Answer: [Vmax]= 6mM/min.
Explanation:
Enzymes are proteins that produce a specific chemical change in a substance, called substrate, resulting in a product. By following the rate of product appearance (or disappearance of the substrate) as a function of time, the kinetics of the reaction is obtained. As the reaction proceeds, the rate of product formation decreases and the substrate is consumed.
The rate [V] indicates the amount of substrate that is converted into product per second. With increasing substrate concentrations [S], the enzyme approaches its maximum velocity [Vmax], but does not reach it. For this reason, there is no specific value of [S] for Vmax. However, a characteristic parameter of the enzyme can be defined using the substrate concentration at which half of the maximum velocity (Vmax/2) is reached. The initial velocity of the reaction is equal to the slope of the zero-time forward curve and is measured before 10% of the total substrate is consumed, so that the substrate concentration can be considered as essentially constant throughout the experiment. So, although it is impossible to measure exactly the substrate concentration that gives Vmax, enzymes can be characterized by the substrate concentration at which the reaction rate is half the maximum rate. This substrate concentration is known as the Michaelis-Menten constant (KM). For enzymes exhibiting simple Michaelis-Menten kinetics this constant represents the dissociation constant of the enzyme-substrate complex (or the inverse of the enzyme-substrate affinity). Thus, Km indicates the affinity of the enzyme for its substrate, since the higher the Km the lower the affinity. And the lower the Km the higher the affinity.
The formula for the calculation is as follows:
[tex]V_{0}= \frac{[Vmax] x [S]}{[S]+[KM]}[/tex]
So if we replace the numbers, we get:
[tex]2mM/min= \frac{[Vmax] x [4mM]}{[4mM]+[8mM]}[/tex]
If we clear the equation, we obtain that [Vmax]= 6mM/min.
According to the phenotypic characters of pneumococcus considered in Griffith's
experiment of transformation, which of the following statements are correct? Choose the
correct option (i) Presence of slime layer (ii) Presence of capsule (iii) Absence of capsule
Answer:
发发发 发斯蒂芬
Explanation:
Presence of capsule. Therefore, option (B) is correct.
What was Griffith's experiment?Griffith's experiment of transformation is a landmark experiment in microbiology conducted in 1928 by British bacteriologist Frederick Griffith. The experiment aimed to determine the nature of the "transforming principle" responsible for transferring genetic material between bacteria.
Griffith used two strains of Streptococcus pneumoniae, one that was virulent and had a polysaccharide capsule (S strain) and another that was non-virulent and lacked the capsule (R strain). He found that when he injected mice with the heat-killed S strain and live R strain, the mice died, and live S strain was found in their blood. This suggested that the R strain had been transformed into the S strain, and the genetic material was responsible for the transformation.
Griffith's experiment provided the first evidence of bacterial transformation and paved the way for future research on the molecular basis of genetics.
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Can you help me with this please?
Match the items of column 'A' with those of column 'B':
Turbidity
Reduces the light in the water column
Natural pollution
Choose
Nitrates and phosphates
Air pollution
Biochemical demand for oxygen
Contaminants into a natural environment
Chemical fertilizers
Pollution
Floating materials
Industrial wastes
Emissions pollution
Reduces the light in the water column
Concentration of hydrogen pH
Eutrophication
CFCS
New species invasion
Temperature
Harmful algal blooms
Oils fats, and foam
Please answer all parts of th Hypertrophication
Toxicity
Volcanic
In crude oil affect eggs and larvae of fish a
Atomic number
a. Hydrocarbon
Answer:
Match the items of column 'A' with those of column 'B':
Turbidity
Reduces the light in the water column
Natural pollution
Choose
Nitrates and phosphates
Air pollution
Biochemical demand for oxygen
Contaminants into a natural environment
Chemical fertilizers
Pollution
Floating materials
Industrial wastes
Emissions pollution
Reduces the light in the water column
Concentration of hydrogen pH
Eutrophication
CFCS
New species invasion
Temperature
Harmful algal blooms
Oils fats, and foam
Please answer all parts of th Hypertrophication
Toxicity
Volcanic
In crude oil affect eggs and larvae of fish a
Atomic number
a. Hydrocarbon
I need help with the work
Answer:
Gw
Explanation:
phổi phải to và dài hơn phổi trái đúng không
không, tùy người thôi
nhưng thường hai phổi bằng nhau
tùy kiểu bạn thở ở mũi nào
The MN blood group in humans is under the control of a pair of co-dominant alleles, M (we will call the frequency of M, p) and N (we will call the frequency of N, q). In a group of 556 individuals, the following numbers of individuals are found for each of the genotypes:
167 MM
280 MN
109 NN
Required:
a. What is the frequency of each allele?
b. What is the value of the Chi-square statistic test to find if in this particular case the genotypic frequencies conform to the Hardy-Weinberg distribution.
c. What is the probability associated with you chi square statistic calculated above?
Answer:
do you know some people have golden blood
What is the biological impact of minimum catch sizes on a population of fish?
a. The population comes to be dominated by smaller, slower-growing individuals.
b. Older, less productive adults are removed, improving the population's health.
It applies a selective pressure for larger, faster growing fish.
d. The fish in the population produce more and healthier eggs to compensate.
C.
Please select the best answer from the choices provided
Ο Α
OB
Ос
OD
Answer:
answer is A.) The population comes to be dominated by smaller, slower- growing individuals
an organism that uses oxygen to break down food to obtain energy
Which of the following words is generally used to describe what managers do as opposed to what leaders do b) Organize c) Inspire O d) Innovate
Acidic amino acids have two -COOH and one -NH2 group per molecule. Select the pair that consists of acidic amino acid
a) Aspartic Acid , Glutamic Acid
b) Lysine , arginine
c) Glycine and alanine
d) Both a and b
Answer:
Its Aspartic acid, glutamic acid
I HOPE ITS RIGHT IF NOT THEN SORRY :)Answer:
a) Aspartic Acid , Glutamic Acid are acidic amino acids
Explanation:
HOPE IT HELPS
Alizarin yellow is a pH indicator that transitions from red to yellow when the pH falls from a value of 11 to below 10. Why is phenol red a better pH indicator than alizarin yellow for detecting a change in pH of broth containing pathogenic bacteria such as E. coli
Answer:
The correct answer is - acidic conditions wouldn't trigger a change in the color of Alizarin yellow.
Explanation:
The growth of E. coli generally occurs at neutral pH, however, its growth is normal at acidic conditions as well. The change in the growth of E. coli is not able to detect by alizarin.
The phenol red turns yellow in the presence of an acid, and the change in pH in an alkaline environment can be detected by the red color of phenol red. Growth of E.coli will grow in pH of 10-12 . But, very slowly. The color change in alizarin is also apparent at pH 10.2 to 12 only.
Which of these is not a component of a molecule of adenosine triphosphate (ATP)?
from brainly
cygilberts
Ambitious
34 answers
1.8K people helped
The right answer is Deoxyribose sugar
Explanation:
ATP consists of adenosine (this is an adenine ring and a ribose sugar) – and three phosphate groups (triphosphate).
Hope this helps
QUESTION 11 1. Through PCR, we have determined the PER3 genotypes for a class of students as follows: H4/H4 = 125 individuals; H4/H5 = 85 individuals and H5/H5=24 individuals. Using the allele's frequencies experimentally derived, calculate the frequency of the H4/H5 genotype that would be expected if the class were a population in Hardy-Weinberg equilibrium. 1. 0.28 2. 0.51 3. 0.19 4. 0.72 5. 0.14 6. 0.24 7. 0.41 QUESTION 12 1. Through PCR, we have determined the PER3 genotypes for a class of students as follows: H4/H4 = 125 individuals; H4/H5 = 85 individuals and H5/H5=24 individuals. Using the genotype frequencies derived assuming that the class were a population in Hardy-Weinberg equilibrium, calculate the number of H4/H4 individuals that would be expected in the class (rounded numbers). 1. 19 2. 57 3. 72 4. 147 5. 171 6. 120 7. 96 QUESTION 13 1. Through PCR, we have determined the PER3 genotypes for a class of students as follows: H4/H4 = 125 individuals; H4/H5 = 85 individuals and H5/H5=24 individuals. Considering the Hardy Weinberg equilibrium and comparing the observed and the expected number of individuals for the three genotypes, calculate the value of the Chi-square statistic 1. 2.69 2. 0.05 3. 28.67 4. 14.59 5. 0.50 6. 22.31 7. 3.84 QUESTION 14 1. Through PCR, we have determined the PER3 genotypes for a class of students as follows: H4/H4 = 125 individuals; H4/H5 = 85 individuals and H5/H5=24 individuals. Considering the Hardy Weinberg equilibrium, which is the (correct) null hypothesis tested by Chi-square? 1. The whole class represents a population that is in Hardy-Weinberg equilibrium 2. The whole class represents a population that may not be in Hardy-Weinberg equilibrium 3. The whole class represents a population that is not in Hardy-Weinberg equilibrium 4. The whole class represents a population that may be in Hardy-Weinberg equilibrium QUESTION 15 1. Through PCR, we have determined the PER3 genotypes for a class of students as follows: H4/H4 = 125 individuals; H4/H5 = 85 individuals and H5/H5=24 individuals. Considering the Hardy Weinberg equilibrium and calculating the Chi-square statistic, do you reject or fail to reject the null-hypothesis? 1. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the appropriate degrees of freedom I conclude that P>0.05. Hence, I reject the null hypothesis. 2. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the appropriate degrees of freedom, I conclude that P>0.05. Hence, I fail to reject the null hypothesis. 3. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the appropriate degrees of freedom I conclude that P<0.05. Hence, I reject the null hypothesis. 4. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the appropriate degrees of freedom I conclude that P<0.05. Hence, I fail to reject the null hypothesis.
According to Hardy-Weinberg, when a population is in equilibrium, it will have the same allelic frequencies generation after generation, meaning that they are stable, they are not evolving.
When any evolutive force is acting on the population, this equilibrium breacks, and allelic and genotypic frequencies change through generations, differing from the expected ones.
A) Option 7 is the correct answer ⇒ 0.41
B) Option 6 is the correct answer ⇒ 120
C) Option 7 is the correct answer ⇒ 3.84
D) Option 1 is the correct answer ⇒ The class represents a population that is in H-W equilibrium
E) Option 1 is correct. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the appropriate degrees of freedom I conclude that P>0.05. Hence, I reject the null hypothesis.
-------------------------------------------
Allelic frequencies in a locus are represented as p and q, referring to the
allelic dominant or recessive forms. The genotypic frequencies after one generation are p² (H0m0zyg0us dominant), 2pq (H3ter0zygous), q² (H0m0zyg0us recessive). Populations in H-W equilibrium will get the same
allelic frequencies generation after generation.
The sum of the allelic frequencies equals 1, this is p + q = 1.
In the same way, the sum of genotypic frequencies equals 1, this is
p² + 2pq + q² = 1
Being
p the dominant allelic frequency,
q the recessive allelic frequency,
p² the h0m0zyg0us dominant genotypic frequency
q² the h0m0zyg0us recessive genotypic frequency
2pq the h3ter0zyg0us genotypic frequency
Situation: Through PCR, we have determined the PER3 genotypes for a class of students as follows:
H4/H4 = 125 individuals;
H4/H5 = 85 individuals;
H5/H5=24 individuals.
⇒ Total number of individuals= 125 + 85 + 24 = 234
⇒ Genotypic frequencies, F(xx):
F(H4/H4) = 125/234 =0.534
F(H4/H5) = 85/234 = 0.363
F(H5/H5) = 24/234 = 0.102
⇒ Allelic frequencies, f(x):
f(H4) = p = F(H4/H4) + 1/2 F(H4/H5) = 0.534 + 0.363/2 = 0.534 + 0.182 = 0.716
f(H5) = q = F(H5/H5) + 1/2 F(H4/H5) = 0.102 + 0.363/2 = 0.102 + 0.182 = 0.284
Questions:
A) According to the theoreticall frame, we know that 2pq is the h3ter0zygous genotypic frequency. So,
F(H4/H5) = 2pq = 2 x 0.716 x 0.284 = 0.408 ≅ 0.41 ⇒ Option 7 is the correct answer.
-------------------------------------------------------------------------------------------------------------
B) According to the theoreticall frame, we know that p² is the h0m0zyg0us genotypic frequency. So,
p = 0.716
p² = 0.5126 ≅ 0.513 ⇒ This is the genotypic frequency.
To calculate the number of individuals carrying this genotype, we need to multiply it by the total number of
individuals.
H4/H4 individuals = p² x total number of individuals = 0.513 x 234 = 120
Option 6 is the correct answer.
-----------------------------------------------------------------------------------------------------------
C) Up to here we know that 2pq = 0.41 and p² = 0.513
Now we need to calculate q ²
q = 0.284, then q² = 0.284² = 0.08
These are the expected frequencies if the population was in H-W equilibrium.
The expected number of individuals with each genotype are:
H4/H4 = 0.513 x 234 = 120 individuals
H4/H5 = 0.41 x 234 = 96 individuals
H5/H5= 0.08 x 234 = 18 individuals
The observed number of individuals with each genotype are:
H4/H4 = 125 individuals
H4/H5 = 85 individuals
H5/H5=24 individuals
X² = ∑ (Observed - Expected)²/Expected)
X² = ((125-120)²/120) + ((85 - 96)²/96) + ((24-18)²/18)
X² = 0.21 + 1.26 + 2 =
X² = 3.47
The clossest option is option 7 = 3.84. The difference might be related to decimals and rounding.
-------------------------------------------------------------------------------------------------------------
D) The correct answer is 1 ⇒ The whole class represents a population that is in Hardy-Weinberg equilibrium
The null hypothesis always predict that populations are in H-W equilibrium.
-----------------------------------------------------------------------------------------------------------
E)
X² = 3.47
Freedom degrees = n - 1 = 3 - 1 = 2
Table p value: 7.82
Significance level, 5% = 0.05
Table value/Critical value = 5.991
5.991 > 0.347
Meaning that the difference between the observed individuals and the expected individuals is statistically significant. Not probably to have differe by random chances. There is enough evidence to reject the null
hypothesis.
Option 1 is correct. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the appropriate degrees of freedom I conclude that P>0.05. Hence, I reject the null hypothesis.
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To determine whether eating sweet snacks (e.g. candy) causes more weight gain than eating oily snacks (e.g. potato chips), you feed two different groups of mice 1g of either candy or potato chips each day for seven days, and you compare the starting weight of each group with the final weight after the week.
Identify the following elements of your snack experiment:
a. iIndependent variable:
b. Dependent variable:
c. Control treatment(s):
d. Experimental treatment(s)
e. Standardized variables:
Answer:
a. the independent variable is the type of food you are giving the mice.
b. the dependent variable is the final weight of each group of mice.
c. the control is the amount of food you give to the mice.
d. the experimental treatment is what you are doing (which is feeding the mice two types of food to see which one causes more weight gain).
e. the standardized variable is the amount of time you are feeding the mice for (seven days for both groups)
Explanation:
I hope this helps ! :)
all dicots are annuals true or false
Answer: The answer is false
What is the reason for having diverse functions for proteins
Answer:
to make the body strong
Ġ Surface area te volume ratio plays a vital role in À. growth rate of organisms B. exchange of materials between organisms and their environment C. the life-span of organisms D. efficiency of various systems in organisms
Answer:
Exchange of materials between organisms and their environment
Explanation:
What is a community?
1 all the animals that live in a habitat
2 a single species that lives in a habitat
3 all the species that live in a habitat
4 a population that lives in a single habitat
Answer:
3. All the species that live in a habitat.
A community is where all the species live in a habitat. Hence the correct option is 1.
A community is an ecological term that encompasses all the different species of organisms that coexist and interact within a specific habitat or geographic area. It includes plants, animals, fungi, and microorganisms that share the same environment and form intricate ecological relationships with each other.
These relationships can be competitive, predatory, symbiotic, or other forms of interactions that influence the dynamics and structure of the community. Understanding the composition and interactions within a community is vital in studying the biodiversity, ecosystem functioning, and overall health of a given habitat.
Hence the correct option is 1.
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In crude oil affect eggs and larvae of fish and increase mortality.
O a.
Hydrocarbon
O b.
PAH
O c.
liquid petroleum
Answer:
its c liquid petroleum 100% surr
What determines the composition of the soil?
1 . original rock
2. underlying rock breaks down
3. minerals that are present in the rock
4.
transported from a different location
Why is the frequency of natural disasters increasing?
1. There are less natural hazards occurring
2. Human population in areas prone to natural hazards has increased
3 . Deaths from natural disasters has decreased in developed countries and increased in developing countries.
Human population in areas prone to natural hazards has increased
Which of the following is not a natural disaster ?
1 . Tsunami
2. Earthquake
3. Terrorism
4. Hurricane
Answer:
terrorism
Explanation:
everyhthing else is natural because hurricanres come from wind and tide overflow tsunamis are caused by earth quakes and earthquakes are natural
The one that follows that is not a natural disaster is the terrorism that is present in Option 3, as the terrorism is done by man while the natural disasters are caused by nature, such as the hurricane and earthquake.
What role does a natural disaster play in the ecosystem?Natural disasters are phenomena caused by nature, such as cyclones, earthquakes, and tsunamis, and they cause significant damage, which may be indirectly caused by man as man causes global warming, climate change, and so on. The terrorism is man-made or artificial, as certain people are involved and harm other people or the country.
Hence, the one that follows that is not a natural disaster is the terrorism that is present in Option 3, as the terrorism is done by man while the natural disasters are caused by nature, such as the hurricane and earthquake.
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What are three ways in which bacteria obtain food
Answer:
The three ways by which bacteria obtain food are photosynthesis, chemosynthesis, and symbiosis.
Explanation:
Glycogen phosphorylase (GP) targets the non-reducing ends of glycogen to cleave glycogen and produce one glucose-1P at a time. GP will do this until it is three glucose molecules from the glucose molecule with the branch point - at which time another enzyme takes over the degradation. Which glucose molecule(s) on glycogen are substrates for GP based on this information
Answer:
Glucose molecules bound together by a-1,4 glycosidic linkages, and they must be >4 glucose molecules away from a branch point.
Explanation:
Glycogen phosphorylase can not degrade the glucose polymer close to the branch point because these sections of the glycogen molecule are to short for the glucose polymer to fit properly into the active site of the GP enzyme. The GP enzyme can therefore only degrade the 'straight' portions of glycogen. To degrade a branch point, a debranching enzyme is required. The debranching enzyme has transferase (cleaves off glucose molecules right before branch point and moves them to the end of another branch) and a-1,6 glycosidic activity which removes the branching glucose.
Glucose molecules are restrained together by a-1,4 glycosidic connections, and they must be >4 glucose molecules missing from a branch issue.
What are Glucose molecules?
Glycogen phosphorylase can not devalue the glucose polymer proximate to the branch pinpoint because these provinces of the glycogen molecule are too quick for the glucose polymer to fit properly into the involved site of the GP enzyme.
The GP enzyme can thus, only impair the 'straight' pieces of glycogen. To degrade a branch point, a debranching enzyme is directed.
The debranching enzyme has transferase (cleaves off glucose molecules correct before the attachment point and carries them to the end of another branch) and a-1,6 glycosidic movement which dismisses the branching glucose.
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Students in a science class tested different plant seeds to determine how long it took each type of seed to fall from the second- story window of the school. They tested 100 seeds of each type in order to find out which type of seed had the the parent plant. The data they collected is shown in the chart below. which of these graphs best represents the students data?
walnut/0.6
maple/4.5
redoak/0.6
ash/2.0
pine/2.4
The correct answer is C.
Explanation
According to the information provided, a bar graph (option C) is needed since this type of graph allows us to express the variation in time, distance, temperature, quantities, weight, among others. In this specific case, a bar graph allows students to analyze the time that each type of seed takes to fall from the second floor. Additionally, this is the only graph that shows the data of the chart. For example, the first bar represents the black walnut and its time is 0,6 seconds, and this value is shown by the light blue bar in the graph, which represent the same seed. According to the above, the correct answer is C.
When thinking about all 4 types of macromolecules discussed, which of the following statements are true? Select all that apply.
Monomers are linked together by hydrolysis.
Monomers are joined together to form functional polymers
Monomers are linked together by dehydration synthesis.
Monomers join together via hydrogen bonds.
If a physicist performs an experiment, who would likely try to replicate it?
The physicist himself
Other physicists
The family of the physicist
No one, because an experiment can never be replicated
Answer:
The physicist himself
Explanation:
Replication in an experiment means to repeat an experiment as many times as possible using the same conditions. This is done to minimize error and ascertain the reliability of the experimental outcome.
An experimenter is responsible for repeating his/her experiments. According to this question, a physicist performs an experiment. The physicist himself should replicate his experiment.
what is the difference of biology and human and social biology?????
Answer:
Man's health (human biology) affects and effects change on society (social biology). Human and social biology scrutinizes the human body, disease, health, nature and the environment's influence on biology.
Answer:
Man's health (human biology) affects and effects change on society (social biology). Human and social biology scrutinizes the human body, disease, health, nature and the environment's influence on biology.
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A stem cell is
A. A specialized cell that can divide limitlessly but cant differentiate
B. A specialized cell that can divide limitlessly.
C. An unspecialized cell that cant divide limitlessly nor differentiate into specialized cell.
Answer:
B
Explanation:
They divide to form more cells called daughter cells under proper conditions
where in an embryo are the instructions located for how to build organs
Answer:In the nucleus
Explanation:
The information for all bodily functions resides in DNA in every cell.
