Test Bank for Lehninger Principles of Biochemistry 7th Edition by Nelson (complete, questions/answers/rationales) | Lehninger Principles of Biochemistry 7th Edition Nelson Test Bank
FULL TEST BANK Lehninger Principles Of Biochemistry Seventh Edition By David L. Nelson (Author) Latest Update Questions And Answers Graded A+
Test Bank For Lehninger Principles of Biochemistry, 7th Edition by LEHNINGER Chapter 1-28
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BB2730 Metabolic Regulation
Metabolism
- Meta- (with, after) bole- (throw) ism means change. Metabolism is all the chemical
processes that occur within a living organism resulting in energy production (catabolism) and
growth (anabolism)
- Main nutrients: carbohydrates, lipids, proteins, minerals and vitamins. Vitamins are
precursors for some of the molecules you need for a reaction → cofactors
- Nearly all energy (90%) comes from carbohydrates and fats; carbon skeletons of amino acids
account for only 10%.
- Amino acids are important as energy source when carbohydrates are insufficient. When
excess amino acids are consumed, they have to be catabolised.
Examples of organs and their metabolic functions
- Brain – coordinates processes in the body
- Lymphatic system – carries lipids
- Adipose tissue – stores lipids
- Skeletal muscle – uses ATP for movement
- Small intestine – absorbs nutrients
- Portal vein – carries nutritious blood from small intestine to liver
- Liver – chemical brain of the body; processes fats, carbohydrates and lipids from diet;
synthesises and distributes lipids, ketone bodies and glucose for other tissues; converts
excess nitrogen (ammonia) to urea
- Pancreas – secretes insulin, digestive enzymes and glucagon
Liver
- The liver receives ingested nutrients from the digestive organs via the portal vein and
processes them for release into the blood
- Certain metabolic activities are only found in the liver
- By synthesising glucose via gluconeogenesis and releasing glucose from breakdown of liver
glycogen, the liver keeps blood glucose stable
- The preferred energy source of hepatocytes is fatty acids not glucose.
- The cellular compartimentalisation has a vital role in metabolic regulation:
o Hepatocytes contain a lot of mitochondria
Cytosol → glycolysis and fatty acid synthesis
Mitochondrial matrix → citric acid cycle and oxidative phosphorylation
Adipose tissue
- White adipose tissue is widely distributed in the body: under skin, in the abdominal cavity
and around the deep blood vessels
- Contributes about 15% of the mass of a human adult and 65% of this as triacylglycerols
- Adipocytes are metabolically very active cells and respond to hormonal stimuli rapidly
- They are important players in the interplay between muscle, liver and heart
- When food enters the body, the fat synthesised is built for adipose tissue. When adipose
tissue mass increases released leptin inhibits feeding and fat synthesis and stimulates fat
oxidation.
,Regulation of food intake
- Cholecystokinin (CCK) and insulin are gastrointestinal hormones that are released by
ingestion of food. They suppress further eating.
- Ghrelin is a peptide hormone secreted by cells lining the stomach it is a powerful appetite
stimulant.
- The stomach releases ghrelin, the pancreas releases insulin and adipose tissue release leptin
to the hypothalamus. Ghrelin stimulates the orexigenic (appetite-stimulating) neurons.
Leptin and insulin stimulate the anorexigenic (appetite-suppressing) neurons.
,Entropy
- Entropy is associated with the randomness or disorder of a system
o Aerobic organisms extract free energy from glucose obtained from their surrounding
by oxidising the glucose with O2, also obtained from the surrounding. The end
products, CO2 and H2O, are returned to the surroundings. The surroundings undergo
an increase in entropy, whereas the organism itself remains in a steady state and
undergoes no change in its internal order.
o Equation: C6H12O6 + 6 O2 → 6 CO2 + 6 H2O
o The atoms contained in 1 molecule of glucose plus 6 molecules of oxygen are more
randomly dispersed by the oxidation reaction and are now present in a total of 12
molecules (6 CO2 + 6 H2O)
Energy coupling
- In a mechanical process:
o The downward motion of an object releases potential energy that can do mechanical
work. The potential energy made available by spontaneous downward motion, an
exergonic process, can be coupled to the endergonic upward movement of another
object
- In a chemical process:
o In reaction 1, the formation of glucose-6-phosphate from glucose and inorganic
phosphate (Pi) yields a product of higher energy than the two reactants. For this
endergonic reaction, ∆G is positive. In reaction 2, the exergonic breakdown of ATP
has a large, negative free-energy change (∆G 2). The third reaction is the sum of
reactions 1 and 2, and the free energy change, ∆G 3, is the sum of ∆G1 and ∆G2.
Because ∆G3 is negative, the overall reaction is exergonic and proceeds
spontaneously.
Glucose + ATP → glucose-6-phosphate + ADP
, Activation energy
- The activation energy is the minimum energy required to initiate a reaction. Enzymes
accelerate reactions by decreasing the activation energy. The combination of substrate and
enzyme creates a reaction pathway whose transition-state energy is lower than that of the
reaction in the absence of enzyme. Because the activation energy is lower, more molecules
have the energy required to reach the transition state.
- The transition state is the reaction intermediate with the highest energy. If the reaction has
enough energetic input to reach this state, the product is formed.
Reaction equilibria
- Consider the following reaction: S ⇄ P. The ∆G of this reaction is given by
[ P]
∆G = ∆G° + RT ln .
[S ]
o ∆G° = the standard free-energy change (kJ mol -1)
o R = the gas constant (kJ mol-1 deg-1)
o T = the temperature in Kelvin
- The equilibrium constant under standard condition, K’ eq, defined as
[ Products] [ P]
K’eq = or K’eq =
[ Reactants] [S ]
- The relation between K’eq and ∆G°’ can be described by this expression:
∆G°’ = -RT ln K’eq
o R = the gas constant (kJ mol-1 deg-1)
o T = the temperature in Kelvin
The equilibrium constant is directly related to the overall standard free-energy change for
the reaction. A large negative value for ∆G°’ reflects a favourable reaction equilibrium but
does not mean that the reaction will proceed at a rapid rate.
- An enzyme cannot alter the laws of thermodynamics and consequently cannot alter the
equilibrium of a chemical reaction. The amount of product formed is the same whether or
not the enzyme is present of not, but the amount of product formed in seconds when the
enzyme is present might take hours to form if the enzyme is absent.
- After a while the rate of product formation levels off. This is because the reaction has
reached equilibrium. Substrate S is still being converted to product P, but P is being
converted to S at a rate such that the amount to P present stays the same → S ⇄ P
o Example
In the absence of enzyme, the forward reaction (k F) for conversion of S into
P in 10-4 s -1 and the reverse constant (kR) for the conversion of P into S is 10-6
s-1. The equilibrium constant K is given by the ratio of these constants:
[ P] k F 10−4
K= = = = 100
[S ] k R 10−6
The equilibrium concentration of P is 100 times that of S, whether or not
enzyme is present. However, it might take a very long time to approach this
equilibrium without enzyme, whereas equilibrium would be attained rapidly
in the presence of a suitable enzyme
- Enzymes accelerate the attainment of equilibria but do not shift their positions. The
equilibrium position is a function only of the free energy difference between reactants and
products
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