I. Energy
Calculate the energy value of food from enthalpy of combustion data
Use the equation: E = mcΔT.
E = Energy
M= Mass of the sample heated (usually water, NOT the product being burned).
C= Specific heat capacity (water = 4.18 J g-1 C-1)
ΔT = Change in temperature of water heated (in either Celcius or Kelvin)
II. Proteins
Draw the general formula of 2-amino acids
*Starting note: IB calls amino acids “2-amino acids”. No clue why, but it’s the same kind you should’ve learned about in bio.
The general form of 2-amino acids: R—CH—COOH
NH2
Describe the characteristic properties of 2-amino acids
Properties of 2-amino acids: ZIMB
Zwitterion- Forms ionic varieties with either:
· A positive charge on the amine end (NH2 à NH3+)
· a negative charge on the carboxyl end (COOH à COO-
o [Note: The double bonded C=O becomes single bonded C-O with two delocalized electron])
Isoelectric Point- pH at which the molecule as a net charge of 0. In solutions too acidic, the charge is +. In solutions too basic, the charge is - Melting Point- High MP around 200-300oC
Buffer Action- Neutralizes pH by accepting H+ in acidic solutions and donating H+ in basic solutions.
Describe the condensation reaction of 2-amino acids to form polypeptides
Condensation Reaction- Reaction between two amino acids, formed when an amine group of one AA and carboxyl group of another AA form a covalent bond, releasing
H2O. (This is the esterification process we learned about in organic chem.)
Dipeptide- molecule composed of two amino acids
Polypeptide- chain of amino acids that, when folded, form proteins
Describe and explain the primary, secondary (a-helix and b-pleated sheets), tertiary and quaternary structure of proteins
Protein Folding- folding of the polypeptide chain to form protein structure. Subdivided into primary, secondary, tertiary, and quaternary structures
Primary- One-dimensional “chain” of amino acids
Bonds involved- Peptide (Covalent)
Secondary- Two-dimensional folding into alpha-helixes and beta-pleated sheets
Alpha-helix- helical structure held together by hydrogen bonds
Beta-pleated sheets- sheets formed by polypeptides folding parallel to one another, held together by hydrogen bonds
Tertiary- Three-dimensional fold of protein chain composed of many different IMF’s:
Covalent bonding- Between side chains
i.e. two –SH groups in cysteine form a disulfide bridge
Hydrogen bonding- between polar groups on the side chain
Salt bridges- ionic bonds formed between –NH2 and –COOH groups
Hydrophobic interactions- non-polar amino acids will migrate towards the center
of the protein away from the polar water molecules
Quarternary- multiple polypeptide chains held together by weak IMFs (like van der waal)
Explain how proteins can be analyzed by chromatography and electrophoresis
Chromatography:
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Electrophoresis:
· R groups of amino acids have different isoelectric points (pH where charge is 0)
· Similar size molecules can be separated using charge in individual AA
· Proteins are placed in a magnetic field, and the positive R-groups are attracted to the negative pole of the magnet, and negative R groups will be attracted to the positive pole
· Movement stops when charge reaches zero (isoelectric point is reached)
· Also works with a pH gradient
List the major functions of proteins in the body
Major functions of proteins in the body- SHIT, Enzyme energy!
Structural- give rigidity or help contract
Hormones- regulate body activities by sending messages between cells
Immunoproteins- aid the immune system (i.e. antibodies)
Transport- move specific molecules around cell
Enzyme- catalyze reactions by lowering activation energy
Energy!- can be metabolized when carbs and fats are low
III. Carbohydrates
Describe the structural features of monosaccharides-
· Carbonyl group (C=O)
· At least two –OH groups
· Empirical formula CH2O
Draw the straight-chain and ring structural formulas of glucose and fructose
Straight chain- carbohydrate structure when dry (powder)
Ring- carbohydrate structure when wet (in aqueous solution)
Drawings of Monsaccharides (KNOW STRAIGHT CHAIN + RING)
Alpha- When the –OH on the rightmost carbon (C1 in glucose, C2 in fructose) is below the plane of the ring
Beta- When the-OH on the rightmost carbon is in line with/above the plane of the ring
Describe the condensation of monosaccharides to form disaccharides and polysaccharides
Condensation- combining 2+ monosaccharides to form disaccharides/polysaccharides. Usually involves the loss of a water molecule
Disaccharides- composed of 2 monosaccharides bonded together.
Examples: Lactose, maltose, sucrose
Polysaccharides- composed of a long chain of monosaccharides
Examples: starch (a-glucose), glycogen (a-glucose), and cellulose (b-glucose)
List the major functions of carbohydrates in the human body
Use of Carbs in the body: PEED
· Precursor to other important biological molecules
· Energy source (glucose)
· Energy storage (glycogen)
· Dietary fibers
Compare the structural properties of starch and cellulose, and explain why humans can digest starch but not cellulose
Structural properties of starch and cellulose
Similarities:
· Polymers of glucose
· 1-4 glycosidic bond is between the 1-carbon and 4-carbon of neighboring glucose molecules
o (amylopectin has additional 1,6 glycocidic bonds)
Differences:
· Starch composed of a-glucose
· Cellulose composed of b-glucose
Why can starch be digested but cellulose not?
Starch is composed of a-glucose, bringing Oxygen atom opposite the CH2OH group (where human enzymes can digest them). Cellulose is composed of b-glucose, bringing Oxygen atom alongside CH2OH where it can only be digested by cellulase (found in fungi, bacteria, and protests)
State what is meant by the term dietary fiber
Dietary fiber- Plant material that cannot be broken down by digestive enzymes, but be broken down by microorganisms in the human gut.
Examples: Cellulose, hemicellulose, lignin, and pectin
Describe the importance of a high fiber diet
Dietary fiber aids in digestion
High fiber diets prevent conditions like Diverticulosis, Irritable bowel syndrome, Diabetes, Constipation, Hemorrhoids , Obesity, and Chron’s disease. (DID CHOC)
IV. Lipids
Compare the composition of the three types of lipids found in the human body
Outline the difference between HDL and LDL cholesterol and outline its importance
HDL cholesterol (Good cholesterol)- High Density Lipoprotein. Since the proportion of protein to lipid is higher, then lipid can move more easily around the bloodstream. It is easier to transport and does not clot as much
LDL cholesterol (Bad cholesterol)- Low Density Lipoprotein. Since there proportion of protein to lipid is lower, the lipids can not move as easily. It is harder to transport, and tends to clot to arterial walls à atherosclerosis (increase risk of heart attack/stroke)
Describe the difference in structure between saturated and unsaturated fatty acids
Type | Molec. Structure | Phys. Structure | Van der Waals | Melting Points |
Saturated | All single bonds C-C | Straight chains | Higher (more atoms à stronger interaction | High (solid at room temp) |
Unsaturated | One C=C (mono-) or several C=C (poly-) | Double-bonds à kinked chain | Lower (fewer atoms à less interaction) | Low (liquid at room temp) |
The longer a fat is, the more van der walls forces and the higher the melting point
Compare the structures of the two essential fatty acids: linoleic (omega-6 fatty acid) and linolenic (omega-3 fatty acid) and state their importance
Omega-_ fatty acids- the number represents how many carbons from the end of the chain that the first double-bond is located. Example, omega-3 has a double bond 3 carbons away from the end.
Essential Fatty Acids
Acid | First C=C | Double Bonds | Found | Symptoms of Lack |
Linoleic C18H32O2 | Omega-6 | C-6, C-9 | Oils, seeds | Dry hair, hair loss, Poor-wound healing |
Linolenic C18 H30O2 | Omega-3 | C-3, C-6, C-9 | Leafy greens, fish | Vision/nerve problems, depression, circulatory problems, arthritis, |
Define the term “iodine number” and calculate the number of C=C double bonds in an unsaturated fat/oil using addition reactions
Iodine Number- mass of I2 needed to satisfy a particular reaction. Used to identify the number of C=C in a particular unsaturated fat.
Using an Iodine Number
Example: Linoleic acid has the formula C18H22O2. What is the iodine number of this acid
Steps
1. Convert so that it is in the molecular formula
C17H31COOH à C18H32O2
2. Determine the number of double bonds
A saturated fat with 18 carbons should have 36 hydrogens. This compound has 32. Therefore, the 4 missing hydrogens suggests 2 double bond
3. Set up a molar ratio
For every 1 mole of C17H31COOH, you’ll need 4 moles of iodine. However, since iodine only comes in the I2 form…
The ratio of C17H31COOH to I2 will be 1:2
4. Set up a mass ratio
Molar mass C17H31COOH = 280 g mol-1
Molar mass I2 = 254 g mol-1 x 2 mol
Ratio will be 280 g to 508 g
5. Find the amount of Iodine needed to react with 100g of the fatty acid
280 g = X .
254 g mol-1 508 g mol-1
0.393700mol = X .
508 g mol-1
ANSWER: 181 grams. The iodine number is 181
Describe the condensation of glycerol and three fatty acid molecules to make a triglyceride
Glycerol- 3-carbon chain with an –OH group on each C
Fatty Acid- carboxylic acid
Ester bonds- Form from the 3 carboxylic acids on the fatty acid chains and alcohol groups in the glycerol, releasing 3H2O molecules.
Explain the enzyme-catalyzed hydrolysis of triglycerides during digestion
Lipases (fat-digesting enzymes) catalyze the reaction:
Triglyceride + 3 H2O à Glycerol + 3 Fatty Acids
Explain the higher energy level of fats as compared to carbohydrates
Fat metabolism occurs more slowly because a greater degree of oxidation is required to convert them into CO2 and H2O (carbohydrates already have one oxygen atom per carbon atom)
The lack of C-O bonds means that more energy will be released per molecule
Explain the important roles of lipids in the body and the negative effects they can have on health
Benefits: HIS STOP
· Hormones
· Insulation
· Store energy
· STructural component of membranes
· Omega-3 fats reduce the risk of heart disease
· Poly-unsaturated fats may lower level of LDL cholesterol
Problems: HO
· Higher risk of heard disease from LDL cholesterol and trnas fats
· Obesity
V. Nutrients
Outline the difference between macro- and micro-nutrients
Micronutrients- Substances required in very small quantities. Make up <0.005% body weight. Include vitamins and trace minerals
Macronutrients- Substances required in much larger quantities. Make up >0.005% body weight. Include proteins, fats, carbs, and minerals
Compare the structures of retinal (Vitamin A), calciferol (Vitamin D), and ascorbic acid (Vitamin C)
Retinol (Vit A)- Relitively heavy oil that is insoluble in water, but highly soluble in glycerol due to its 1 hydroxyl group and non-polar alkene/methyl groups
Calciferol (Vit D)- Heavy powder that is insoluble in water and highly soluble in glycerol (due to its 1 hydroxy group and alkane/alkene/methyl groups
Ascorbic Acid (Vit C)- Relatively light solid that is soluble in water (due to its 4 hydroxyl groups) and insoluble in glycerol
Deduce whether a vitamin is water- or fat-soluble from its structure
Water soluble vitamins- are defined by a high proportion of polar molecules such as –OH and N (Examples: Vitamins B1 and C)
Fat soluble vitamins- are defined by a lack of polar molecules and an abundance of long hydrocarbon chains (Examples: Vitamins D and K)
Deduce the causes and effects of nutrient deficiencies and suggest solutions
Malnutrition- lack of proper or adequate nutrients in stable food source
MALNUTRITION TABLE OF DOOM
Nutrient | Deficiancy | Symptoms |
Vitamin A | Night blindness | Lack of vision Decreased pathogen resistance |
Vitamin B3 | Beriberi | Swelling in limbs Weight loss |
Vitamin C | Scurvy | Tooth loss Skin spots |
Vitamin D | Rickets | Bending of the bones |
Iodine | Goiter | Inflammation of thyroid |
Protein | Kwashikoror | Expanded belly Anorexia Lack of energy |
Iron | Anemia |
|
Solutions to fight malnutrition:
· Providing food rations high in vitamins and minerals
· Fortifying food with nutrients
· Genetic modification of food
· Nutritional supplements
· Selenium supplements
VI. Hormones
Outline the production and function of hormones in the body
Hormones- chemicals that control and regulate standard body functions
Hormone production- excreted by the endocrine glands in response to stimuli and released directly into bloodstream
Compare the structures of cholesterol and the sex hormones
Cholesterol- steroid with 3 6-C rings and one 5-C ring and a long hydrocarbon chain
Sex Hormones- steroid with 3 6-C rings and one 5-C ring (like cholesterol), but lack the characteristic long chains present in cholesterol because sex hormones are derived from progesterone.
Describe the mode of action of oral contreceptives
Oral contraceptives- birth control taken in the form of a swallowed pill
Method for oral contraceptive function- pill releases estrogen and progestin which prevent ovulation [so the egg cannot fertilize] and developing of the endometrium (uterus) lining [so any fertilized eggs cannot bind]
Method of mini-pill- progestin only pills that prevent ovulation , prevent endometrial thickening, and maintain cervical mucus.
Outline the use and abuse of steroids
Anabolic steroids- chemicals that speed tissue regeneration, which can be used to increase rate of muscle growth
Use of steroids: help with wasting illnesses, regain muscle tissue
Abuse of steroids: help increase sports performance, blood/liver/heart problems, development of opposite sex characteristics (malesàfemales, femalesàmales), impotence, bladder problems
VII. Enzymes
Describe the characteristics of biological catalysts (enzymes)
Enzymes- Proteins acting as catalysts by increasing the rate of chemical reaction without undergoing any permanent chemical change
Enzymes increase the rate by lowering the activation energy of the products
Activity depends on tertiary and quaternary structure
Differences between inorganic and organic compounds (SSS)
Specificity- Inorganic are less specific (work under high temp or any PH)
- Organic are more specific (require certain temp/pH)
Structure - Inorganic are structured from metals
- Organic are structured from proteins
State - Inorganic can be any state (solid/liquid/gas) and function the same
- Organic only work in aqueous solution
Describe the relationship between substrate concentration and enzyme activity
- Low concentrations = linear relationship
o Enzymes have more active sites than substrate molecules, so there will be a linear increase in rate
- As concentration increases, the rate will slow down because more active sites are being used up
o Eventually, the point of saturation is reached when the rate will not increase any further because all active sites are in use and none are opening up
Determine Vmax and the value of Km by graphical means and explain the significance
- Vmax- Maximum rate of reaction at the point of saturation (horizontal asymptote)
- Only way to increase Vmax is to increase enzyme concentration.
- Km- Substrate concentration when rate is at 1/2 Vmax (Michaelis-menten constant)
o Representation of how readily substrate-binding occurs
Explain the mechanism of enzyme action, including enzyme substrate complex, active site, and induced fit model
Mechanism of Enzyme action
- Enzyme + Substrate à Enzyme Substrate Complex à Enzyme + Product
- Substrate binds to the active site (location of binding)
- Active site is specific to certain molecules
- Rarely does substrate fit perfectly to site
o Induced fit theory makes up for this factor
o States: Enzyme will change its shape slightly to encompass the particle
o Inexact fit puts stress on certain bonds of the substrate
§ This lowers the Ea necessary to break the bond
- Enzyme name represents substrate (ex. Protease breaks down proteins)
Compare competitive and non-competitive inhibition
Competitive Inhibitors- Similar in shape to substrate. Binds to active site to prevent substrate)
- Vmax stays the same, but Km↑ (showing that substrate cannot bind as well in the presence of this inhibitor)
- When Concentration of substrate ↑, the effect of inhibitors ↓
Noncompetitive Inhibitors- Bind to enzyme at location different from active site.
· Vmax ↓ but Km stays the same
· When Concentration of substrate ↑, the effect of inhibitors does not change
State and explain the effects of heavy-metal ions, temperature changes, and pH changes on enzyme activity
Heavy metals- Act as non-competitive inhibitors by reacting with sulfhydral groups, turning –SH à -SMetal, which interferes with the tertiary structure. Also interferes with the side chains. (This is why lead and mercury poisoning are so dangerous)
Temperature- Increasing the temperature will ↑ the rate of enzyme catalyzed reactions as more reactants reach activation energy. Temperatures too high à breaking of H-bonds, altering the structure of the protein and denaturing it.
pH- Change in pH interferes with acidic and basic molecules on the side chain, altering the tertiary structure. Each enzyme has an optimal pH, at which rate is maximum
VIII. Nucleic Acids
Describe the structure of the nucleotides and their condensation polymers (nucleic acids or polynucleotides)
Structure of nucleotides: sugar, phosphate, nitrogenous base
Phospate- Negatively-charged molecule (either phosphate ion or phosphoric acid).
Sugar- 5-carbon sugar (ribose in RNA, deoxyribose in DNA)
Base- Heterocyclic shape with either one ring (pyrimidines- cytosine, thymine, and uracil) or two (purines- adenine and guanine)
Nucleotides compose nucleic acids through covalent phosphodiester bonds between the 5’carbon bound Phosphate and the 3’ carbon
Base sequence- code for traits
Distinguish between the structures of DNA and RNA
Chemical differences between RNA and DNA
- Sugars- RNA uses ribose. DNA uses deoxyribose
- Bases- RNA uses uracil, DNA uses thymine
- Strands- RNA is single, DNA is double
Explain the double helical structure of DNA
Nucleotide bonding- Bases H-bond to one another, forming the double strand and stabilizing the large molecule
Note: Other forces will have a smaller role in stabilizing (Dip-dip, hydrophobic, VDW)
Complementary base pairing- Electronegative atoms on each base will only correspond to one other base. For example, adenine will bind to thymine and cytosine to guanine.
Benefits of the structure- Weak H-bonds separate easily, which allows the sequence to be copied. Strong enough to hold DNA together
Helix- Result of repelling negative phosphate groups
Strands of DNA- One codes for necessary information (sense strand). One is inactive complement (antisense strand).
Describe the role of DNA as the repository for genetic information, and explain its role in protein synthesis
Role of DNA as a repository of information- Base sequence is transcribed à mRNA. mRNA polymerase binds to promoter region, adds ribonucleotides to strand, and lets go once a terminator is reached (He’ll be back…). mRNA is translated à protein as each 3-base sequence codes for specific amino acid, which is added to the chain.
Outline the steps involved in DNA profiling and state its use.
· DNA extracted from blood cells and cut into fragments by restriction enzymes
· PCR replicates copies of the fragments thousands of times
· Segments of DNA separated into bands by electrophoresis
o Negatively charged phosphate groups attracted to positive pole
o Larger molecules move more slowly through the sieve, while smaller molecules move farther
· Banding patterns transferred onto a nylon membrane
· Radioactive DNA probe finds specific sequence
· Film placed next to x ray to display fingerprint
· Banding patterns are unique to an individual, as they are passed down by parents
PURPOSE- Paternity suits and forensic investigations
IX. Respiration
Compare aerobic and anaerobic respiration of glucose in terms of oxidation/reduction and energy released
Respiration- Breakdown of controlled breakdown of energy-rich molecules into usable form
Respiration (for chemists)
What is reduced? Glucose à Pyruvic acid à Pyruvate à Lactate or Ethanol (Anaerobic only), NADH à NAD+
What is oxidized? NAD+ à NADH, O2 à CO2 + H2O
Energy differences b/w Aerobic + Anaerobic- Aerobic releases 40% of stored energy. Anaerobic releases 2% of stored energy.
Outline the role of copper ions in electron transport and iron ions in oxygen transport
Role of Iron in biological molecules
Hemoglobin composed of 4 polypeptides, each with its own HEME group (complex ion) containing Fe2+. Hydrophobic environment allows oxygen to bind to Fe2+ without oxidizing it, in a process called oxygenation. 4 HEME groups means each Hemoglobin can bind to 4 O2.
Danger of Carbon-monoxide- CO binds more tightly to Fe2+, which causes a lack of oxygen and asphyxiation if CO is not displaced.
Role of Copper in biological molecules
Cytochromes are proteins containing iron and copper that are reduced by electrons (in the electron chain) and re-oxidized as they pass along the electron transport chain (HEME groups are the receptors of the electrons). In the last cytochrome (cytochrome oxidase), copper receives the electron and binds them to O2, forming water.
Danger of Cyanide- CN- binds to the cytochrome oxidase and backs up the electron transport chain, slowing respiration
