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Complete the following sentence: "The interior of living cells is more
than the exterior because more
ions are expelled than ions are taken in by the sodium-potassium pump."
Select one:
O a. electropositive. Nak
O b. electronegative, Na.
O C. electronegative, Na, K
O d. electropositive, Na+, K+
Question 6
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The high concentration of protons in the inner mitochondrial space relative to the mitochondrial matrix represent
O
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Answer:
i really really need the brainly points
Explanation:
sry for this answer, i need the answers for myself
What is microbiology?
Microbiology is the study of microscopic organisms, such as bacteria, viruses,
Two species have homologous structures. What do these homologous structures show about the evolutionary relationship between the two species?
The outbreak has rebounded in at least 30 US states in recent weeks, with the three most populous states -- California, Texas and Florida -- seeing a surge in new cases, with the highest daily number of new cases since the outbreak began.
Answer:
whats the question then?
Explanation:
Which one of the following would be inhibited by a well-designed antiviral drug? Cell wall synthesis Viral binding to human cells Virus assembly outside of the infected cell Translation of host cell RNAs
Viral binding to human cells is inhibited by the antiviral drug.
Well-designed antiviral drug inhibited Viral binding to human cells so that the virus can't get the place of attachment and unable to use the cell's machinery for its growth and multiplication. In this way, the humans can be prevented from having the viral infection. There are some other mechanisms also used by the antiviral drug to inhibit the growth of virus in the human body such as uncoating of virus and synthesis of new viral components.
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hey everyone,what is greenhouse effect
Answer:
The greenhouse effect is a natural process that warms the Earth's surface.
When the Sun's energy reaches the Earth's atmosphere, some of it is reflected back to space and the rest is absorbed and re-radiated by greenhouse gases.
The absorbed energy warms the atmosphere and the surface of the Earth causing green house effect .
Answer:
The greenhouse effect occurs when radiation from a planet's atmosphere warms the planet's surface to a higher temperature than it would be without the atmosphere. Radiatively active gases emit energy in all directions from a planet's atmosphere.
OAmalOHopeO
Which of the following combinations of phylum and description is correct? (and why?)
a) Nematoda − roundworms, internal skeleton
b) Porifera − gastrovascular cavity, coelom
c) Echinodermata - radial symmetry as a larva, coelom
d) Platyhelminthes − flatworms, gastrovascular cavity, no body cavity
The phylum and description that is correct among the options is Platyhelminthes − flatworms, gastrovascular cavity, no body cavity
Flatworms belongs to the phylum platyhelminthes. They are said to have no true body cavity, but they do have bilateral symmetry. For the lack of a body cavity, it is called acoelomates. They also have an incomplete digestive system. That is, the digestive tract has only one openingsFrom the above we can therefore say that Option d is the correct option that gives the true description of the phylum platyhelminthes
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Given that the intracellular concentration of potassium is 150 mEq/L, how would the potassium equilibrium potential be affected if the extracellular concentration of potassium is changed from 5.0 to 3.5 mEq/L
Answer:
The potassium equilibrium potential would increase, meaning that more K+ would be leaving the cell.
Explanation:
Let us assume that the only ion transported through the cell membrane is K+. We need to use the Nernst equation to know the destiny of the ion.
Nernst equation:
E = 58 millivolts/z. [Log₁₀ (C-out/C- in)
Where,
• E = Equilibrium potential
• 58 millivolts/z = Constant
• z = Ion charge + positive or negative symbol
• C-out = Ion concentration out of the cell
• C-In = Ion concentration inside the cell
By convenience, in the Nerts equation, the bigger concentration value corresponds to the numerator and the smaller concentration value to the denominator.
Now let us see the provided values,
• z = Ion charge + positive or negative symbol ⇒ +1 ⇒ K+
• C-out = Ion concentration out of the cell ⇒ 5 mEq/L
• C-In = Ion concentration inside the cell ⇒ 150 mEq/L
E = 58 millivolts/z. [Log₁₀ (Ion in/Ion out)
E = 58 millivolts/+1. [Log₁₀ (150 mEq/L / 5 mEq/L)
E = 58 millivolts (Log₁₀ 30)
E = 58 millivolts (1.477)
E = 85.67 millivolts
85.7 mV is the absolute value of equilibrium potential.
E = 58 millivolts/z. [Log₁₀ (Ion in/Ion out)
E = 58 millivolts/+1. [Log₁₀ (150 mEq/L / 3.5 mEq/L)
E = 58 millivolts (Log₁₀ 42.85)
E = 58 millivolts (1.63)
E = 94.65 millivolts
94.7 mV is the absolute value of equilibrium potential.
If the extracellular concentration of potassium is changed from 5.0 to 3.5 mEq/L, there will be an increase in the membrane potential from 85.7 to 94.7 mV. The increase in the equilibrium potential will result in more potassium diffusing out of the cell, turning the cell interior less positive than before.
The potassium equilibrium ability might increase, which means that greater K+ might be leaving the cell.
Let us expect that the handiest ion transported via the cell membrane is K+. We want to apply the Nernst equation to recognize the future of the ion.
Nernst equation:
[tex]E = 58 millivolts/z. [Log₁0 (C-out/ -in)[/tex]
Where,
E = Equilibrium ability.58 millivolts/z =Constant.z=lon charge + advantageous or terrible symbol. C - out = Ion awareness out of the cell. C-ln= ion awareness in the cell.For convenience, withinside the Nerts equation, the larger awareness fee corresponds to the numerator and the smaller awareness fee to the denominator. Now allow us to see the supplied values,
[tex]z=lon charge + effective or terrible ⇒+1 ⇒ K+\\C - out = lon awareness out of the cell ⇒5 mEq/L\\C-ln= lon awareness withinside the cell ⇒150 mEq/LE = 58 millivolts/+ 1.[Log 10 ( 50mEq / L / 5mEq/L)\\E = 58millivolts (Log30)\\E = 58 millivolts (1.477)[/tex][tex]E = 85.67 millivolts\\85.7 mV is absolutely the fee of equilibrium capability.\\E = 58 millivolts/z. [Log10 (lon in/lon out)\\E =58 millivolts/+1. [Log 10 (a 100and 50 mEq/L / 3.five mEq/L)\\E =58 millivolts (Log10 42.85)\\E = 58 millivolts (1.63)\\E = 94.65 millivolts94.7 mV is absolutely the fee of equilibrium capability.[/tex]
What happens if the extracellular attention of potassium is modified from 5.0 to 3.5?If the extracellular attention of potassium is modified from 5.0 to 3.5mEq/L, there can be a growth withinside the membrane capability from 85.7 to 94.7 mV. The growth withinside the equilibrium capability will bring about extra potassium diffusing out of the cell, turning the cell indoors much less high quality than before.
Thus it is clear from this that the potassium equilibrium potential is affected.
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As water is cooled from 4° C to 0° C, its density
A. stays the same
B. decreases and increases
C. increases
D. decreases
Answer:
I believe that it increases (becoming more dense)
Explanation:
Well 0°C is freezing point so I think that in that state it will become a solid from a liquid and freeze into ice.
Starch and protein digestion in a single stomach?
Answer:
Explanation: Protein digestion occurs in the stomach and the duodenum through the action of three main enzymes: pepsin, secreted by the stomach, and trypsin and chymotrypsin, secreted by the pancreas. During carbohydrate digestion the bonds between glucose molecules are broken by salivary and pancreatic amylase.
In the oceans on either side of the Isthmus of Panama are 30 species of snapping shrimp, 15 species on the Pacific side and 15 different species on the Atlantic side. Species live at different water depths. Morphological and genetic data show that Atlantic and Pacific species that live at similar depths are sister species. The sister species on each side of the isthmus cannot interbreed because the water in the canal is fresh water, not salt water, and provides a barrier to reproduction. A sea-level, salt-water canal between the two oceans has been proposed to make transport across the isthmus easier. Which of the following outcomes is the most likely result if such a canal were built?
A. greater percentage of difference in DNA sequences between sister species that inhabit deep water than between sister species that inhabt stalow water
B. greater percentage of difference in DNA sequences between sister species that inhabit shallow water than between sister species that inhabit deep water
C. similar percentages of difference in DNA sequences between all pairs of sister species
D. greater percentage of difference in DNA sequences between Atlantic species than between Pacific species
Answer:
The options of this question are wrong, you can find the correct options by navigating on the web. The options of this question are as follow:
1) The sister species will continue to diverge from each other.
2) None of the sister species will interbreed with each other.
3) The Atlantic and Pacific shrimp will continue to live in their respective oceans and not enter the new canal.
4) Shallow-water species from the two oceans that are sister species would be more likely to interbreed with each other than would be deep-water species.
Answer:
4) Shallow-water species from the two oceans that are sister species would be more likely to interbreed with each other than would be deep-water species
Explanation:
In evolutionary biology, sister species are defined as descendant species formed when one species splits during the course of evolution. Moreover, adaptation refers to the evolutionary process of adjustment of organisms to the environment, which is usually due to natural selection. During the course of evolution, organisms under different environments must change to adapt to their environments. In this case, it is expected that sister species that live in similar environmental conditions (i.e., shallow-water species) exhibit fewer phenotypic differences, being therefore more likely to interbreed with each other.
As you read in this chapter, fungi have long formed symbiotic associations with plants and with algae. How may these two types of associations lead to emergent properties in biological communities
Answer:
Fungi show symbiotic association with algae and plants. With plants, they thrive as endophytes in a form of the symbiotic association. This symbiotic association results in the emergence of novel characteristics in the world of biology.
The lichens function as a tool in finding the quality of air, as they grow in the environment containing good air quality. The tolerance towards heat is another characteristic. Some of the endophytes are witnessed in the plants, which grow in very hot conditions.
At such conditions, no fungi or plant can thrive, however, in the symbiotic association, they possess the tendency to thrive. If one tries to separate them, it results in the death of both.
Explanation:
supuie ) Prokaryotes have nucleus that is without : i) nuclear ii) membrane nucleous iii) nucle
Answer:
the ans is
NUCLEOUS MEMBRANEor
MEMBRANE nucleus u can call it either way
Answer:
Nucle
Explanation:It is said that Prokaryotes have nucleus that has nuclear,membrane and nucleous
what evidence shows that biological molecules on earth form naturally?
Explanation:
La siguiente entrada tiene como objetivo realizar una breve explicación sobre las moléculas biológicas lipídos y carbohidratos, las cuales son muy diversas ya que están formadas por carbono, lo cual hace que puedan formar muchos tipos de enlaces. Esta capacidad permite que las moléculas orgánicas adopten muchas formas complejas, como son las cadenas, las ramificaciones y los anillos.
Las moléculas biológicas son grandes polímeros que sintetizas para poder enlazarse con otras subunidades mucho mas pequeñas conocidos como monómeros. Las cadenas de subunidades estan unidas por enlaces covalentes los cuales se forman por deshidratación, estas cadenas pueden romperse por hidrólisis. La moléculas biologicas más importantes son los carbohidratos, lípidos, proteínas y ácidos nucleicos.
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
What variable did the experimenter change (independent variable), and what variable did the experimenter observe (dependent variable)?
Answer:
An independent variable can be controlled by the experimenter to observe it's affect on dependent variable which cannot be changed manually. Rather the dependent variable changes due to the effect of independent variable.
Answer:
The experimenter changed the carbon dioxide concentrations in the two flasks and observed the temperature change over time.
Explanation:
plato answer
What else is produced during the replacement reaction of silver nitrate and potassium sulfate?
2AgNO3 + K2SO4 Ag2SO4 + ________
KNO3
2KNO3
K2
2AgNO3
Answer:
2KNO3
Explanation:
Pls mark it brainliest
hope it helps u
Answer:
The answer is B.) 2KNO3
Explanation:
In a hydrogen ion pump, the energy is used to join small molecules together
to
make larger ones. Which factor most likely has the greatest effect on the
number of molecules mitochondria can produce?
Answer: The number of H+ ions moving down the channel
Explanation:
What happened to the California Condors?
A. They went extinct in 2002 due to destruction of their habitat.
B. There were only 149 left in the 1980's but even with a captive breeding
program they went extinct in 2010.
C. They went extinct in the 1700's but remains from a well preserved skeleton
contained enough to DNA to begin cloning them.
D. In 1985 there were only 9 left, but a captive breeding program has increased
their number to 300.
Answer:
B
Explanation:
California Condors have been in decline about as long as European Settlements began to spread across North America. These birds have been on the U.S endangered species list since 1967 and were near extinction when their captive-breeding program began. California Condors also mature and reproduce slowly.
M. magneticum can only survive in low-oxygen environments, which are typically found near the bottom of bodies of water.
a. True
b. False
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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Water droplets are pulled towards earth by ______________.
How are human affiliative and aggressive behaviors different from what is seen in nonhuman primates? Provide specific examples.
Answer:
Humans are less aggressive as compared to non-human primates.
Explanation:
Human affiliative and aggressive behaviors are different from nonhuman primates because humans do more friendly and peaceful acts and are less aggressive in anger as compared to non-human primates which are less friendly and more aggressive in anger. Due to more aggressive behaviour of non-human primates causes more damaged to other animals and humans as compared to humans who is less damaging and aggressive.
what is gestation period?
Answer:
gestation period is the period of development during the carrying of the embryo or fetus inside viviparous animals
How are oxygen and carbon dioxide exchanged between the alveoli and the
capillaries?
A. Endocytosis
B. Osmosis
C. Simple diffusion
5
D. Active transport
Answer:
B. Osmosis
Explanation:
Osmosis is the process in which oxygen and carbon dioxide exchanged occur between the alveoli and the capillaries because the oxygen enters the body and the carbondioxide gas leaves the cell through a semi-permeable membrane and we know that Osmosis is a process in which smaller molecules moves from higher concentration to lower concentration through semi-permeable membrane of the cell.
Indicate whether each statement is true or false: 1. Compliance is the tendency for blood vessel volume to increase as blood pressure decreases. True 2. Blood vessels with a large compliance exhibit a small increase in volume when the pressure increases a small amount. True 3. Venous compliance is approximately 24X greater than arterial compliance, so as venous pressure increases the volume of veins greatly increases. True
Answer:
1. False
2. False
3. True
Explanation:
1. Compliance is the capacity of a container to increase in size to allow it hold more content. Blood vessel, arteries and veins expand (increase in volume) to be able to accommodate a surge in blood flow, which is as a result of an increase in pressure of the blood from the heart pumping of the blood
Therefore, compliance in the tendency for blood vessel volume to increase as the blood pressure increases not decrease
The statement is false
2. A large compliance is indicative of being highly sensitive to changes in pressure
Compliance, C = ΔV/ΔP
From the above equation, a blood vessel with a large compliance, exhibit a large increase in volume when the increase in pressure is small
Therefore, the statement 'Blood vessels with a large compliance exhibit a small increase in volume when pressure increases a small amount; is false
3. The compliance of the vein ranges from 10 to 20 times (30 times in some literature) greater than arteries. A factor which can be affected by the vascular smooth muscle contraction or relaxation
Therefore, the statement, 'venous compliance is approximately 24 times larger than arterial compliance, so as venous pressure increases the volume of veins greatly increases' is true
Which of these is a covalent bond in which the electrons are not evenly shared?
Answer:
polar covalent bond
Explanation:
Imagine you found S. aureus to be resistant to Penicillin by Kirby Bauer analysis, but susceptible to Penicillin treatment in liquid culture (in other words, a MIC was determined). Which of the following are possible explanations for this inconsistency?
a. The concentration of Penicillin was higher in the Penicillin antibiotic disk than that used in liquid culture treatment.
b. The Penicillin disks used for Kirby Bauer analysis were expired/no longer active.
c. The concentration of bacteria was lower on the Mueller Hinton plate for Kirby Bauer analysis than that used in liquid culture treatment
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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In this excerpt, a reader can conclude that Lizzie is playful based primarily on her
Answer:
based on Lizzie's words
