Answers
Answer:
Denaturation and renaturation are properties of protein but in some cases denaturation is irreversible and protein is unable to renature.
The primary sequence of a protein determines which portion of protein fold closely together in order to form three-dimensional conformation and the conformations are held by hydrogen bonds.
Denaturation of protein in-vitro is generally obtained by increasing temperature. Heat or high temperature disrupts hydrogen bonds that held between protein folds because high temperature increases the kinetic energy and the bonds are disrupted due to rapid heating. Disruption of hydrogen bonds will change the 3D structure of the protein and denatured proteins lose its function.
Chemicals are also used to disrupt hydrogen bonds such as alcohol.
So, high temperature and chemicals denature protein by affecting the hydrogen bonds holding 3D shape of primary sequence in protein and the change in protein becomes irreversible.
Related Questions
What happens during the electron transport chain phase of cellular respiration?
Answers
There is the formation of 36 ATP in cellular respiration which will be used to perform functions in the body.
The electron transport chain is also known as oxidative phosphorylation and in this, the electron transport chain forms a hydrogen gradient or proton gradient across the inner membrane of the mitochondria because the electron transport chain takes place in mitochondria and that is why it is called as powerhouse of the cell as all the ATP or large amount of ATP is generated by ETC and they generate around 36 ATP from one chain and this chain has series of proteins like ferredoxin, cytochrome, a and b, and phycocyanin, plastoquinone through which the electrons are passed on from each protein molecules and continue to do the redox reaction and the energy which is released from these reactions is caught by the formation of protein gradient which then undergoes the process of chemiosmosis to form the ATP. The protons are passed from the F1 F0 channel and using the ATP synthetase enzyme they synthesize ATP at the end of the electron transport chain electrons are transferred to the molecular oxygen which gets split in half and takes the H+ to form the water.
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HELP ME ASAP!!!!
Select the correct answer.
What is the difference between sedimentary and gaseous cycles?
A.
gaseous cycles involve elements that are abundantly found in the gaseous state
о В.
gaseous cycles involve elements that are essential for life
C. sedimentary cycles do not involve the biosphere
D.
sedimentary cycles do not pass through the atmosphere
Answers
Answer:
I think A. gaseous cycles involve elements that are abundantly found in the gaseous state
Explanation:
Answer:
A. gaseous cycles involve elements that are abundantly found in the gaseous state
If the enzyme sucrase has a Vo of 0.40 mM per minute when [S] = 1.25 mM, and a Vo of 1.0 mM per minute when [S] = 5.0 mM, what is its Vmax? A. 0.50 mM per minute B. 1.0 mM per minute C. 2.0 mM per minute D. 5.0 mM per minute E. 10.0 mM per minute
Answers
If the enzyme sucrase has a Vo of 0.40 mM per minute when [S] = 1.25 mM, and a Vo of 1.0 mM per minute when [S] = 5.0 mM, its Vmax is option B 1.0 mM per minute
To determine the Vmax (maximum velocity) of the enzyme sucrase, we can use the Michaelis-Menten equation, which describes the relationship between the substrate concentration ([S]) and the reaction rate (Vo) of an enzyme-catalyzed reaction.
The Michaelis-Menten equation is given by:
Vo = (Vmax * [S]) / (Km + [S])
Where:
Vo = Reaction rate (velocity)
Vmax = Maximum velocity
[S] = Substrate concentration
Km = Michaelis constant (substrate concentration at half of Vmax)
We are given two sets of data:
At [S] = 1.25 mM, Vo = 0.40 mM per minute
At [S] = 5.0 mM, Vo = 1.0 mM per minute
Let's use the first set of data to determine Km:
0.40 mM per minute = (Vmax * 1.25 mM) / (Km + 1.25 mM)
Now, using the second set of data:
1.0 mM per minute = (Vmax * 5.0 mM) / (Km + 5.0 mM)
By solving these two equations simultaneously, we can find the values of Vmax and Km. However, we need to make some assumptions. One assumption is that the substrate concentration ([S]) is much higher than the Km value, which allows us to simplify the equation:
Vmax = Vo
Using this assumption, we can see that the Vo at [S] = 5.0 mM is higher than the Vo at [S] = 1.25 mM. Therefore, the Vmax would be 1.0 mM per minute, which corresponds to option B.
Therefore the correct option is B
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what is the overall outcome and the light reactions in photosynthesis
Answers
Answer:
The plant will react with the sunlight and the cells will start to grow more causing to change.
What is a stomata?
A. the little mouths on the leaf of a plant where they take in CO2,and release 02
B. Where the plant takes in water
Answers
A. the little mouths on the leaf of a plant where they take in CO2,and release 02
stoma=mouth
Those structures are located on the backside of the leaf
In a diploid species of plant, the genes for plant height and fruit shape are syntenic and separated by 18 m.u. Allele D produces tall plants and is dominant to d for short plants, and allele R produces round fruit and is dominant to r for oval fruit. The expected percentages of progeny phenotypes are listed below:
tall oval = 41%
short round = 41%
tall round = 9%
short oval = 9%
(a) What proportion are expected to be tall and produce round fruit?
Answers
In a diploid plant species, the proportion of progeny expected to be tall and produce round fruit is 32%.
In a diploid species of plant, the genes for plant height and fruit shape are syntenic and separated by 18 m.u. The expected percentages of progeny phenotypes are tall oval = 41%, short round = 41%, tall round = 9%, and short oval = 9%.To determine the proportion of progeny expected to be tall and produce round fruit, we need to consider the alleles for each trait. Allele D produces tall plants (dominant) and allele d produces short plants (recessive). Allele R produces round fruit (dominant) and allele r produces oval fruit (recessive).From the given percentages, we can deduce that tall ovals and short ovals combined make up 9%. Since tall oval is 41% and short oval is 9%, the proportion of progeny expected to be tall and produce round fruit is 41% - 9% = 32%.Therefore, the proportion of progeny expected to be tall and produce round fruit is 32%.For more questions on diploid plant
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Which describes a function of plasmodesmata? They allow molecules to move between cells. They store glucose made during photosynthesis. They store water and collect poisons. They provide rigid support for the cell.
Answers
Answer:
They allow molecules to move between cells
Explanation:
Plasmodesemata are channels or tunnels in the cell wall of plant cells that connect the cytoplasm of adjacent cells.
1. Which gene mutation (insertion, deletion, or substitution) appears to have the biggest impact on an amino acid sequence? Explain your answer.
Answers
Answer:
A frameshift mutation caused by an insertion or deletion of one or more nucleotides in the coding sequence of a gene has the most significant impact on the amino acid sequence of the resulting protein. This is because the genetic code is read in groups of three nucleotides (codons), and an insertion or deletion shifts the reading frame and changes the codon sequence downstream of the mutation. As a result, all the amino acids downstream of the mutation are different from those in the original protein sequence.
You are working on research involving competition between animals. Which of the following resources do you not need to measure?
A. Light
B. Space
C. Water
D. Food
Answers
Have a nice day :)
Compare and contrast eukaryotic and prokaryotic cells. Give at least three examples for each.
Answers
Answer:
Prokaryotic cells are unicellular organisms, which reproduce through binary fission. In some cases few prokaryotic organisms also reproduce by budding. Prokaryotic cells have a cell envelope, which generally consists of a capsule, cell wall, cytoplasm, plasma membrane, cytoplasm region or nucleiod region, ribosome, plasmids, pili and flagella.
Identify the organ of the digestive system.
A)stomach
B)small intestine
C)large intestine
D)liver
Answers
Environment: Clean Forest Phenotype Frequency Allele Frequency Genotype Frequency Environment: Polluted Forest Moths Released G1 G2 G3 G4 G5 Typica 490 186 148 114 77 40 Carbonaria 510 367 617 763 974 1331 Total 1000 553 765 877 1051 1371 Phenotype Frequency Color Initial Frequency Frequency G5 Typica Light 0.49 0.03 Carbonaria Dark 0.51 0.97 Allele Frequency Allele Initial Allele Frequency G5 Allele Frequency q d 0.70 0 p D 0.30 Genotype Frequency Moths Genotype Color Moths Released Initial Frequency Frequency G5 Number of Moths G5 q2 Typica dd Light 490 0.49 2pq Carbonaria Dd Dark 420 0.42 p2 Carbonaria DD Dark 90 0.09
Answers
The data shows a shift in moth populations in response to pollution, with a decrease in the Typica (light) phenotype and an increase in the Carbonaria (dark) phenotype, indicating natural selection and adaptation to the polluted environment.
The given data represents the phenotypic and genotypic frequencies of moths in two different environments: a clean forest and a polluted forest. The moths are categorized into two phenotypes: Typica (light) and Carbonaria (dark). The initial frequencies indicate the starting proportions of each phenotype and allele.
In the clean forest, the Typica phenotype has a frequency of 0.49, while Carbonaria has a frequency of 0.51. This suggests that the two phenotypes were initially present in approximately equal proportions. However, in the polluted forest, the frequencies have significantly changed. The Typica phenotype has decreased to a frequency of 0.03, while Carbonaria has increased to 0.97.
The allele frequencies indicate the frequencies of the two alleles, "q" and "p." The initial allele frequency of "q" is 0.70, indicating that the "d" allele was more common initially. The G5 allele frequency of "q" is 0, suggesting that the "d" allele has become very rare in the polluted forest.
The genotype frequencies provide information about the combinations of alleles in the moths. The G5 genotype frequencies for Typica are 2pq (light, heterozygous) and p^2 (light, homozygous). For Carbonaria, the G5 genotype frequency is p^2 (dark, homozygous).
Overall, the data indicates a significant shift in the moth populations in response to the pollution in the forest. The Carbonaria phenotype, which is better camouflaged in the polluted environment, has become more prevalent, while the Typica phenotype has decreased. This is a classic example of natural selection and adaptation in response to changes in the environment.
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help asap giving branlist
Answers
Answer:
symbiotic relationship
Explanation:
Residual parent material refers to the *weathered rock* and *soil* that remains in its place of origin, while *transported parent material* is material that has been carried and deposited by natural agents such as water, wind, or glaciers.
The impact of these different types of parent material on *soil formation* can be significant. Residual parent material tends to contribute to the formation of soils with characteristics similar to the parent rock. The weathering process breaks down the rock into smaller particles, allowing for the development of soil horizons and the release of minerals that influence soil fertility. In contrast, transported parent material can introduce a diverse range of materials to a given area, leading to variations in soil composition, texture, and fertility. The transportation process can mix different types of sediment, resulting in the formation of heterogeneous soils with varying properties.
Answers
Residual parent material is the rock and soil that is still located in its original location and is weathered. Transported parent material is material that is carried and deposited by natural agents such as water, wind, or glaciers. The effect of these different types of parent material on soil formation can be significant.
Residual parent material usually contributes to the development of soils with qualities similar to the parent rock. The breakdown of rock into smaller particles through weathering enables the formation of soil horizons and the release of minerals that influence soil fertility.
The physical and chemical characteristics of soils derived from this form of parent material are often similar and homogeneous, allowing for predictable fertility and properties. Transported parent material, on the other hand, can introduce a diverse range of materials to a given area, resulting in variations in soil composition, texture, and fertility.
The transportation process can blend different types of sediment, resulting in the formation of heterogeneous soils with different characteristics. Transported parent material can also be exposed to weathering processes, altering its original composition and resulting in soils with varying degrees of fertility.
While transported parent material can increase the diversity of soil properties in an area, it can also make it difficult to predict soil characteristics, particularly in areas where different types of sediment have been deposited. The relationship between parent material and soil development is complex, with both the initial characteristics of the material and the surrounding environment influencing soil properties.
In general, residual parent material contributes to the development of homogeneous soils with predictable properties, while transported parent material can result in more diverse and heterogeneous soils.
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comparing mitosis and meiosis biology please answer ASAP!
Answers
Mitosis
Purpose: To make new cells
Number of Cells Produced: 2
Rounds of Cell Division: 1
Haploid or Diploid: Diploid
Daughter cells identical to parent cells: Yes
Daughter cells identical to each other: Yes
Meiosis:
Purpose: To make gametes
Number of Cells Produced: 4
Rounds of Cell Division: 2
Haploid or Diploid: Haploid
Daughter cells identical to parent cells: No
Daughter cells identical to each other: No
Lab Intro - The Metric System
Exercise A - Measurement or distance
Someone help please
Answers
The metric system is a system of measurement that is utilized throughout the world. The fundamental units of the metric system are meter, gram, and liter. In this system, prefixes are used to represent measurements that are bigger or smaller than the fundamental units.
The prefix “kilo-,” for example, indicates a measurement that is a thousand times bigger than the fundamental unit. Lab Intro - The Metric System Exercise A - Measurement or Distance The metric system is used to measure length or distance.
The standard unit of length in the metric system is the meter, which is represented by the symbol “m.” The metric system is used to measure distances or lengths that vary from millimeters (mm) to kilometers (km).In order to determine measurements in the metric system, the fundamental units are combined with prefixes.
For example, one centimeter (cm) is equal to 1/100th of a meter, while one millimeter (mm) is equal to 1/1000th of a meter. One kilometer (km), on the other hand, is equal to a thousand meters (m).
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What is hidaya's punnett squares
Answers
A Punnett square allows the prediction of the percentages of phenotypes in the offspring of a cross from known genotypes. A Punnett square can be used to determine a missing genotype based on the other genotypes involved in a cross.
hope it helps...
discuss how you would go about teaching about, in and for the environment to address the challenge of biodiversity in school
Answers
Teaching about and for the environment, specifically addressing the challenge of biodiversity, is a crucial topic to raise awareness and empower students to become environmentally responsible individuals.
Some steps to consider when designing a curriculum or lesson plan for teaching about biodiversity in schools:
Establish learning objectives: Clearly define the goals you want to achieve through your teaching. These could include developing an understanding of biodiversity, its importance, and the threats it faces, as well as promoting an appreciation for nature and inspiring actions for conservation.
Provide foundational knowledge: Start by introducing the concept of biodiversity and explaining its significance in maintaining healthy ecosystems. Cover topics such as species diversity, genetic diversity, and ecosystem diversity. Use visual aids, videos, and interactive activities to engage students and make the information more accessible.
Explore local biodiversity: Help students understand how biodiversity relates to their own surroundings. Conduct field trips to nearby parks, nature reserves, or even the school grounds to observe and identify different species. Encourage students to document their findings through sketches, photographs, or field notes.
Study ecosystems and habitats: Explore various ecosystems such as forests, wetlands, coral reefs, or grasslands. Discuss the different species that inhabit these ecosystems, their interdependencies, and the role they play in maintaining ecological balance. Use case studies and real-life examples to illustrate the effects of habitat destruction, climate change, pollution, and invasive species on biodiversity.
Promote citizen science: Encourage students to actively participate in citizen science projects that contribute to biodiversity monitoring and conservation efforts. Provide guidance on how to identify and report species sightings, collect data, or engage in community initiatives. Online platforms like Naturalist and eBird can be valuable resources for students to contribute to real-world scientific research.
Discuss conservation strategies: Introduce students to various conservation approaches and strategies such as protected areas, habitat restoration, captive breeding programs, and sustainable practices. Explore success stories of conservation efforts and discuss the challenges faced in preserving biodiversity. Encourage students to think critically about potential solutions and their own roles in protecting the environment.
Encourage hands-on activities: Engage students in practical activities that foster a connection with nature and promote biodiversity conservation. Examples could include creating a school garden or wildlife-friendly spaces, participating in tree planting campaigns, or organizing awareness campaigns in the local community.
Foster a sense of responsibility: Help students understand the impact of individual actions on biodiversity. Discuss sustainable practices such as reducing waste, conserving energy and water, supporting local and organic food choices, and minimizing the use of harmful chemicals. Encourage students to make informed choices and be environmentally responsible citizens.
Remember to adapt the curriculum to suit the age group, cultural context, and available resources. By providing an engaging, hands-on learning experience and fostering a sense of responsibility, you can inspire the next generation to become stewards of biodiversity and the environment.
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A scientist is conducting research to see whether dogs or cats are more popular. He goes to a dog park and asks all of the owners which pet they prefer. 100% of those polled stated that dogs are their favorite animal.
This experiment is an example of:
A) personal bias
B) experimental bias
C) cultural bias
D) a controlled experiment
Answers
Answer:
C) Cultural bias
Explanation:
A cultural biased experiment is one which is based on the cultural views or perceptions. In this experiment where a scientist is trying to see whether dogs or cats are more popular, he conducted a questioning assessment from all the dog owners in a dog park.
He found that 100% (all) of them preferred dogs to cats. This preference is due to the to the culture of the dog owners who have a mentality of dogs being their best choice without exploring possibilities from other pet animals. The scientists goes to the dog park knowing that the dog owners will favour dogs over cats. Hence, it is a cultural biased experiment.
A symbiotic relationship in which one species benefits and the other is harmed is known as
Answers
(as the species that is benefiting is kind of like a parasite)
Answer:
If this is for "McGraw hill" the answer is commensalism.
Explanation:
Watching your fish tank, you see many similarly sized fish species represented. You notice the members of one species spending more time near the top of the tank, while another closely related species spends more time on the bottom of the tank. This is an example of...
Answers
Answer:
The correct answer is - Resource partitioning.
Explanation:
Resource Partitioning is the method of differentiating in which there is differentiating in the resource and the source of nutrition so that there is no or less competition among the closely related species.
In this case, one of the species of the same size fishes spends more time on the surface of the tank, and the other spends on the bottom of the tank. This shows the example of Resource partitioning to avoid competition.
.
Which organ system is the xylem and phloem of the plant part of? A)Digestive system
B)Endocrine system
C)Respiratory system
D)Vascular system
Answers
Answer:
C is correct ans
Explanation:
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1a. Conflict with farmers is one cause of declining populations of predators, like
lions and wolves. *
O True
False
Answers
In cystic fibrosis, an important protein destined for the plasma membrane never makes it beyond the Golgi complex. This protein is necessary for Na and Cl- transport and is critical for maintaining mucous secretions of the appropriate consistency. Devise an explanation for why this may occur.
Answers
Answer:
The correct answer is - Golgi apparatus marks or tags on this protein as incorrect or dysfunctional tags which then prevents it from transported to the membrane.
Explanation:
The role of the Golgi apparatus is to process and package the proteins produced and come from ER and transfer them to the plasma membrane in order to transfer them to their target sites.
The Golgi apparatus marks or tags the proteins that are incorrect or dysfunctional, in order to prevent these proteins from transporting to the plasma membrane and harm the membrane. In CF, the case is similar as the Golgi apparatus tags these proteins as incorrect or dysfunctional.
The process in which plants absorb nitrates through their roots is called
a. nitrification.
b. denitrification.
c. assimilation.
d. ammonification.
Answers
Based on the information in the table, which trees are most similar to each other?A red maple trees and Texas red oak treesB red maple trees and box elder treesC Texas red oak trees and American beech treesD Northern California black walnut trees and horse chestnut trees
Answers
Based on the table, we can see that the most similar trees are the Red maple and the Box elder, because both belong to the genus Acer.
This means B is the right answer.
How many fundamental forces exist in nature?
two
three
four
five
Answers
Answer:
four
Explanation:
Answer:
Four
Explanation:
They are; gravity, electromagnetism, the weak nuclear force and the strong nuclear force.
You are breeding alligators to be reintroduced in the wild. One of the female alligators is albino. Would it be possible to get normal colored alligators from this female? Identify the genotypes of each individual and show your work in
the Punnett squares below.
Answers
Answer:
It is possible for normal coloured alligators to be produced. If the albino trait is recessive and normal is dominant, the male would need 1-2 dominant alleles to produce a normal coloured alligator. It is also possible for a normal coloured alligator to be produced if albino is dominant and normal is recessive. In this circumstance, the female must be h e t e r ozygous, but the male can be either h o m ozygous or h e t e r o zygous to produce normal coloured offspring.
First punnet square genotypes:
Female - albino
Male - albino/normal (can be either one)
Second punnet square genotypes:
Female - Normal
Male - albino/normal (can be either one)
For her research, Ananya frequently accesses a new database that contains functional data relating to the human genome. This database was likely created during what phase of the ENCODE project?
Group of answer choices
pilot phase
analysis phase
sequencing phase
technology development phase
Answers
The ENCOD project was a collaborative effort aimed at identifying all functional elements in the human genome. It consisted of several phases, including the pilot phase, technology development phase, sequencing phase, and analysis phase.
What is ENCOD?
The ENCODE (Encyclopedia of DNA Elements) project was a collaborative research effort aimed at identifying all functional elements in the human genome. The project was launched in 2003 with the goal of creating a comprehensive catalog of functional elements, such as genes, regulatory elements, and chromatin modifications, and to understand how these elements work together to control gene expression and cellular function.
Since the database that Ananya accesses contains functional data, it was likely created during the analysis phase of the ENCODE project. During this phase, the raw data generated in the sequencing phase was analyzed and interpreted to identify functional elements such as genes, regulatory regions, and chromatin modifications.
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What happens when a positive regulator (proto-oncogene) is overactivated?
Choose 1 answer:
(Choice A) It allows uncontrolled cell division to occur.
(Choice B) It promotes the growth of normal cells.
(Choice C) It is able to attack cancer cells more effectively.
(Choice D) It reduces the likeliness of tumor development.
Answer: It allows uncontrolled cell division to occur.
Answers
When a proto-oncogene is overactivated occurs uncontrolled cell division because the cell lacks apoptotic mechanisms (Option A).
What is a proto-oncogene?A proto-oncogene is a gene that when expressed at a faulty time and level and may cause uncontrolled cell proliferation, leading to the formation of tumors.
Apoptosis, also known as programmed cell death is a mechanism that prevents uncontrolled cell proliferation as a consequence of mutations that trigger protooncogene activation.
In conclusion, when a proto-oncogene is overactivated occurs uncontrolled cell division because the cell lacks apoptotic mechanisms (Option A).
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