bacteria prevented the transformation of genetic virulence than might have concluded that RNA is the genetic material as transformation does not occur because the RNAase degraded the RNA.
But if the genetic material is DNA then RNAase will not work because RNAase can not degrade DNA and then DNA will pass from virulent heat killed strain to nonvirulent strain and will cause transformation.
Why does it make biochemical sense that chaperones recognize hydrophobic surface area? What catastrophic event are chaperones meant to prevent in cells?
Answer:
Molecular chaperons in the cells helps in protein folding. These are the group of proteins that have functional similarity and they also assist protein folding.
They have the ability to prevent the non specific binding and aggregation by the binding of the non native proteins.
Molecular chaperons helps in recognizing the hydrophobic surfaces of the unfolded proteins because they themselves are hydrophobic in nature and will combine to the hydrophic binding and bonding.
This helps in guiding the protein to folding.
Chaperones recognize hydrophobic surface areas to prevent protein aggregation during folding, which could lead to cellular dysfunction. They help proteins fold correctly and prevent catastrophic events such as the formation of protein aggregates.
Explanation:Chaperones recognize hydrophobic surface areas because many proteins require assistance in the folding process to prevent them from aggregating during folding. Hydrophobic regions on the protein's surface tend to be exposed and can interact with other hydrophobic regions, leading to protein aggregation. Chaperones bind to these hydrophobic regions and help the protein fold correctly by preventing aggregation. This prevents catastrophic events such as the formation of protein aggregates, which can lead to cellular dysfunction and disease.
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Both crystal violet and safranin are basic stains and may be used to do simple stains on Gram-positive and Gram-negative cells. This being the case, explain how they end up staining Gram-positive and Gram-negative cells differently in the Gram stain.
Answer:
Gram positive cells retains purple stains of crystal violet, gram-negative cells takes up red or pink stain of safranin
Explanation:
Gram staining is a simple method of distinguishing between Gram-positive cells and Gram-negative cells. The method make use of the following steps
Heat fix cells on a glass slide and then:
1. Stain cell with crystal violet for 1 min
2. Rinse with water
3. Decolourize with alcohol for 30 seconds and wash with water
4. flood cells with iodine as mordant
5. Flush with water
6. Counterstain with safranin for 1 min and wash with water
7. View under the microscope
Gram-positive cell possesses a thick peptidoglycan in its cell wall which helps retain the purple colour of the crystal violet, While Gram-negative cells will loose out this colour when flushed with alcohol but takes up the red or pink colour when stained with safranin.
Crystal violet stains Gram-positive cells purple, while safranin stains Gram-negative cells pink in the Gram stain.
Explanation:Crystal violet and safranin are used in the Gram stain to differentiate between Gram-positive and Gram-negative cells. Crystal violet is a primary stain that binds to the thick peptidoglycan layer of Gram-positive cells, making them appear purple.
Safranin, a secondary counterstain, is applied after decolorization and stains the decolorized Gram-negative cells pink. The differences in cell wall structure and the interaction of the stains with the cell walls result in the differential staining of Gram-positive and Gram-negative cells.
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Have your partner focus on an object on the far side of the room (e.g., the chalkboard or a chart) for 1 minute. Then have your partner switch focus to an object in your hand (e.g., a pencil). Watch your partner's pupils carefully as the point of focus is changed from far to near.
Answer:
Eye changes or adjusts itself according to the focus of our eye.Explanation:
Pupil changes its shape when the eye focuses on near or far object. When we observe something far away, the pupil gets widen. But when we look at something near to us, pupil get smaller. So, pupil adjust itself according to the focus of our eye. If the partner switches from far to near, the pupil changes from wide to small. Pupil changes its shape according to bright as well as dim light also.
Case Study: Parents of a 3-year-old boy noticed that their son had a waddling gait, fell frequently and had difficulty getting up again, and was not able to run. By age five, there was progressive muscular weakness and muscle wasting, and weakness of the trunk muscles led to abnormal posture. By age 9 he was confined to a wheelchair. Contractures appeared, first in the feet, as the gastrocnemius muscles tightened.1. This hereditary X-linked recessive disease characterized by progressive muscular weakness is ______.2. What does dystrophy mean? Why is this term used to describe this case?
Answer:
Muscular dystrophy
Explanation:
Muscular dystrophy is an inheritable genetic condition that involves mutations in genes that encodes the production of muscle proteins to build and to preserve healthy proteins. The disease is characterised by progressive weakness and loss of muscle mass.
Dystrophy is referred to as a disorder in which the a tissue or an orgasm progressively wastes away. Thé dystrophy term is used because the disease involves the gradual and progressive wasting away of the muscle which is a type of tissue causing muscular weakness allowing the child to experience these symptoms.
in general terms, what two general factors in a medium affect small speed of sound?
Please explain! I will give brainliest!
Answer:
Density of the medium, temperature of the medium, and stiffness of the medium.
Explanation:
Medium
Medium has a huge effect of the speed of sound. When most people discuss the “speed of sound” they are talking about the propagation of sound waves through the medium of “Air”. For anyone who has gone underwater and listen to people talking above it is likely that one would notice the muted an “odd” way that voices sound underwater. This is because the “medium” of water greatly bends, distorts and changes the speed of sound wave.
There is a whole aspect of science that measure and defines the effect of different mediums (gaseous and liquid) on the speed of sound. This is called Fluid Dynamics. Underwater communication is possible if you understand how this wave propagation as well as another important factor (pressure).
Because of elasticity of materials sound will, as a rule of thumb, generally travel faster in solids than in liquids and faster in liquids than in gases.
♦ Temperature
Temperature has a large effect on the speed of sound. Not as much as the “Medium” does, but far more than anything else. Temperature affects the speed of sound because temperature can affect the “elastic” qualities of different mediums. At the very basics lower temperatures will decrease the speed of sound while higher temperatures will increase the speed of sound, all other factors being equal.
♦ Pressure
Pressure is the final factor that has a significant impact on the speed of sound. The effect of pressure on the speed of sound is due to the materials inertial properties. In short, the more pressure that is applied to the material or medium the denser it becomes and the greater the “inertia” becomes. This makes any interactions between particles slower. Therefore the speed of sound throughout the medium is slowed due to the greater pressure.
Answer:
Elasticity and density
Explanation:
The speed of sound depends on elasticity and density of the medium through which it is travelling. The sound travels faster in liquids than gases. The sound travels faster in solids than in liquids. The lesser the elasticity and higher the density , the sound travels slower in a medium.
Speed = elasticity / density
The blood of a normal healthy adult rat contains 5 million red blood cells (RBCs) per microliter of blood. Each deciliter of blood contains 15 grams of hemoglobin. Each RBC contains 250 million molecules of hemoglobin (a fairly close approximation for a healthy adult retired breeder female rat). Suppose you have 8.56 mL of blood drawn from a healthy female retired breeder rat, then how many RBCs are in this blood sample?
Answer:
There are 4.28×10^10 RBCs in the blood sample.
Explanation:
Concentration of blood = 5×10^6 RBC/10^-6 L = 5×10^12 RBC/L of blood
Volume of blood sample = 8.56 mL = 8.56/1000 = 0.00856 L
Number of RBCs = concentration of blood × volume of blood sample = 5×10^12 RBC/L × 0.00856 L = 4.28×10^10 RBCs
"Suppose that two parents are both heterozygous for sickle cell anemia, which is an autosomal recessive disease. They have eight children. Use the binomial theorem to determine the probability that three of the children have sickle cell anemia and five of the children are healthy. Round your answer to the nearest tenth."
If two parents are both heterozygous for sickle cell anemia, the probability that three of the children have sickle cell anemia and five of the children are healthy is approximately 0.1 or 11.0%.
What is the probability of sickle cell anemia?The probability of having 3 children with sickle cell anemia and 5 healthy children out of 8 total children is explained below.
The probability of any individual child having sickle cell anemia with heterozygous parents is 1/4; the probability of any individual child being healthy is 0.75.
Using the binomial theorem,
P(3 children with sickle cell anemia and 5 healthy children)
= [tex]P^8[/tex]₃ ×[tex](1/4)^3[/tex] ×[tex](3/4)^5[/tex]
= P(3 children with sickle cell anemia and 5 healthy children) = 0.110 that is nearly 11%.
Hence, if two parents are both heterozygous for sickle cell anemia, the probability that three of the children have sickle cell anemia and five of the children are healthy is approximately 0.1 or 11.0%.
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Final answer:
The probability that three out of eight children of two heterozygous parents for sickle cell anemia have the disease and five are healthy is approximately 20.7%.
Explanation:
The question posed involves the application of the binomial theorem to a genetic problem involving sickle cell anemia, an autosomal recessive disease. To find the probability that three of the children have sickle cell anemia and five of the children are healthy when both parents are heterozygous for the trait, we can use the binomial probability formula:
P(X=x) = ⁿCₓ × (p)ˣ × (q)ⁿ⁻ˣ
Where 'n' is the total number of children (8), 'x' is the number of children affected with sickle cell anemia (3), 'p' is the probability of having sickle cell anemia (1/4), and 'q' is the probability of being healthy (3/4). The term 'ⁿCₓ' represents the binomial coefficient, which can be calculated using combinations.
P(3 have sickle cell) = ⁸C₃ × (1/4)³ × (3/4)⁵
Calculating further:
P(3 have sickle cell) = 56 × (1/64) × (243/1024)
P(3 have sickle cell) = 56 × (243/65536)
P(3 have sickle cell) = 13608/65536
Converting to a percentage and rounding to the nearest tenth gives us approximately:
P(3 have sickle cell) = 20.7%
Imagine a genomic researcher who is analyzing the genome of different types of cats. She finds that a particular sequence in the North American Bobcat genome is exactly homologous to a sequence found in the common house cat, while all other sequences in those two genomes differ at many nucleotides. (The most recent common ancestor between bobcats and house cats is estimated to be about 6.8 million years, plenty of time for mutation to generate DNA sequence variation.)
Which of the following could explain the identical sequence in these otherwise differing genomes?
01. The sequence encodes a ribosomal protein that is critical for life and can not be easily mutated while retaining function.
02. The sequence is contained in a virus that has infected germline cells in both species.
03. The sequence is from an intron of a gene that encodes a muscle protein.
Provide a rationale for why each of the three statements above are correct or incorrect.
Answer:
Is 1. The sequence encodes a ribosomal protein that is critical for life and can not be easily mutated while retaining function.
Explanation:
A 35-year-old man presents with anemia, neutropenia, thrombocytopenia, myeloblasts with the presence of Auer rods, and one or two distinct nucleoli and promyelocytes. Cytochemistry examination demonstrates peroxidase and Sudan black B (SBB) positive and TdT terminal eoxynucleotidyl transferase (TdT) negative. This hematologic picture is consistent with: A) Acute lymphoblastic leukemia (ALL) B) Acute myeloblastic leukemia (AML) C) Chronic myelocytic leukemia (CML) D) Chronic lymphocytic leukemia (CLL) E) None of the above
Answer:
B) Acute myeloblastic leukemia (AML
Explanation:
It is a type of cancer that affects the blood as well as the bone marrow. It causes a situation whereby there is excessive immature white blood cells thus affecting the production of matured or normal white and red blood cells, blood platelets through the interference of the myeloid cells.
It is treatable medical condition that requires medical diagnosis. It requires laboratory tests with imaging always needed. It is also known as Acute myelogenous leukemia, acute granulocytic leukemia and acute nonlymphocytic leukemia.
Answer: The hematologic picture is consistent with Acute myeloblastic leukemia (AML). Therefore the correct option is B.
Explanation:
Leukaemia is a type of cancer that affects the blood forming tissues such as the bone marrow, leading to excessive or over production of abnormal blood cells ( usually the white blood cells). They occur in two categories; it can be Acute leukaemia or Chronic leukaemia.
There are different forms of acute leukaemia which is classified according to the lineage ( myeloid or lymphoid). They include:
- Acute Myeloblastic Leukaemia (AML)and
- Acute Lymphoblastic Leukaemia ( ALL).
In AML, the following Laboratory diagnosis differentiates it from other leukaemia, these includes:
- myeloblasts with the presence of Auer rods, and one or two distinct nucleoli and promyelocytes, and
-peroxidase and Sudan black B (SBB) positive and TdT terminal eoxynucleotidyl transferase (TdT) negative. Therefore the correct option is B
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Linda wrote a few steps to describe how carbon circulates between the atmosphere and living organisms. Step 3 is missing.
Step 1: Carbon enters the atmosphere as carbon dioxide from respiration and combustion
Step 2: Plants use carbon dioxide to make glucose
Step 3:
Step 4: Animals exhale carbon dioxide
Which of the following best describes step 3?
a. Dead animals are buried underground and form fossil fuels.
b. Carbon present in glucose is converted into oxygen by animals.
c. Carbon present in glucose is transferred into plant-eating animals.
d. Dead animals decompose and recycle the carbon in the form of carbon dioxide.
Answer:D
Explanation:decomposition of dead plants and animals.
Answer:
c
Explanation:
edit: yup its c i just did the test and got it right
what color would the ecoli cell appear under the microscope following a gram stain? explain why with a clear explanation of the gram staining process and the peritnent cellular componests that cause cells to stain
Answer:E. Coli will appear pink in color under the microscope following gram staining
Explanation:
The Gram stain is a differential technique that is commonly used for the purposes of classifying bacteria. The staining technique distinguishes between two main types of bacteria (gram positive and gram negative) by imparting color on the cells.
Being Gram-negative bacteria, E. coli have an additional outer membrane that is composed of phospholipids and lipopolysaccharides. The presence lipopolysaccharides on the outer membrane of bacteria gives it an overall negative charge to the cell wall. Because of these properties, E. coli does not retain crystal violet during the Gram staining process.
20. During filtration, which of the following does NOT enter the bowman's capsule from the bloodstream?
A. glucose
B. water
C. ions
D. plasma proteins
E. amino acids
Answer: option D.
Plasma proteins.
Explanation:
Filtration is the transfer of water and solutes like glucose, ions, amino acids from the plasma to the renal tubules in the renal corpsicule plasma . Renal corpsicule help to filter blood. The fluid present in the blood in the glomerulus is pass to the Bowman's capsule to form glomerular filtrate , plasma protein do not enter Bowman's capsule. The filtrate is processed to form urine in the nephron.
An individual suffers a blood clot in an artery that delivers blood to his leg. The leg begins to take on a blue hue, becomes colder than the rest of his body and he experiences numbness in the leg.
He is most likely experiencing:
a) anemic hypoxia
b) ischemic hypoxia
c) hypoxic hypoxia
d) histotoxic hypoxia
e) allergic hypoxia
Answer: Anemic hypoxia.
Explanation:
Anemic hypoxia is a medical condition that result when few haemoglobin are present in the blood leading to decreased ability of the blood to carry oxygen.
Oxygen is essential for human proper body functioning.
This is normally cause by lack of red blood cells ( haemoglobin) that carry oxygen.
The symptoms of anemic hypoxia are;
Skin color change to blue or cherry red.
Numbness
Cold hands and feet.
Headache
Weakness.
From the questions, the symptoms are similar to anemic hypoxia.
Restriction enzymes, which are extensively used in molecular biology, are products of bacterial "immune" system. Since bacterial genomes span several million base pairs (E. coli > 4 million bps), and the presence of restriction site within a genome is more than likely, how does a bacteria manage to protect itself from innate REs?
Explanation:
In the term of disease by intracellular microscopic organisms, they have the ability and repeated inside phagocytic cells, which causes the circulating antibodies to be distant to intracellular bacteria The natural safe reaction against these microscopic organisms is intervened basically by phagocytes and NK cells Innate immunity additionally arrives in a protein substance structure, called innate humoral immunity. for example, the body's supplement system and substances called interleukin-1 (which causes fever) and interferona) There are abiotic and biotic factors that keep population size of species under control Provide two examples that affect (or have affected) population size of native species in WA (one example for density dependent and one for density independent). You need to pick two real examples, provide specific species names and habitats/ecosystems where those species occur. You need to use different examples from the ones I used in lectures
b) Explain why species that overlap a great deal in their fundamental niches have a high probability of competing. Now explain why species that overlap a great deal in their realized niches and live in the same area, probably do not compete significantly.
c) What are two important pieces of ecological knowledge that you could apply to the decisions you make about your own life? Mainly from an ecological point of view, but it is good that you make a connection with other aspects of your life (personal or professional).
Answer:
Answer A: A key example of biotic and abiotic protagonists is the presence of rocky structures that exist in the mountains of southern Argentina (Patagonia) that thanks to them exist the ideal temperatures for the procreation of animals such as the Patagonian condor and the possibility of form nests in these rocky structures that secure their young.
Answer B:
The explanation for this is that they have populations that increase in number and have their niches realized and constant resources necessary to develop the species in this way, considering the dominant species in the area and the one that predominates, in the case of fundamental niches. , they are necessary and even obligatory niches for many populations, which if populations do not go to them, they die, extinguish, or decrease their number, that is why in the face of this reality made up of a fundamental niche since it is fundamental to persist life of the population.
Answer c:
Two key principles are: one, maintain the equilibrium, that is why one seeks to have an environmentalist position preserving nature, and two considering that we are part of that balance but that does not give us the right to break it with environmental contamination, intervention of natural food systems and many other factors.
Explanation:
It can be desirable to produce eukaryotic proteins in prokaryotes such as E. coli. To do this several DNA sequences must be joined or cloned together. Place the labels in the appropriate position to allow for the expression of the insulin protein in a prokaryotic cell. cDNA is DNA that is synthesized from the mature mRNA from eukaryotic cells using the enzyme reverse transcriptase.
5' end_____ _____ _____ _____3' end
1. Poly A signal sequence
2. Genomic clone of insulin
3. Rho terminator
4. -95 CAT box
5. -10 TATA
6. -25 TATA box
7. cDNA of insulin
8. -35 Sequence
Final answer:
To express human insulin protein in E. coli, assemble a DNA sequence that contains a -35 Sequence, a -10 TATA box for bacterial promoter activity, followed by the cDNA of insulin, and ends with a Rho terminator for transcription termination.
Explanation:
To generate a recombinant version of the human insulin protein in a prokaryotic cell, such as E. coli, we need to assemble a proper expression cassette. The cassette needs to have a eukaryotic gene sequence converted to cDNA, combined with prokaryotic promoter and terminator sequences to allow expression in the host cell.
The correct order for cloning the insulin cDNA for expression is:
-35 Sequence-10 TATA box (also known as Pribnow box)cDNA of insulinRho terminator or a similar prokaryotic terminatorThe -35 Sequence (-35 Sequence) and the -10 TATA box (-10 TATA) are part of the bacterial promoter necessary for initiating transcription. The cDNA of insulin (cDNA of insulin) is inserted downstream of the promoter, allowing it to be transcribed. Finally, a termination sequence such as the Rho terminator is necessary to stop transcription.
Why these specific elements? The -35 and -10 sequences are binding sites for the RNA polymerase in prokaryotes, which begins the process of transcription. The cDNA of insulin, obtained through reverse transcription of the mature mRNA, excludes any introns that would be present in the genomic clone (thus, we do not use the Genomic clone of insulin or the Poly A signal sequence which are found in eukaryotic expression systems). The Rho terminator is a sequence that facilitates the termination of transcription in prokaryotes.
Two strains of true-breeding maize that both produce ears of corn with white kernels are crossed and found to produce F1 plants that all make ears of corn with red kernels. If these F1 plants are backcrossed to one of the parents, what proportion of the offspring should have white kernels
Answer:
50% or 1/2
Explanation:
Since the two parental strains are true-breeding, that is, homozygous; it means that the red kerneled ears of corn offspring produced are heterozygous and one of the parental strains must have been homozygous recessive and the other homozygous dominant.
Assuming the ears of corn's colour is coded for by the allele A, it means that one of the parents has the genotype AA while the other has aa with the offspring having genotype Aa.
AA and aa genotypes produce white ears of corn while Aa genotype produce red ears of corn.
Thus, if Aa is backrossed to either of AA and aa:
Aa x AA = AA (white), AA (white), Aa (red) and Aa (red)
Aa x aa = Aa (red), Aa (red), aa (white) and aa (white)
Hence, the proportion of the offspring with white kernels will be 50% in each case.
g Imagine two different sigma factors with different promoter recognition sequences. What would happen to the overall gene expression profile in the cell if one sigma factor were artificially overexpressed?
Answer:
Transcription is affected via the modulation of the concentrations of the different types of holoenzymes, so saturated promoters are only weakly affected by sigma factor competition. However, in case of overlapping promoters or promoters recognized by two types of sigma factors, we find that even saturated promoters are strongly affected. Active transcription effectively lowers the affinity between the sigma factor driving it and the core RNAP, resulting in complex cross-talk effects. Sigma factor competition is not strongly affected by non-specific binding of core RNAPs, sigma factors and holoenzymes to DNA. Finally, we analyze the role of increased core RNAP availability upon the shut-down of ribosomal RNA transcription during the stringent response. We find that passive up-regulation of alternative sigma-dependent transcription is not only possible, but also displays hypersensitivity based on the sigma factor competition.
In cattle, coats may be solid white, solid black, or black-and-white spotted. When true-breeding solid whites are mated with true-breeding solid blacks, the F1 generation consists of all solid white individuals. After many F1×F1 matings, the following ratio was observed in the F2 generation: 12/16 solid white 3/16 black-and-white spotted 1/16 solid black Part A How many gene pairs are involved in the inheritance of cattle coat color?
Answer:
Two locus
Explanation:
Let assume the gene for white (1st locus) be W i.e. ww = recessive
here, Let the alleles for the 2nd locus be B and b.
White: W_B_, W_bb
Black&White: wwB_
Black: wwbb
This is dominant epistasis. In dominant epistasis, where the dominant allele of the 1st locus (W) masks or hide the expression of the 2nd locus. When the two alleles in the 1st locus are recessive (ww), the alleles in the 2nd locus can be expressed(B and b).
Two locus gene pairs are involved in the inheritance of cattle coat color
Assumption of Generation:Let us assume the gene for white (1st locus) be W i.e. ww = recessive here, Let the alleles for the 2nd locus be B and b.
White: W_B_, W_bbBlack&
White: wwB_Black: wwbb
This represents dominant epistasis. In this, the dominant allele of the 1st locus (W) should mask or hide the expression of the 2nd locus. At the time when the two alleles in the 1st locus are recessive (ww), so the alleles in the 2nd locus could be expressed(B and b).
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Some renewable energy resources, such as solar energy and hydropower, have several important advantages over nonrenewable resources. Which is an advantage of renewable energy resources?
Answer:
A,B,F
Explanation:
I just took the test and got it correct.
Renewable energy resources such as solar and hydropower are sustainable, enable decentralized power generation, and help in combating climate change by reducing greenhouse gas emissions.
One significant advantage of renewable energy resources such as solar energy and hydropower over nonrenewable resources is their ability to be replenished within a short timeframe. This makes them sustainable for long-term use. Solar energy, for example, is essentially inexhaustible, as it comes from the sun, which is expected to continue shining for another 4-5 billion years.
Renewable energy also facilitates local, decentralized control over power, allowing homes, businesses, and isolated communities to generate electricity with technologies such as solar panels without being dependent on a grid or centralized power plant. This is particularly beneficial for remote locations where connecting to the power grid is impractical or too expensive.
You purchase two identical houseplants and place them side by side on your windowsill. You water both plants equally. One plant, plant a, you leave alone. On the other plant, plant b, you inject florigen into the cells of the apical meristem. Which of the following would you expect to occur? (Select all answer options that apply.)
a. While flowering, the leaves will stop growing on plant b because florigen inhibits leaf meristems.
b. In subsequent years, plant b will only generate flowers in the same place.
c. Plant b will produce a flower at its apical meristem.
d. Once the flower is gone, plant b will grow taller, but only from growth of lateral meristems.
e. Both plants will produce flowers at the same time, but plant b will have more of them.
f. Once the flower is gone, the apical meristem will develop again, and plant b will continue to grow from that stem.
Answer: Option E.
Both plants will produce flower at the same time but plant B will have more of them.
Explanation:
Florigen is a protein or hormone like molecule that help to control or boost flower production in plants. It is a flowering hormone. It is produced in plants leaves and acts on apical meristem or growing tips. The hormone is normally injected at the growing tips or apical meristem where flowers growth will be activated. Since plant B is injected with florigen,it will further trigger flower production on the plant than plant A .
Answer: C
Explanation:
Which of the following processes directly require ATP? Choose all that apply.
a.Diffusion of calcium into the motor neuron at the neuromuscular junction.
b.Release of cross-bridge (interaction) between actin and myosin.
c.Movement of calcium ions back into the sarcoplasmic reticulum after contraction.
d.Conformation change of troponin resulting in the movement of tropomyosin off the actin active sites.
Answer:the following processes directly require ATP includes:
B (Release of cross-bridge (interaction) between actin and myosin.)
C (Movement of calcium ions back into the sarcoplasmic reticulum after contraction.)
Explanation:
Muscle contraction occurs in various parts of the body to ensure proper body functioning. This process requires the release of calcium ion and the use of ATP ( Adenosine Triphosphate) as source of energy at various levels for the process to take place. The distinct role of ATP in muscle contraction includes:
-ATP is directly required as it causes detachment from actin after power stroke when it binds at one of the reactive sites of myosin. This explains option B (Release of cross-bridge (interaction) between actin and myosin.)
-it powers the pump that transports calcium ions back into the sarcoplasmic reticulum after contraction. This explains option C (Movement of calcium ions back into the sarcoplasmic reticulum after contraction)
- it activates the myosin head so it can bind to actin and rotate by the action of ATpase enzyme.
The following processes which directly require ATP include
Release of cross-bridge (interaction) between actin and myosin.Movement of calcium ions back into the sarcoplasmic reticulum after contraction.The myosin head has to move in order for the release of cross-bridge
between them to be achieved. The myosin head movement brings about
movement of the actin also.
The movement of calcium also requires the use of ATP as it helps in the
pumping of the calcium ions into the sarcoplasmic reticulum.
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The main purpose of the chloride channel proteins on the apical surface of the intestinal epithelial cells is to create an osmotic gradient that ultimately causes __________ to move through the intestinal epithelium and into the intestinal lumen to assist in the formation of mucus. These are the same chloride channel proteins, adversely affected in cystic fibrosis.
Answer:
Water.
Explanation:
Cystic fibrosis may be defined as the medical condition in which the excess mucus is produced by the body that might affects the lungs and pancreas as well.
This occurs due to the abnormal functioning of chloride channel proteins. The main function of this protein is the transport of the water molecule that helps in the mucus formation. The water moves through the epithelial cells and maintain the role in epithelial fluid and fluid transport.
Thus, the answer is water.
Black hair in rabbits is determined by a dominant allele, B. White hair results when a rabbit is homozygous for the recessive allele, b. Two heterozygote rabbits mate and produce a litter of three offspring.a.What is the probability that the offspring are born in the order white-black-white
Answer:
The answer if 3/64 or 4.69%
Explanation:
B-: Black; bb: white
Two heterozygote rabbits: 2^2 = 4 possible gametes.
P: Bb x BbF1: 1/4 BB; 2/4 Bb; 1/4 bbthere are 3/4 black and 1/4 whiteThe probability that their offspring are given birth following the order white - black - white is:
1/4 white * 3/4 black * 1/4 white = 3/64 = 4.6875%Final answer:
The probability that two heterozygous rabbits will produce offspring in the order of white-black-white is 3/64.
Explanation:
The question pertains to the probability of obtaining offspring with specific phenotypes in a certain order when two heterozygous rabbits mate. Since black hair in rabbits is determined by a dominant allele B, and white hair is the result of being homozygous for the recessive allele b, a heterozygous rabbit (Bb) can produce both black and white offspring. When two heterozygous (Bb) rabbits mate, each offspring has a 3/4 chance of being black (genotypes BB or Bb) and a 1/4 chance of being white (genotype bb).
To calculate the probability of the offspring being born in the order white-black-white, we can use the multiplication rule of independent events:
Probability of first offspring being white (bb): 1/4.
Probability of second offspring being black (BB or Bb): 3/4.
Probability of third offspring being white (bb): 1/4.
Multiplying these probabilities together:
(1/4) x (3/4) x (1/4) = 3/64
So, the probability that the offspring will be born in the order white-black-white is 3/64.
25] Discuss what your analysis above indicates about the applicability of the Hardy-Weinberg criteria to this population. Which assumptions, if any, of the Hardy-Weinberg criteria are violated
Answer: The Hardy–Weinberg principle, also known as the Hardy–Weinberg equilibrium, model, theorem, or law, states that allele and genotype frequencies in a population will remain constant from generation to generation in the absence of other evolutionary influences.
Explanation: the population is in Hardy-Weinberg equilibrium for a gene, it is not evolving, and allele frequencies will stay the same across generations. There are five basic Hardy-Weinberg assumptions: no mutation, random mating, no gene flow, infinite population size, and no selection.
Fill in the blanks, your answers should be lower case. The process of makes an RNA copy from a DNA template. The RNA is made by the enzyme which makes the RNA from (3 or 5)' to (3 or 5)'. The process of makes a protein from an mRNA. The protein is made by the which catalyzes the peptide bond between each amino acid. In bacterial cells these two processes occur in the cytoplasm. However, in eukaryotes the process occurs in the nucleus and the process occurs in the cytoplasm
The process of transcription makes an RNA copy from a DNA template.
The RNA is made by the enzyme RNA polymerase which makes the RNA from (3 or 5)' to (3 or 5)'.
The process of Translation makes a protein from an mRNA.
The protein is made by the peptidyl transferase in Ribosome which catalyzes the peptide bond between each amino acid.
In bacterial cells these two processes of transcription and translation occur in the cytoplasm.
However, in eukaryotes the process of transcription occurs in the nucleus and the process of translation occurs in the cytoplasm
Explanation:
The process of translation and transcription comprises the central dogma of molecular biology. By these two process the information stored in the genes flows into the proteins.
The process of creating an RNA copy from a DNA template is known as transcription, conducted by the enzyme RNA polymerase. The translation of an mRNA into a protein is facilitated by the ribosome. While both processes occur in the cytoplasm in bacteria, in eukaryotes, transcription occurs in the nucleus and translation in the cytoplasm.
Explanation:The process of transcription makes an RNA copy from a DNA template. The RNA is made by the enzyme RNA polymerase which makes the RNA from 3' to 5'. The process of translation makes a protein from an mRNA. The protein is made by the ribosome which catalyzes the peptide bond between each amino acid. In bacterial cells, these two processes occur in the cytoplasm. However, in eukaryotes, the transcription process occurs in the nucleus and the translation occurs in the cytoplasm.
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In a car accident, the injuries to the driver include a fractured femur. Part of the broken end of the diaphysis of the femur is visibly penetrating through the skin in the thigh area. Which of the following properly classifies this fracture?1. Displaced, complete, open (compound)
2. Nondisplaced, complete, open (compound)
3. Displaced, complete, closed (simple)
4. Nondisplaced, incomplete, closed (simple)
5. Displaced, incomplete, open
Answer:
1. Displaced, Complete, Open
Explanation:
Displaced fracture - in this type of bone fracture the bone breaks into two or more parts, moves and thus the parts become misaligned.
Complete fracture - in this type of fracture the parts of bone become completely separated from rest one.
Open fracture - it is compound fracture where skin breaks and part of the broken bone protrudes out of the skin.
. The Vmax of a glucose transport into a certain preparation of red blood cells is determined to be 1206nmol glucose/s without ATP present in the buffer. The Vmax of a glucose transport into the same preparation of red blood cells is determined to be 1158nmol/s when ATP is present in the buffer. However, when cells that do not express any glucose transporter are probed for rate of glucose transport, Vmax is infinite. Explain these data.
Answer:
During a process of facilitated diffusion, presence ATP increase the rate of glucose transport through a protein membrane of a red blood cells. Vmax will then be on increased.
Explanation:
The rate of glucose transport in to the certain sample of blood is a facilitated diffusion process. It is understandable to be on increase through a protein membrane with the presence of enzyme to speed up the rate of the biological reaction. This is reflected when it was initially observed that with the enzyme ATP presence in the buffer, the Vmax of a glucose transport into the certain preparation of red blood cells is determined to be 1206nmol glucose/s(considerably low).
When ATP is added, the Vmax is on increase. That is, the Vmax of a glucose transport into the same preparation of red blood cells is determined to be 1158nmol/s. Then Vmax rises to infinity, when cells that do not express any glucose transporter are probed for rate of glucose transport.
It can be further explained that glucose moves into the blood through the permease in the membrane between the cell and the blood.Thus, ATP is used as an energy source to drive Na+ out of the cell, resulting in glucose transport from the intestine to the blood.
Membrane proteins must have an asynchronous distribution on the cell membrane for the system to function. This is an example of the membrane synthetic apparatus determining where in a membrane a protein should be localized. The Na+K+ ATPase must be localized to the membrane between the cell and our blood.
The data reflects that glucose transport via GLUT1 is highly efficient and typically passive, but slightly inhibited by ATP. Without GLUT1, glucose is minimally transported across the membrane, emphasizing the necessity of transport proteins. The minor reduction in Vmax with ATP indicates possible regulatory mechanisms.
The data indicates different behaviors for glucose transport under various conditions:
Vmax of glucose transport without ATP is 1206 nmol/s, showing a high rate of passive, protein-mediated transport via GLUT1.When ATP is present, Vmax decreases slightly to 1158 nmol/s, possibly due to competition or inhibition of the transport process.Without glucose transporters, Vmax is infinite, suggesting that glucose does not pass through the membrane efficiently without these transporters, highlighting the necessity of proteins like GLUT1 for effective glucose transportation.The data helps illustrate passive transport via GLUT1, a glucose transporter protein, and how ATP impacts the transport rate.
Many chemotherapy drugs target rapidly dividing cells, such as cancer cells. Why might cancer cells divide, and therefore evolve, more quickly than other cells
Answer:
The faster that cancer cells divide, the more likely it is that chemotherapy will kill the cells, causing the tumor to shrink. They also induce cell self-death or apoptosis. Chemotherapy drugs that kill cancer cells only when they are dividing are called cell-cycle specific.
Explanation:
Imagine that zika virus has a 1% incidence in the population. A test for the virus has a 3% false positive rate and no false negative rate. If someone takes the test and gets a positive result, what is the chance that they are infected?
Answer:
There is 97% Chance that the person is infected.
Explanation:
According to the question, the test for the virus has 3% false positive and there is no false negative rate.
Therefore if someone takes the test and gets a positive result, the chances of the person to be infected ( True positive ) is
100 - 3 = 97%
Note: False positive result is a result that indicates that a given condition is present when it is not actually present.