The endoplasmic reticulum bound enzyme that hydrolyzes glucose-6-phosphate to glucose in liver is: glucose-6-phosphatase.
Glucose-6-phosphatase (G6Pase), an enzyme found mainly in the liver and the kidneys, plays the important role of providing glucose during starvation. Unlike most phosphatases acting on water-soluble compounds, it is a membrane-bound enzyme, being associated with the endoplasmic reticulum.
Liver cells contain a membrane bound enzyme called glucose-6-phosphatase for glycogenolysis by glucagon especially during starvation when free glucose is required. As glucagon enters the liver cells it activates the enzyme glucose-6-phosphatase which then acts on glucose-6-phosphate and hydrolyzes it. As glucose-6-phosphate is hydrolyzed, it results in the formation of a phosphate group and a free glucose. The free glucose thus formed is transported from the liver cell to other tissues by specific glucose transport membrane protiens.
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C. 38 ATP
<span>In the presence of oxygen, one
glucose molecule has the energy to make up to 38 ATP. The ATP production is
determined by the following steps, (-2 ATP) glycolysis preparatory phase, (7-9
ATP) glycolysis pay-off phase, (5 ATP) oxidative decarboxylation of pyruvate
and (20 ATP) Krebs cycle. One glucose which has 38 ATP hence was the summation
of all the process mentioned that took place. All these process take place under the
cellular function of cellular respiration. </span>
Answer:
The answer is B.
Explanation:
If warmer temperatures last longer there will be no need for the white fur
Answer:
<em>The correct option is A. dihybrid</em>
Explanation:
A cross is drawn to generate the possible outcomes of traits being passed from parents to offsprings.
A monohybrid cross can be described as a cross in which only one trait is studied. For example, just studying the trait for eye colour.
A dihybrid cross can be described as a cross in which two traits are studies in a single cross. For example, studying the traits for eye colour and skin colour.
Answer:
Frequency of allele A1- 0.41
Explanation:
In Hardy weinberg equilibrium,
P refers to the dominant allele
q refers to the recessive allele
The allele frequency will be p+q=1
The genotypic frequency is- P²+q²+2pq=1
P²= genotype of dominant trait ( A1A1)- 77
2pq= genotype of heterozygotes (2pq)- 65
q²= genotype of recessive trait (A2A2)- 123
Total number of offsprings= 77+ 65+ 123
= 265
Now to calculate allele frequency of A1=
= 77/265 + 1/2( 65//265)
= 0.290+ 0.122
= 0.413
Thus, 0.41 is correct.
