Biochem

Question 1

food

Answer 1

bread: contains complex carbs. contains starch, the major source of carbohydrates in food consisting of glucose polysaccharides amylose (α1,4 glycosidic bonds) and amylopectin (α1,4 and α1,6 glycosidic bonds)
Honey: contains monosaccharides glucose, fructose
Coffee: contains sucrose disaccharide (depends on roast)
Milk: contains lactose (galactose and glucose disaccharide)
sugar: consists of sucrose (glucose and fructose disaccharide)

Question 2

complex carbs are disassembled into simple sugars with the help of enzymes

Answer 2

an ezyme alpha-amylase and glucose polysaccharide it breaks complex carbs such as starch to dextrins( a glucose oligosaccharide) and maltose (glucose disaccharide).
then enzymes such as dextrinase breaks down dextrin with addition of a water molecule to a glucose a simple monosaccharide. then enzymes such as maltase breaks down maltose with addition of a water molecule to a glucose a simple monosaccharide. then enzymes such as lactase breaks down lactose with addition of a water molecule to a glucose a simple monosaccharide and galalactose. then enzymes such as sucrase breaks down sucrose with addition of a water molecule to a glucose a simple monosaccharide and fructose

• in our body, we can only distribute monosacchrides

Question 3

from the digestive system

Answer 3

monosaccharides are transported to the cells of the small intestine and then to blood with the help of proteins, glucose transporters. Disaccharides and dextrins cannot be transported

Question 4

Lactose

Answer 4

Lactose intolerance is an example of a condition caused by
inefficient hydrolysis of disaccharides.
Lactose is a disaccharide found in milk, “milk sugar”
Lactose is hydrolyzed to monosaccharides by betagalactosidase, a.k.a. lactase
Mammals are unlikely to encounter lactose after they are
weaned, so the level and activity of lactase are low in adult
Non-hydrolysed lactose moves through the digestive tract to the
colon, fermented and large quantities of CO2, H2 and acids are
produced
Lactose intolerance: normal condition in adult humans.
More common in Asians

how to help people with lactose intolerance: Removal of lactose or hydrolysis of lactose can help people with lactose intolerance

Question 5

From the blood, glucose enters cells in our body via different
transporters located at the cell surface

Answer 5

.

Question 6

production of ATP

Answer 6

glucose is used to recycle ADP to ATP, which is used to drive other processes, e.g. movement of organelles (1 ATP molecule per 8.1 nm)

Question 7

cells recycle ATP from ADP via

Answer 7

glycolysis

Question 8

Glycolysis is a set of 10 reactions in the cytosol

Answer 8

glycolysis can be split in 2 parts: 1 preparatory phase and 2 pay off phase

Question 9

Preparatory phase

Answer 9

The first five steps are regarded as the preparatory (or investment) phase because they consume energy to convert the glucose into two three carbon sugar phosphates. where 2 ATP molecules are lost in this molecule

Question 10

glucose transporters

Answer 10

can transport glucose into and out of the cells with the same efficiency. the purpose of the first reaction of glycolysis is to trap glucose in the cell. the first reaction is catalysed by hexokinase, which phosphorylates glucose to prevent its transport from the cell. The reaction is highly spontaneous and irreversible under cellular conditions. It acts to keep glucose in the cell, because cells can transport glucose across the membranes but lack transporters for D-Glucose-6-phosphate. the hexokinase transports the phosphate of ATP; which converts to ADP, to the monosoccharide to form D-Glucose-6-phosphate

Question 11

in the second reaction,

Answer 11

phosphoglucose isomerase converts
glucose 6-phosphate to fructose 6-phosphate. The reaction is reversible and is necessary to prepare the molecule for the next steps – phosphorylation and cleavage. in this phase an atp is lost

Question 12

in the third reaction.

Answer 12

phosphofructokinase phosphorylates the molecule second time to produce Fructose-1,6-bisphosphate, where also an ATP was lost in the reaction. the reaction is irreversible under cellular conditions.

Question 13

in the fourth reaction

Answer 13

aldolase cleaves fructose-1,6-bisphosphate and generates two similar products; Dihydroxyacetone phosphate and Glyceraldehyde-3
phosphate. the reaction is reversible and the products are similar but not identical

Question 14

in the last preparatory reaction

Answer 14

triosephosphate isomerase
converts dihydroxyacetone into glyceraldehyde-3 phosphate. As
a result, the preparatory phase leads to generation of two
identical molecules of glyceraldehyde-3 phosphate (note, that 1
molecule of glyceraldehyde-3 phosphate is produced by
aldolase).

Question 15

in the payoff phase

Answer 15

two identical
cleavage products
are converted into
pyruvate with the
release of energy
using the same set
of reaction

Question 16

start of pay off phase

Answer 16

starts with oxidation of Glyceraldehyde 3-
phosphate to 1,3-Bisphosphoglycerate with the help of
dehydrogenase.NAD+ is converted to NADH+ H+ in the process.
NAD+ - Nicotinamide adenine dinucleotide, a coenzyme found in all living cells

Question 17

stage 2 of pay off phase

Answer 17

3-Phosphoglycerate Transfer from 1,3-Bisphosphoglycerate to ADP by
phosphoglycerate kinase results in generation of ATP

Question 18

Conversion of 3-Phosphoglycerate to 2-Phosphoglycerate

Answer 18

by phosphoglycerate mutase

Question 19

Dehydration of 2-Phosphoglycerate to Phosphoenolpyruvate

Answer 19

by enolase and water is released. the loss of the water molecule from 2-phosphoglycerate causes a redistribution of energy within the molecule

Question 20

Transfer of the phosphoenolpyruvate with ADP to
ATP and pyruvate by pyruvate kinase hence results in generation of ATP

Question 21

summary

Answer 21

glucose +2ATP + 2ADP +2Pi + 2NADH+ —-> 2PYRUVATE + 4ATP + 2 NADH + 2H+ + 2H2O

GLUCOSE + 2ADP + 2Pi + 2NAD+ —-> 2 PYRUVATE + 2ATP + 2NADH + 2H+ + 2H2O

Question 22

Glycolysis releases only a fraction of energy
stored in glucose

Answer 22

glycolysis energy yield = 5%
glycolysis—> 2x pyruvate and 2x acetylcoA —> CO2 + H2O

2x pyruvate and 2x acetylcoA —> CO2 + H2O; TRICARBOXYLIC
ACID CYCLE and full oxidation Energy yield
= 95 %

Question 23

why is glycolysis important?

Answer 23

• It is anaerobic (can take place in
  the absence of oxygen)
  so can provide ATP for muscle
  during strenuous exercise.
  and

It provides the ATP extremely
rapidly (suitable for muscle
contraction during sprints).

Oxidative phosphorylation occurs in special organelles
called mitochondria. Glycolysis is used as a primary
mechanism for ATP production in cells that don’t have
mitochondria (e.g. in erythrocytes).

Even under aerobic conditions, glycolysis is the major starting
point for carbohydrate metabolism

Question 24

Energy Supply During Exercise

Answer 24

ATP: up to 4 seconds. High jump, power
lift, shot put, tennis serve
* Phosphocreatine: up to 10 seconds.
sprints, American football line play
* Glycolysis: up to 1.5 minutes. 200-400 m
race, 100 m swim. Cancer tissues switch to glycolysis, because the growing tumour has insufficient access to blood vessels and has
to use alternative strategy to obtain energy. Positron Emission Tomography is used to detect cancer
tumours by monitoring the glucose uptake
* Oxidative phosphorylation: race beyond
500 meters

Question 25

Levels of AMP, ADP and ATP determine the energy state of the cells

Answer 25

Low energy state, a lot of a (3) ADP, low levels of (1) ATP ADP. In a very low energy state, 2 ADP molecules are converted into 1 ATP molecule and 1 AMP molecule resulting in accumulation of AMP. High energy state, a lot of a ATP, low levels of ADP

Question 26

In a low energy state,

Answer 26

, the speed of glycolysis is enhanced via an increase in the activity of phosphofructokinase, a rate limiting enzyme. Phosphofructokinase is activated by ADP and AMP

Question 27

An increase in fructose 6-phosphate levels leads to generation of the fructose 2,6 bisphosphate, which allosterically activates phosphofructokinase

Answer 27

The activity of phosphofructokinase 2 is hormonally stimulated when the blood glucose level is high.

Question 28

Glycolysis rates depend on the transport of glucose to the cytosol. The efficiency of the transport is determined by the numbers of glucose transporters at the cell surface

Answer 28

Glucose transporters are stored in vesicles, which deliver the transporters to the cell surface when glucose levels increase

Question 29

Summary of the key enzymes in the regulation of the glycolysis

Answer 29

Hexokinase (step 1): Glucose + ATP -> glucose-6-phosphate + ADP is inhibited by glucose-6-phosphate (product inhibition) Phosphofructokinase (step 3): Fructose-6-phosphate + ATP -> fructose-1,6-biphosphate + ADP is activated by ADP and AMP, inhibited by ATP and NADH Pyruvate kinase (step 10): Phosphoenolpyruvate + ADP -> pyruvate + ATP is inhibited by ATP

Question 30

Liver cells store excess glucose when it is not needed for other cells by expressing glucokinase instead of hexokinase

Answer 30

Hexokinase is ubiquitous, the same end product: glucose-6-phosphate , high affinity to glucose, and initial step in glycolysis. glucokinase is in liver cells, low affinity to glucose, traps only excess glucose to store as glycogen

Question 31

balance sheet of glycolysis

Answer 31

glucose + 2ADP 2Pi + 2NAD+ -----> 2 PYRUVATE + 2 ATP + 2NADH + 2H+ + 2H2O the problems: 1 cells need NAD+ for glycolysis to continue and cells need to prevent accumulation of NADH and pyruvate. cells solve these problems differently depending on whether they have access to oxygen. we must consider: 1 aerobic conditions (abundant oxygen) 2 anaerobic conditions (limiting oxygen)

Question 32

in the presence of oxygen

Answer 32

pyruvate is transported from the cytosol to the mitochondrial matrix by the pyruvate translocase. in the mitochondrial matrix, pyruvate is converted into the acetyl-CoA by pyruvate dehydrogenase. this process is called pyruvate decarboxylation and is the entry to the TCA cycle. Pyruvate + CoA + NADH+ -----> acetyl-CoA + CO2 + NADH+ H+ the pyruvate dehydrogenase complex (PDH) is a large, highly integrated complex of three kinds of enzymes,

Question 33

E1 E2 E3

Answer 33

associate with each other to form the pyruvate dehydrogenase complex.

Question 34

pyruvate dehydrogenase is a multienzyme complex

Answer 34

multienzyme complexes are groups of noncovalently associated enzymes that catalyse two or more sequential steps in a metabolic pathway. advantages: 1. enzymatic rates are limited by a frequency with which enzymes collide with their substrates. when a series of reactions occur within a multienzyme complex, the distance that substrates must diffuse between active sites is minimised. 2 the channeling of metabolic intermediates between successive enzymes in a metabolic pathway reduces the opportunity for these intermediates to react with other molecules. 3 the reactions can be co-ordinately controlled pyruvate dehydrogenase complex deficiency causes diseases: pyruvate dehydrogenase complex deficiency caused by mutations in genes coding for the E1 subunit is a type of metbaolic disease. the signs appear in infancy and include brain abnormalities, low weight, high levels of lactate, extreme tiredness, rapid breathing and other symptoms. people often pass away in the first years of life.

Question 35

regeneration of NAD+ FROM NADH the anaerobic pathway: Fermentation

Answer 35

louis Pasteur originally defined fermentation as respiration without air. fermentation is a process which extracts energy as ATP but does not consume oxygen or change the concentrations of NAD+ or NADH. In yeasts, regeneration of NAD+ from NADH in the absence of oxygen is coupled to formation of ethanol and CO2 ( the anaerobic or fermentation pathway in yeast).

Question 36

in skeletal muscle during exercise

Answer 36

regeneration of NAD+ from NADH is coupled to the anaerobic fermentation of lactate.

Question 37

The primary role of enzymes

Answer 37

is to enhance rates of reactions, so that they are compatible with the needs of an organism.

Question 38

What is Enzyme Kinetics?

Answer 38

The rate at which an enzyme works.

Question 39

Why Study Enzyme Kinetics?

Answer 39

* to understand enzyme catalytic power * when and why it works best * factors that affect activity (inhibitors) * these are important for biology * are of economic importance for industry* are central to pharmaceutical design

Question 40

Enzyme Reaction Velocity is AffectedBy

Answer 40

Substrate Concentration

Question 41

The Michaelis-Menten Constant - KM

Answer 41

KM the substrate concentration at which the enzyme reaction velocity is half its maximal rate

Question 42

Enzymes have Different KM Values*

Answer 42

enzyme lysozyme has the substrate Hexa-N-acetylglucosamine. a KM of 6 and turnover/sec of 0.5 enzyme Penicillinase has the substrate Benzylpenicillin a KM of 50 and aturn over/sec of 2000 enzyme carbonic anhydrase has the substrate CO2 a KM of 8000 and a turnover/sec of 600,000 turnover represents the number of molecules of substrate converted to a product per second and is directly related to Vmax

Question 43

What is the Meaning of KM?

Answer 43

1) The substrate concentration required for significant catalysis to occur 2) A value related to the rate constants of the steps of catalysis

Question 44

When does KM allow estimation of binding strength?

Answer 44

Km= k-1+ k2/k1 If the rate k-1 >> the rate of k2 ES dissociates rapidly to E and S and product (P) is not formed. - low KM indicates strong binding (greater affinity)- high KM indicates weak binding (lesser affinity). The value of KM is inversely related to the affinity of the enzyme for its substrate. V0 = Vmax [S]/ ([S] + KM) Case 1: when [S] is very low and smaller than KM, V0 is proportional to [S] Case 2: when [S] is very large and greater than KM, V0 is equal to Vmax

Question 45

Problems with the Michaelis-Menten Equation:

Answer 45

Vmax is approached asymptotically, which is difficult to measure.As KM is Vmax/2, KM is difficult to measure.

Question 46

Perfect Kinetics in Enzymes:

Answer 46

A kinetically perfect enzyme is one for which: - its catalytic rate is limited only by the rate at which it encounters substrate - in these cases, catalysis rates reflect those of diffusion. Enzymes such as superoxide dismutase, acetylcholine esterase, triosephosphate isomerase have achieved ‘kinetic perfection’. Why is acetylcholine esterase no surprise? If we can measure the efficiency of enzymes, we can engineer them to become more efficient

Question 47

Competitive Inhibition

Answer 47

Inhibitor molecules bind the active site of enzyme. Inhibitor competes with binding of substrate, thus reducing rate of catalysis. Degree of inhibition depends on concentration of substrate and inhibitor.

Question 48

Non-Competitive Inhibition

Answer 48

Inhibitor molecules bind at another site of the enzyme. This alters the structure of the active site, affecting the capacity of the enzyme to convert substrate to product (affects transition state?) It does not compete with binding of substrate.

Question 49

Allosteric Enzymes - inhibition

Answer 49

Allosteric enzymes contain regulatory and active sites - Small molecules can bind to regulatory sites. These cause conformational changes that reduce catalysis at active site. - Allosteric enzymes exist in two conformations. Some BUT NOT ALL enzymes show feedback inhibition, where the products formed are their own inhibitors. Where feedback inhibition is found, the inhibition is often to the first enzyme in a pathway. Allosteric Enzymes Can Show Cooperativity Allosteric enzymes also have multiple active sites that display cooperativity. Activity at one active site increases the activity at others Allosteric Enzymes do not show Michaelis-Menten Kinetics. Allosteric enzymes show sigmoidal kinetics (resembles an S).

Question 50

Allosteric Feedback Inhibition of ATCase

Answer 50

Aspartate transcarbomoylase (ATCase) catalyses the first step in pyrimidine biosynthesis. ATCase is inhibited by cytidine triphosphate, the final product in the pathway. ATCase has separate catalytic (C) and regulatory (R) subunits: they form the protein complex: C6R6

Question 51

Zymogens – inactive precursors of enzymes

Answer 51

Many enzymes are active after folding to their 3-D shape. Other enzymes are made as inactive precursors, or zymogens. A once-off cleavage event will activate the enzyme

Question 52

Enantiomers

Answer 52

are mirror images of each other

Question 53

Two sugars that differ only in the configuration around one carbon atom are called

Answer 53

epimers

Question 54

Anomers

Answer 54

are epimers which differ in the position of H and OH groups around the C1 atom in the cyclic forms of the monosaccharides. In solution, glucose predominantly exists as α and β anomers and each molecule of glucose continuously converts from one form to another via mutarotation

Question 55

Amylose

Answer 55

s a linear polymer of glucose glucose units are linked in a linear way with α(1→4) glycosidic bonds; * water insoluble, non-sweet * hydrolyzed slowly * dense Amylopectin is a branched polymer of glucose formed by plants store energy -contains 2,000 to 200,000 glucose units! - not soluble in water. - Glucose units are linked in a linear way with α(1→4) glycosidic bonds. Branching takes place with α(1→6) bonds occurring every 24 to 30 glucose units. Linear amylose and branched amylopectin form starch Depending on the plant, starch generally contains 20 to 25% amylose and 75 to 80% amylopectin (higher in medium-grain rice till 100% in waxy rice). Plants store starch within specialized organelles called amyloplasts. Plants also form polymers of glucose, cellulose, providing structural support. Cellulose * is formed by glucose units linked in a linear way via β(1→4) glycosidic bonds * has no taste * cannot be digested by animals * can be broken down by some bacteria, fungi * component of dietary fiber

Question 56

Glucose and glycogen

Answer 56

is the major form of sugar transported to different parts of the body in animals Glycogen is the polymer of glucose used by animal cells to store energy * is similar to amylopectin * formed when excess glucose is eaten * branching occurs every 8 to 12 glucose units * is stored in the liver and muscle tissue as an "instant" source of energy

Question 57

Polysacharides

Answer 57

1 Heteropolysaccharides * contain two or more different monosaccharide units; * most naturally occurring contain only two different ones and are closely associated with lipid or protein; * examples: pectin, lignin, glycoproteins, glycolipids, and mucopolysaccharides 2 Homopolysacharides * formed by at least six identical monosaccharides; * have a well-defined chemical structure * examples: glycogen, starch, cellulose

Question 58

thioredoxin (below), all polar amino acid side chains are shown as ball-and-stick models. The spatial arrangement of these amino acids illustrates a common feature of

Answer 58

globular proteins. thioredoxin is an antioxidant protein

Question 59

The general structure of an amino acid includes:

Answer 59

carboxyl group

Question 60

You are given an amino acid in solution at pH 13. What are the most likely ionisation states of its amino and carboxyl groups?

Answer 60

NH2 and COO-

Question 61

An enzyme substrate:

Answer 61

must interact with the active site for catalysis to take place

Question 62

The difference in free energy between the substrates and products of a reaction:

Answer 62

indicates whether a reaction is likely to occur spontaneously. If the difference in the free energy of substrates and products of a reaction is negative (i.e. the free energy of products is lower than that of substrates), then the reaction is favourable and can occur 'spontaneously'. However, recall that without an enzyme present, even favourable/spontaneous reactions may take a very long time to occur!

Question 63

'Enzyme kinetics’ is the study of:

Answer 63

the catalytic properties of enzymes.

Question 64

The Michaelis-Menten constant of an enzyme:

Answer 64

is high when an enzyme-substrate complex dissociates easily to form free enzyme and substrate. The Michaelis-Menten constant, KM, of a reaction can estimate the dissociation constant of the enzyme-substrate complex. A high KM indicates weak binding of the enzyme to its substrate, whereas a low KM indicates strong binding of the the enzyme to its substrate.

Question 65

When studying an enzyme-catalysed reaction, you add equal concentrations of enzyme, substrate, a competitive inhibitor, and a non-competitive inhibitor. In theory, what do you think will happen?

Answer 65

The non-competitive inhibitor will inhibit the reaction to a greater extent. In theory, the non-competitive inhibitor will have a greater effect on slowing the rate of the reaction because it binds to a site on the enzyme that is separate from the active site, whereas the competitive inhibitor will need to compete with the substrate in order to exert its inhibitory effects on the enzyme.

Question 66

The ‘cooperativity’ exhibited by some allosteric enzymes is a useful property for:

Answer 66

regulating biochemical pathways. This is true because allosteric enzymes often have multiple active sites and the property of cooperativity means that the activity at one active site can readily increase or decrease the activity at other active sites of the same enzyme. This is useful for the measured control of biochemical pathways that involve regulatory enzymes with multiple active sites because cooperativity helps the enzyme to become more (or less) effective more quickly.

Question 67

C4H8O4 is:

Answer 67

a tetrose monosaccharide

Question 68

Which one of the following carbohydrates is less related to the others? Question 12Select one: a. Cellulose b. Amylopectin c. Glucose d. Glycogen e. Amylose

Answer 68

Glucose is a 6-carbon monosaccharide whereas all other molecules listed here are large branched or non-branched polymers of many glucose units (i.e. polysaccharides).

Question 69

Glycolysis

Answer 69

utilises and generates ATP.

Question 70

Why is glycolysis important?

Answer 70

Primary means of energy production in cells with no mitochondria There are several reasons that glycolysis is a very important pathway. One of them is that glycolysis provides essential energy to cells that do not have mitochondria, such as red blood cells. (Recall that mitochondria are important for producing large quantities of energy via oxidative phosphorylation). Without glycolysis, such cells would not survive.

Question 71

When energy levels in the cell are low:

Answer 71

AMP and ADP levels activate phosphofructokinase. When ATP levels are low, higher AMP and ADP levels in contrast can act to allosterically activate phosphofructokinase (the key regulatory enzyme of glycolysis), to ultimately enhance the rate of glycolysis so that more energy can be produced via catabolism.

Question 72

When lactate is generated from pyruvate in skeletal muscle cells undergoing intense exercise:

Answer 72

it can be converted back to glucose via the liver. As part of the Cori Cycle, skeletal muscle cells can release lactate into the bloodstream which can then be recycled back to glucose via gluconeogenesis in the liver.

Question 73

the conversion of one molecule of fructose 1,6 biphosphate into 2 molecules of pyruvate result in the net synthesis of

Answer 73

2 NADH and 4 ATP

Question 74

the conversion of 1 molecule of glucose to 2 lactate

Answer 74

results in net synthesis of 2 NADH AND 2 ATP

Question 75

the 10 reactions of glycolysis take splace in the

Answer 75

cytoplasm

Question 76

in stage 1 of glycolysis two atp molecules

Answer 76

are consumed per molecule of glucose

Question 77

glyceraldehyde 3 phosphate is

Answer 77

oxidised and phosphorylated to form 1,3-biphosphoglycerate

Question 78

phosphofructokinase

Answer 78

the most importaant control element in glycolysis is inhibited by ATP and citrate and it is activated by AMP and fructose 2,6-biophosphate

Question 79

how much atp is generated from glucose 6 phosphate

Answer 79

3

Question 80

how much atp is generated from digydroxyacetone phosphate in the process of glycolysis to lactate

Answer 80

2

Question 81

how much atp is generated from glyceradehyde 3 phhosphate in the process of glycolysis to lactate

Answer 81

2

Question 82

how much atp is generated from fructose in the process of glycolysis to lactate

Answer 82

2

Question 83

how much atp is generated from sucrose in the process of glycolysis to lactate

Answer 83

4