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2 years ago

Glycolysis that starts with glycogen instead of glucose can be considered to have a higher energy yield because?

1 answer:

8 0

Glycolysis that starts glycogen instead of glucose cab be considered to have a higher energy yield because phosphorolysis reaction cleave bonds with phosphate instead of water.

Glycogen phosphorylase is a pyridoxal phospahte enzyme that transform glycogen to glucose 1- phosphate , a normal intermediate of glycolysis. Glycogenolysis is the biochemical pathway in which glycogen.

The process is under the regulation of two key enzyme phosphorylation kinase and glycogen phosphorylation. Glycogen degradation is initiated by the phosphorlyase action. Glucose is made up of many connected glucose molecules called glycogen. Glycogen is broken down to release glucose into the bloodstream to used as fuel for the cells.

To learn more about Glycolysis here

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Explain how the physical environment affected the way mesopotamians viewed the world.

Aleks [24]

The people lived in Mesopotamia are known as Mesopotamians.
The physical environment and the natural resources found in Mesopotamia changed the life of Mesopotamians. <span>They viewed the world because the environment is very suitable for agriculture, which freed up time for other activities like art, religion etc. and thus it makes them the modern society.</span>

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4 years ago

Why was Wegener’s theory rejected?

Fynjy0 [20]

He could not explain why the continents were moving.

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3 years ago

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attashe74 [19]

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3 years ago

Select ALL the correct answers.

tekilochka [14]

Answer:

(b) The interquartile range of B is greater than the interquartile range of A.

(d) The median of A is the same as the median of B.

Explanation:

Given

$A = \{1, 4, 2, 2, 3, 1, 1, 2, 1\}$

$10th\ run = 9$

So:

$B = \{1, 4, 2, 2, 3, 1, 1, 2, 1,9\}$

Required

Select all true statements

(a) & (d) Median Comparisons

$A = \{1, 4, 2, 2, 3, 1, 1, 2, 1\}$ $B = \{1, 4, 2, 2, 3, 1, 1, 2, 1,9\}$

$n = 9$ $n = 10$

Arrange the data:

$A = \{1, 1, 1, 1, 2, 2, 2,3, 4\}$ $B = \{1,1,1,1,2,2,2,3,4,9\}$

$Median = \frac{n + 1}{2}th$

$Median = \frac{9 + 1}{2}th$ $Median = \frac{10 + 1}{2}th$

$Median = \frac{10}{2}th$ $Median = \frac{11}{2}th$

$Median = 5th$ $Median = 5.5}th$ --- average of 5th and 6th

$Median = 2$ $Median = \frac{2+2}{2} = 2$

Option (d) is correct because both have a median of: 2

(b) & (c) Interquartile Range Comparisons

$A = \{1, 1, 1, 1, 2, 2, 2,3, 4\}$ $B = \{1,1,1,1,2,2,2,3,4,9\}$

$n = 9$ $n = 10$

First, calculate the lower quartile (Q1)

$Q_1 = \frac{n + 1}{4}th$ [Odd n] $Q_1 = \frac{n}{4}th$ [Even n]

$Q_1 = \frac{9 + 1}{4}th$ $Q_1 = \frac{10}{4}th$

$Q_1 = \frac{10}{4}th$ $Q_1 = 2.5$

$Q_1 = 2.5th$

This means that:

$Q_1 = 2nd + 0.5(3rd - 2nd)$ $Q_1 = 2nd + 0.5(3rd - 2nd)$

$Q_1 = 1 + 0.5(1- 1)$ $Q_1 = 1+ 0.5(1 - 1)$

$Q_1 = 1$ $Q_1 = 1$

Next, calculate the upper quartile (Q3)

$Q_3 = \frac{3}{4}(n + 1)th$ [Odd n] $Q_3 = \frac{3}{4}(n)th$ [Even n]

$Q_3 = \frac{3}{4}(9 + 1)th$ $Q_3 = \frac{30}{4}th$

$Q_3 = \frac{30}{4}th$ $Q_3 = 7.5th$

$Q_3 = 7.5th$

This means that:

$Q_3 = 7th + 0.5(8th- 7th)$ $Q_3 = 7th + 0.5(8th- 7th)$

$Q_3 = 2 + 0.5(3- 2)$ $Q_3 = 2+ 0.5(4 - 2)$

$Q_3 = 2.5$ $Q_3 = 3$

The interquartile range is $IQR = Q_3 - Q_1$

So, we have:

$IQR = 2.5 - 1$ $IQR = 3 - 1$

$IQR = 1.5$ $IQR =2$

(b) is true because B has a greater IQR than A

(e) This is false because some spread measures (which include quartiles and the interquartile range) changed when the 10th data is included.

The upper quartile and the interquartile range of A and B are not equal

8 0

3 years ago

Blue Genotype BB Yellow Genotype YY What would be the Genotype for Blue and Yellow? A. BY, or YB B. GG C. Not Possible D. BuYe

Fittoniya [83]

Answer:

A. BY, or YB

Explanation:

This question is depicting a phenomenon in inheritance called CODOMINANCE, which is a type of non-mendelian inheritance in which two alleles of a gene both express themselves in that gene. In this question, the allele for blue color (B) is codominant with the allele for yellow color (Y).

This means that an individual that has a blue phenotype will possess the genotype BB while an individual that has a yellow phenotype will possess the genotype YY. Hence, in an heterozygous state in which both alleles combine to produce a simultaneous blue and yellow phenotype, the genotype will be BY or YB.

7 0

3 years ago