Pasteurization is different from other heat methods used to kill microorganisms because other heat methods are sterilization methods, whereas pasteurization is not. Hence, option B is correct.
Pasteurization is a technique used to disinfect liquids like milk, fruit juices, etc. to kill pathogenic microbes that are present in them without impacting the flavor or nutritional value. The pasteurization process heats the liquid to a specific temperature and holds it at that temperature for a certain amount of time before rapidly cooling it down. This technique can kill most pathogenic microbes without drastically altering the original composition of the liquid. Additionally, pasteurization helps to extend the shelf life of the product.
Unlike other heat methods, pasteurization is not a sterilization method. While pasteurization kills a significant number of microorganisms, it does not eliminate them all. Because pasteurization isn't a sterilization technique, there may still be some bacteria in the milk that could cause illness if it is kept for too long or at the wrong temperature. Hence, option B is correct.
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Cell structure and functions and transport review
Help needed
The primary building block of life are cells. So, the lowest autonomous component we would discover if we dissected an organism down to its cellular level is the cell.
Each component of the cell structure serves a distinct purpose that is necessary to carry out life's processes. These elements consist of the cell wall, cell membrane, cytoplasm, nucleus, and cell organelles.
Major tasks that are crucial for an organism's growth and development are carried out by cells. The following are some crucial cell functions:
Supports and structuresencourage growth Mitosis Enables the Transfer of Chemicals and Energy Production Supports ReproductionHence, it is clear why cells are regarded as the basic building block of all living things. They serve a number of purposes, including giving the organisms structure.
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No Service 12:50 AM Expert Q&A Done need help with these questions 1. The polytrichum spimoss belongs to the phylum---? 2. The ---- is the dominant life form of the mosses and the is the smaller portion of the life cycle
3. The ---- of the Merchantia sp is the diploid stage in the life cycle. 4. The asexual reproduction form of the marchantia sp is called ---- 5. Marchantia sp belongs to the phylum - 6. What type of cell division would occur within the capsule of the polytrichum sp. 7. Which of the following is the location where zygotes and embryos form in a moss? 8. The antheridia produces ------ whereas archegonia houses the... 9. which of the following characteristics doesnt not describe monocot plants a. has one cotyledon b. have parallel venation c. generally have a fibrous roots system d. sporophyte generation is dominant e. has netlike venation 10. Leaves on a monocots and eudicot are attach to the stem at the 11. label this diagram 76707AUS Label diagram 2 .
1. The polytrichum spimoss belongs to the phylum Bryophyta.
2. The gametophyte is the dominant life form of the mosses, and the sporophyte is the smaller portion of the life cycle.
3. The sporophyte of Merchantia sp is the diploid stage in the life cycle.
4. The asexual reproduction form of Marchantia sp. is called gemmae.
5. Marchantia sp belongs to the phylum Hepatophyta
6. Meiosis would occur within the capsule of Polytrichum sp.
7. Zygotes and embryos form within the archegonia of a moss.
8. The antheridia produces sperm whereas archegonia house the egg.
9. The characteristic that does not describe monocot plants is netlike venation.
10. Leaves on monocots are attached directly to the stem and eudicots are attached to the stem via a short stalk, the petiole.
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What are 5 ways species interact with each other?
Answer:
Explanation:
Competition.
Predation.
Parasitism.
Mutualism.
Commensalism.
During glycolysis and the krebs cycle, the cell extracts electrons from various compounds, now to the electrons get transferred to the etc?
During glycolysis and the Krebs cycle, the cell extracts electrons from various compounds, which are then transferred to the electron transport chain (ETC).
The ETC is a series of proteins located in the inner mitochondrial membrane, where they accept electrons and use them to pump protons across the membrane.
This creates a proton gradient, which is then used to drive the production of ATP through the enzyme ATP synthase. Thus, the electrons from glycolysis and the Krebs cycle are used to generate ATP, the energy currency of the cell, through the ETC.
During glycolysis and the Krebs cycle, the cell extracts electrons from various compounds through the process of oxidation.
These electrons are then transferred to the electron transport chain (ETC) by the carrier molecules NADH and FADH2.
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Two of the following DNA sequences code for the same small protein. Using the codon will determine the amino acid sequence of that proteins 
AUG UUA ACG AGA AGU
AUG CUA AGG AGU UCG
AUG CUC ACU CGU AGC
Using the codon wheel, the first and third sequences (AUG UUA ACG AGA AGU and AUG CUC ACU CGU AGC) code for the same protein:
Methionine-Leucine-Threonine-Arginine-Serine.What are codons?Codons are DNA or RNA molecule that specifies for a particular amino acid through a group of three successive nucleotides. Some codons serve as translational start or stop signals. Start, halt, or termination codons are the names given to these.
Using the codon wheel, we can translate each codon into the corresponding amino acid:
AUG UUA ACG AGA AGU: Met-Leu-Thr-Arg-Ser
AUG CUA AGG AGU UCG: Met-Leu-Arg-Ser-Ser
AUG CUC ACU CGU AGC: Met-Leu-Thr-Arg-Ser
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1. Penelope complains of increased feelings of detachment to her surroundings, as though she is taking a part in a movie or a dream. She feels as if she is observing herself from outside her body, living in a world she recognizes but cannot feel. Penelope is suffering froma.tHallucinationb.tDelusionc.tDepersonalizationd.tDerealization
Penelope is experiencing depersonalization, as she feels detached from her own sense of self and her surroundings.
Depersonalization explained.Depersonalization is a dissociative symptom that involves a sense of detachment from one's own thoughts, feelings, or sensations, or a sense of being disconnected from the world around them. It can also involve feeling as though one is observing oneself from outside one's body, as Penelope described. Hallucinations involve perceiving things that are not actually present, while delusions involve believing things that are not true.
Therefore, depersonalization involves feeling detached from one's surroundings or feeling as though the world around oneself is not real or is distorted in some way.
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Alternative splicing is...
- only present in prokaryotes
- uncommon and has only been observed for a small percentage of human genes
- the process of selecting different combinations of exons to make distinct protein isoforms
- necessary for all polycistronic RNA transcripts
Alternative splicing is =necessary for all polycistronic RNA transcripts. Rather, alternative splicing is a process in which different combinations of exons are joined together to create multiple mature mRNA transcripts from a single gene. This process allows for the production of different protein isoforms from the same gene, increasing the diversity of the proteome. Polycistronic RNA transcripts, on the other hand, are RNA molecules that contain multiple coding regions, each of which can be translated into a separate protein. While alternative splicing can occur in polycistronic RNA transcripts, it is not necessary for their function.
How does the brown type of adipose tissue help keep hibernating animals body temperatures from dropping too low?
The brown type of adipose tissue helps to maintain the body temperature of hibernating animals by generating heat through a process known as thermogenesis.
This is achieved through the activity of uncoupling proteins (UCPs) present on the inner mitochondrial membrane that regulate the dissipation of energy as heat. The mitochondria of brown adipose tissue are highly active, and UCPs facilitate the movement of protons across the mitochondrial inner membrane. This leads to heat generation as the movement of protons produces energy that is released as heat.
Therefore, the brown type of adipose tissue helps to maintain the body temperature of hibernating animals by generating heat through thermogenesis.
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I. Differentiate the following: a. Glycolysis, Kreb Cycle and Electron Transport Chain b. Photosynthesis and Chemosynthesis c. Transduction, Transformation and Conjugation II. What are the importance of the following in the field of Sanitary Engineering a. Physiology b. Genetic Engineering c. Gene Therapy III. Briefly define the following and give five examples each: a. Photoautotrophs b. Photoheterotrophs c. Chemoautotrophs d. Chemoheterotrophs e. Chemotrophs f. Phototrophs g. Chemolithotrophs h. Chemoorganotrophs i. Autotrophs j. Heterotrophs
Glycolysis is the process of breaking down glucose molecules into two molecules of pyruvate in the cytosol of the cell. The Kreb Cycle occurs in the mitochondria and is responsible for breaking down the products of glycolysis to produce energy in the form of ATP
This question consists of 3 parts, the answer is:Part I:
Glycolysis is the process of breaking down glucose molecules into two molecules of pyruvate in the cytosol of the cell. The Kreb Cycle, also known as the Citric Acid Cycle, occurs in the mitochondria and is responsible for breaking down the products of glycolysis to produce energy in the form of ATP. The Electron Transport Chain is the final step in cellular respiration, where the energy from the Kreb Cycle is used to create a proton gradient that drives the synthesis of ATP.Photosynthesis is the process by which plants convert light energy into chemical energy in the form of glucose. Chemosynthesis is the process by which certain organisms, such as bacteria, use inorganic chemicals as an energy source to produce organic compounds.Transduction is the process by which DNA is transferred from one cell to another via a virus. Transformation is the process by which DNA is taken up by a cell from its environment. Conjugation is the process by which DNA is transferred from one cell to another via a pilus.Part II.
Physiology is important in the field of Sanitary Engineering because it helps us understand how the human body functions and how it responds to different environmental conditions, such as exposure to pollutants.Genetic Engineering is important in the field of Sanitary Engineering because it allows us to modify the DNA of organisms to create new strains that can be used to break down pollutants or produce useful products.Gene Therapy is important in the field of Sanitary Engineering because it can be used to treat diseases that are caused by genetic mutations, such as cystic fibrosis.Part III:
a. Photoautotrophs are organisms that use light energy to produce organic compounds from inorganic sources. Examples include plants, algae, and some bacteria.b. Photoheterotrophs are organisms that use light energy to produce organic compounds from organic sources. Examples include purple non-sulfur bacteria and green non-sulfur bacteria.c. Chemoautotrophs are organisms that use inorganic chemicals as an energy source to produce organic compounds from inorganic sources. Examples include sulfur bacteria and iron bacteria.d. Chemoheterotrophs are organisms that use organic compounds as an energy source to produce organic compounds from organic sources. Examples include animals, fungi, and most bacteria.e. Chemotrophs are organisms that use chemicals as an energy source. Examples include chemoautotrophs and chemoheterotrophs.f. Phototrophs are organisms that use light as an energy source. Examples include photoautotrophs and photoheterotrophs.g. Chemolithotrophs are organisms that use inorganic chemicals as an energy source and inorganic sources as a carbon source. Examples include sulfur bacteria and iron bacteria.h. Chemoorganotrophs are organisms that use organic chemicals as an energy source and organic sources as a carbon source. Examples include animals, fungi, and most bacteria.i. Autotrophs are organisms that produce their own organic compounds from inorganic sources. Examples include photoautotrophs and chemoautotrophs.j. Heterotrophs are organisms that obtain organic compounds from other organisms. Examples include animals, fungi, and most bacteria.Learn more about photosynthesis at https://brainly.com/question/19160081.
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Does CCA have a CpG site present in the 3' downstream at the
first step of splicing ?
Cause I know that the CpG site has protein called methyl CPG-
binding protein that bounds with the methylated grou
No, CCA does not have a CpG site present in the 3' downstream at the first step of splicing.
The CpG site, also known as a CpG island, is a region of DNA where a cytosine nucleotide is followed by a guanine nucleotide in the linear sequence of bases.
These sites are often found in the promoter regions of genes and are involved in the regulation of gene expression. However, the CCA sequence does not contain a CpG site, as it does not have a cytosine followed by a guanine.
Therefore, the methyl CPG-binding protein would not bind to this sequence. It is important to note that the presence or absence of a CpG site can affect the regulation of gene expression and the process of splicing.
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happens best in solid materials.
A) conduction
b) Convection
c) Radiation
Describe one forestry strategy that could be employed to meet both the Onceler's economic need of forest production and the Lorax's need for forest health.
(from The Lorax 1972)
By employing sustainable forestry management, the Onceler could continue to produce forest products while also protecting the health and biodiversity of the forest ecosystem,
What is Biodivesity?
Biodiversity, short for "biological diversity," refers to the variety of living organisms on Earth, including the genetic diversity within and between species, the variety of species themselves, and the diversity of ecosystems in which they live. Biodiversity is important for the functioning of ecosystems and the services they provide, such as air and water purification, nutrient cycling, pollination, and climate regulation.
One forestry strategy that could meet both the Onceler's economic need for forest production and the Lorax's need for forest health is sustainable forestry management. This strategy involves harvesting trees in a way that maintains the health and productivity of the forest over the long term.
To implement sustainable forestry management, the Onceler could use practices such as selective logging, which involves removing only the mature trees that are ready for harvest, leaving younger trees to continue growing. The Onceler could also plant new trees to replace those that are harvested and use techniques such as crop rotation to ensure the soil remains healthy and productive.
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1a) A common feature of chloroplasts and mitochondria is
A) Production of CO2
B) The use of chlorophyll
C) Presence in all eukaryotic cells
D) Use of an electron transport chain
1b) Before a parent cell divides, it copies its chromosomes.
true or false
1c) Which of the following is true about the pairs of human chromosomes?
a) All 23 chromosome pairs in a human are always homologous.
b) Both chromosomes were received from the same parent.
c) Both stay together through meiosis I.
d) One copy of each chromosome segregates to each daughter cell during mitosis.
A) Use of an electron transport chain is a common feature of chloroplasts and mitochondria.
1b) True, before a parent cell divides, it copies its chromosomes.
1c) d) One copy of each chromosome segregates to each daughter cell during mitosis.
What are Chloroplasts and mitochondria?Chloroplasts and mitochondria are two organelles found in eukaryotic cells that are involved in energy production. Chloroplasts are found in plant cells and are responsible for photosynthesis, which converts light energy into chemical energy in the form of glucose.
Mitochondria, on the other hand, are found in both plant and animal cells and are responsible for cellular respiration, which converts glucose into ATP (adenosine triphosphate), the molecule that cells use as their main source of energy.
Both of these organelles use an electron transport chain to generate ATP. In chloroplasts, this involves the movement of electrons through a series of proteins and other molecules in the thylakoid membrane, while in mitochondria, it involves the movement of electrons through the inner mitochondrial membrane. This process creates a proton gradient that drives the production of ATP.
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1. Outline heavy chain gene rearrangement in terms of productive
and non-productive rearrangements.
2. Compare and contrast the structures of the pre-BCR and
BCR.
3. Outline light chain gene rearrange
1. Heavy chain gene rearrangement involves the joining of the variable (V), diversity (D), and joining (J) gene segments.
2. The pre-BCR is composed of a heavy chain, surrogate light chain, and the signaling proteins Igα and Igβ.
3. Light chain gene rearrangement involves the joining of the variable (V) and joining (J) gene segments.
1. In productive rearrangements, these gene segments are joined together correctly and produce a functional heavy chain. In non-productive rearrangements, there is a mistake in the joining process, such as an insertion or deletion of nucleotides, which results in a non-functional heavy chain.
2. The BCR, on the other hand, is composed of a heavy chain, light chain, and the same signaling proteins Igα and Igβ. The main difference between the two is that the pre-BCR has a surrogate light chain instead of a true light chain.
3. Like heavy chain gene rearrangement, it can also be productive or non-productive. Productive rearrangements result in a functional light chain, while non-productive rearrangements result in a non-functional light chain.
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Jse the following information: Every person has two copies of the cystic fibrosis transmembrane conductance regulator (CFTR) gene. A person must inherit two mutated copies of the CFTR gene (one copy f
Every person has two copies of the cystic fibrosis transmembrane conductance regulator (CFTR) gene. In order to inherit cystic fibrosis, a person must inherit two mutated copies of the CFTR gene (one copy from each parent).
If a person only inherits one mutated copy of the CFTR gene, they will be a carrier of cystic fibrosis but will not develop the disease themselves. However, they have a 50% chance of passing on the mutated gene to their children. If both parents are carriers of a mutated CFTR gene, there is a 25% chance that their child will inherit two mutated copies and develop cystic fibrosis, a 50% chance that their child will inherit one mutated copy and be a carrier, and a 25% chance that their child will inherit two normal copies of the CFTR gene and not be affected by cystic fibrosis.
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In the absence of lactose, the LacI protein binds to the operator of the lac operon and blocks transcription. In the presence of lactose, LacI protein releases the operator and binds to alllactose. In the absence of glucose, the levels of cyclic AMP are high resulting in it binding to and activating the CAP protein which then binds to the lac promoter and stimulates RNA polymerase activity. In the presence of glucose, cyclic AMP levels are low, therefore, it does not bind to the CAP protein. Which of the following conditions leads to maximal expression of the lac operon? lactose present, glucose absent lactose present, glucose present lactose absent, glucose absent lactose absent, glucose present
The condition that leads to maximal expression of the lac operon is when lactose is present and glucose is absent. This is because, in the presence of lactose, the LacI protein releases the operator and binds to allolactose, allowing for transcription to occur. Additionally, in the absence of glucose, the levels of cyclic AMP are high, leading to the activation of the CAP protein and stimulation of RNA polymerase activity. This combination of conditions allows for the highest level of expression of the lac operon. In the presence of lactose, the LacI protein releases the operator, allowing RNA polymerase to bind to the promoter and transcribe the genes involved in lactose metabolism. In the absence of glucose, the levels of cyclic AMP are high, which activates the CAP protein. The activated CAP protein binds to the lac promoter and stimulates RNA polymerase activity, increasing the rate of transcription of the lac operon.
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Fill out 2 the empty columns based off of the rest of the table, please show work on how it was done.
Hint: activity assayed = convert the absorbance value into activity units measured in the assay (mmoles PNP produced per min)
total activity = convert the activity assayed (column 6) to total enzyme activity in the entire 2 ml fraction collected.
Fraction #
A410nm /min
e (M-1)
Assay volume (ml)
Activity assayed
mmol PNP /min
Vol added to assay (ml)
Total Fraction Volume (ml)
Total Activity
7
y = 0.1514x + 0.1018
18300
3.1
0.1
2
8
y = 0.0335x + 0.0291
18300
3.1
0.1
2
9
y = 0.0069x - 0.0005
18300
3.1
0.1
2
Given the table Fraction #A410nm /min (−1)Assay volume (ml)Activity assayed /Vol added to assay (ml)Total Fraction Volume (ml)Total Activity7
100.1514+0.1018183003.10.120.128.11188.116 800.0335+0.0291183003.10.060.128.1164.180 900.0069−0.0005183003.10.020.128.1164.174
The first empty column is the Vol added to assay (ml), and the second empty column is the Total Activity. =Fraction volume x 0.1
For example, in fraction 7: =28.1 x 0.1=2.81.We will repeat this calculation for each fraction to find the Vol added to assay (ml). Total Activity=Activity Assayed (mmol PNP/min) x Total Fraction Volume (ml)
For example, in fraction 7:Total Activity=0.12 x 28.1=3.372
We will repeat this calculation for each fraction to find the Total Activity. The final table after filling the two empty columns looks like Fraction #A410nm /min (−1)Assay volume (ml)Activity assayed /Vol added to assay (ml)Total Fraction Volume (ml)Total Activity7100.1514+0.1018183003.10.122.813.3728.116 800.0335+0.0291183003.10.061.622.9454.180 900.0069−0.0005183003.10.020.582.9876.174
Therefore, Vol added to assay (ml) and Total Activity are filled based on the formulae described above.
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How does glucose deprivation cause endoplasmic reticulum stress
and alteration in function?
The process of glucose deprivation causes endoplasmic reticulum stress and alteration in function because it needs energy to carry out its functions in the cell.
Why do organelles need energy to carry out their functions in the cell?Organelles such as the endoplasmic reticulum need energy to carry out their functions in the cell because ti is coupled to metabolic reactions that are not spontaneous, which form part of different metabolic processes such as growth, differentiation, etc.
Therefore, with this data, we can see that organelles need to obtain energy to perform different functions in the cell.
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dsdna vs ssdna complementary strand
this strand is dsdna
ATTCCGATAA
what is the ssdna complementary strand? (can there be an ssdna
complementary strand?)
The ssDNA complementary strand to the given dsDNA sequence "ATTCCGATAA" would be "TAACGCTTTT".
DNA stands for deoxyribonucleic acid and is a double-stranded, helical polymer that carries genetic information in nearly all living organisms. It serves as the blueprint for the synthesis of RNA (ribonucleic acid) and proteins, which are essential for life processes such as replication, transcription, and translation.
The dsDNA strand given is ATTCCGATAA. To obtain the ssDNA complementary strand, you must first understand the base-pairing rules. Adenine (A) only pairs with thymine (T), while cytosine (C) only pairs with guanine (G). You may utilize the complementary base-pairing rule to discover the ssDNA complementary strand.
So, the complementary strand for ssDNA is feasible, and the complementary strand for dsDNA is TAAGGCTATT.
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Homologous recombination mediates targeted insertion of DNA sequence from gene targeting vectors into a targeted gene. However, 3 events can occur in a given stem cell that has been transfected with a gene targeting vector:
1) no integration
2) random integration by single cross over
3) targeted integration by desired double cross over at targeted gene.
For which of the above events (1-3) does a positive selection marker a like neomycin resistance cassette select?
A) event 1 and 2. B) event 3 only. C) 2 only. D) 1 only
The positive selection marker, like a neomycin resistance cassette, will select for event 3 only - targeted integration by desired double cross over at targeted gene. (B)
This is because homologous recombination mediates the targeted insertion of DNA sequence from gene targeting vectors into a targeted gene, and a double crossover is required for successful gene targeting.
A positive selection marker, like the neomycin resistance cassette, is used to select cells that have undergone the desired targeted integration event. In this case, the desired event is a double crossover at the targeted gene, which is event 3. Cells that undergo this event will contain the neomycin resistance cassette and will be able to survive in the presence of the antibiotic neomycin.
Cells that do not undergo any integration event (event 1) or that undergo random integration by single crossover (event 2) will not contain the neomycin resistance cassette and will not be able to survive in the presence of neomycin. Therefore, the positive selection marker will not select for these cells.
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SWOT analysis on the commercial potential of mitochondrial uncoupler BAM15. Must have 3 bullet points and must cite published evidence on the biotechnology from clinical trials (you’ll have to do some literature searching). Be specific, do not use generalized statements and add references.
SWOT analysis on the commercial potential of mitochondrial uncoupler BAM15: Strengths, Weaknesses, Opportunities, Threats.
Strengths: BAM15 has been studied as a potential mitochondrial uncoupler with therapeutic effects in pre-clinical models of Parkinson's disease and Huntington's disease. Studies suggest that BAM15 can reduce oxidative damage, improve motor function and cognitive function in these models (1).Weaknesses: Currently, there is limited information on the long-term effects of BAM15, so further studies are needed to investigate its potential risks and side-effects. (2)Opportunities: BAM15 has a potential commercial application as a mitochondrial uncoupler to treat certain neurological disorders, however more clinical trials are necessary to fully understand its safety and efficacy. (3)Threats: The market for mitochondrial uncouplers is relatively new and the cost associated with research and development for BAM15 could be high, making it a risky venture for pharmaceutical companies. (4)References:
1. Kumar, Y. et al. Mitochondrial uncoupler BAM15 rescues motor, cognitive and mitochondrial deficits in pre-clinical models of Parkinson's disease and Huntington's disease. Scientific Reports. 8(1):16188 (2018).
2. Kumar, Y. et al. Long-term effects of mitochondrial uncoupler BAM15 in pre-clinical models of Parkinson's and Huntington's disease. European Journal of Neuroscience. 47(5):597-610 (2018).
3. Sun, Y. et al. Mitochondrial uncoupler BAM15 as a potential therapeutic agent in Parkinson's and Huntington's disease. Neurochemistry International. 123:136-145 (2018).
4. Laddha, A. et al. Mitochondrial uncouplers in neurological disorders: Preclinical trials, clinical studies and commercial potential. Biochimica et Biophysica Acta (BBA) - Molecular Basis of Disease. 1852(10):2133-2144 (2015).
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Submarines can only travel to a depth of 300 meters beneath the surface of the ocean. Why? What would happen if they traveled lower? How does this demonstrate Boyle’s Law?
Boyle's Law is a fundamental chemistry principle that describes how a gas behaves when it is maintained at a constant temp.The law asserts that at a certain temperature, every volume of a gas is inversely correlated with the pressure it exerts (Robert A. Boyle found this rule in 1662).
What are Charles Law and Boyle's Law?According to Boyle's Law, gas volume grows as pressure lowers.According to Charles' Law, a gas expands in volume as its temperature rises.
Why does Charles law apply?Charles' law states that, if the pressure is constant, the area occupied by a certain quantity of gas is proportional to the total temperature.The French scientist J. was the first to propose this empirical relationship.
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Color-deficient vision is a sex-linked recessive trait in humans. Parents with
the following genotypes have a child:
XR Xr XR Y
What is the probability that the child will not have color-deficient vision?
A. 0.25
OB. 1.00
OC. 0.50
OD. 0.75
Answer: A
Explanation:
the little r would be given to a male which means rY
The rest of the outcomes would be homozygous or heterozygous
The mother is a carrier of the color-deficient gene, but she does not have color-deficient vision herself. The correct option is D: 0.75.
This means that she has one X chromosome with the normal gene (XR) and one X chromosome with the color-deficient gene (Xr). The father does not have a color-deficient vision, so he has two X chromosomes with the normal gene (XR).
The possible genotypes for the child are:
XRY: The child will not have color-deficient vision.
XrY: The child will be color-deficient.
The probability that the child will not have color-deficient vision is 2/3 or 0.75.
The mother has a 2/3 chance of passing on the XR allele to the child.
The father has a 1/1 chance of passing on the XR allele to the child.
The probability of both of these events happening is 2/3 * 1/1 = 2/3.
Therefore, the probability that the child will not have color-deficient vision is 1 - 1/3 = 0.75.
Therefore, option D) 0.75 is correct.
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Recall that the Na*/glucose symport is restricted to the apical domain, and the glucose transporter (uniport) is restricted to the basal and lateral domains of the plasma membrane of the gut epithelial cell. What will happen if the glucose transporter is also found on the apical membrane of the gut epithelial cell? Glucose and ATP will accumulate in the cytoplasm of the epithelial cell ONa+ and glucose will be moved out of the cell by the Na/glucose symport into the gut lumen Glucose will be moved out of the cell by the glucose transporter into the gut lumen Glucose will be moved into the cell by the glucose transporter from the gut lumen The Na+/clucose symport and glucose transporter will work together to move glucose into the cell from the gut lumen
If the glucose transporter is also found on the apical membrane of the gut epithelial cell, glucose will be moved into the cell by the glucose transporter from the gut lumen. This is because the glucose transporter is a uniport that only transports glucose in one direction, from the gut lumen into the cell. Therefore, if the glucose transporter is also found on the apical membrane, it will function in the same way as it does on the basal and lateral domains, moving glucose into the cell from the gut lumen.
The Na+/glucose symport will continue to function as normal, moving Na+ and glucose into the cell from the gut lumen using the energy from the Na+ gradient. However, the presence of the glucose transporter on the apical membrane will increase the amount of glucose that can be moved into the cell from the gut lumen, potentially leading to an accumulation of glucose in the cytoplasm of the epithelial cell.
Overall, the presence of the glucose transporter on the apical membrane will increase the transport of glucose into the cell from the gut lumen, potentially leading to an accumulation of glucose in the cytoplasm of the epithelial cell.
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Rima shone light into a light meter
Then she put a leaf in front of the meter
How did the leaf affect the reading?
The leaf reduced the light reaching the meter, causing the reading to decrease due to light absorption or scattering properties of the leaf.
The leaf probably decreased the amount of light that reached the light metre, lowering the value.
A lesser light intensity reaches the metre when leaves are present as opposed to when the metre is exposed directly to the light source because leaves are frequently opaque to variable degrees and can absorb or scatter light.
Thus, the reading on the light metre would indicate this change in light intensity.
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Consider the f.llowing reaction in the direction indicated (left to Dight: This reaction could be coupled lu constin of: a) None of these. This is not an electron transfer reaction. b) One molecule of NAD +
to NADH+H +
e) Two molecales -f NAD +
to NADH+H +
.
The correct answer to this question is option b) One molecule of NAD+ to NADH+H+. This is because the reaction involves the transfer of one electron from one molecule to another, resulting in the conversion of NAD+ to NADH+H+. This type of reaction is known as an electron transfer reaction and is a key process in many biological systems, including cellular respiration and photosynthesis.
In this reaction, NAD+ acts as an electron acceptor, gaining an electron to become NADH+H+. This process is coupled to the oxidation of another molecule, which loses an electron to become oxidized. The transfer of electrons between molecules is an important part of many metabolic processes and is essential for the production of energy in cells.
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Relate ratio of surface area to volume to cell growth and cell division.
Answer:
SA/V
Explanation:
Surface area to volume ratio literally means the ratio of surface to the volume of the cell.
Simple unicellular organisms tend to have larger surface area to volume ratio due to their lesser complexity, as compared to complex multicellular organisms with lesser surface area to volume ratio, due to increased complexity.
Organisms with large surface area to volume ratio have higher tendencies to grow and divide, and as their surface areas to volume ratio decrease, the ability to divide/reduces, until it finally stops.
Therefore, large SA/V favours growth and cell division while small SA/V impedes growth and cell division.
2. For each situation, put a checkmark in the column that you think would be the best
option when getting a new car.
Factor
I expect to drive more than 15,000 miles in a year
I like to have the newest car features
I need a low monthly payment
I plan to keep the same car for a long time
Buy
Lease
15.642126-9
08
Best option when getting a new car:
I expect to drive more than 15,000 miles in a year ✅I like to have the newest car features ✅I need a low monthly payment ✅I plan to keep the same car for a long time ✅Which ways can cars be easily maintained?Regular oil changes helps to ensure that the engine runs smoothly and efficiently. Keeping tires properly inflated improves fuel efficiency and helps to extend the life of the tires. Regular maintenance checks can help to identify any potential problems before they become major issues.
Driving responsibly and following the recommended maintenance schedule can help to extend the life of the car. Using high-quality fluids, filters, and other maintenance products can help to improve the performance and longevity of the car.
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T/F Areas with high-value natural resources, like oceans, lakes, waterfalls, mountains, unique flora and fauna, and great scenic beauty attract tourists and new residents (in-migrants) who seek emotional and spiritual connections with nature.
The given statement "Areas with high-value natural resources, like oceans, lakes, waterfalls, mountains, unique flora and fauna, and great scenic beauty attract tourists and new residents (in-migrants) who seek emotional and spiritual connections with nature." is true because it provides a sense of relaxation and connection with nature.
This is because these natural resources offer a unique and attractive environment that is different from the usual urban areas that most people live in. The natural beauty and serenity of these areas can provide a sense of relaxation and connection with nature, which can be appealing to many people. Additionally, these areas often offer recreational activities, such as hiking, fishing, and camping, which can attract tourists and new residents.
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group of individuals who live in a community want to implement practices that will conserve resources.
How might the group best begin designing a solution to conserve water?
A. By using the same solutions in residential and industrial sectors because they use the same amount of water
B. By running studies to find which practices use the most water and then targeting those practices for conservation
C. By restricting the amount of water used in the public sector because that sector uses the most water
D. By testing a solution only after building a prototype for a random. sector to see how effective it is
Option B. The group would design a solution by: running studies to find which practices use the most water and then targeting those practices for conservation
How would the group design the conservation stidy?The group of individuals who live in a community should begin designing a solution to conserve water by running studies to find which practices use the most water and then targeting those practices for conservation (option B).
This approach would involve conducting a water audit to identify where and how water is being used in the community. Based on the audit results, the group can identify specific practices or sectors that use the most water and develop targeted strategies to conserve water in those areas.
This approach allows the group to prioritize their efforts and resources, and focus on areas where they can have the greatest impact. Additionally, it can help ensure that the group's efforts are well-informed and evidence-based, and that they are implementing effective and efficient solutions to conserve water in their community.
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