IB Chemistry/Human Biochemistry

Option C: Human Biochemistry

C.1 Diet (2h)

In a calorimeter experiment, the food is completely combusted. The energy released is used to heat a sample of water. The heat energy can be calculated using this equation: H = m cp T H is the heat released (J) m is the mass of water heated (g) cp is the specific heat capacity of water (4.18 J g-1 K-1) T is the change in water temperature (oC or K)

The value calculated by this equation will typically need to be divided by the number of moles, or the number of grams of the food which has been burnt.

Example: 0.547 g of sucrose, C12H22O11, was completely combusted in a food calorimeter. The heat evolved was equivalent to increasing the temperature of 250 g of water from 22.22 °C to 30.81 °C.

Calculate the calorific value of sucrose (in kJ g-1 and kJ mol−1) given the specific heat capacity of water in Table 2 of the Data Booklet. Using the formula H = m cp T H is the heat released (J) m is the mass of water heated (250 g) cp is the specific heat capacity of water (4.18 J g-1 K-1) T is the change in wa ter temperature (8.59 oC or K) H = 8.98 kJ

This is due to combustion of 0.547 g of sucrose. Dividing by this mass gives us: 16.4 kJ g-1. 1 mole of sucrose has a mass of 342 g, so the molar enthalpy change is 5612 kJ mol-1. (342 g x 16.4 kJ g-1 or 8.98 kJ ÷ 1.60 x 10-3 mol.)

Questions

03NS3 C1. (a) State how genetically modified food differs from unmodified food. [1] (b) List two benefits and two concerns of using genetically modified crops. [4] C2 (d) 1.00 g of sucrose, C12H22O11, was completely combusted in a food calorimeter. The heat evolved was equivalent to increasing the temperature of 631 g of water from 18.36 °C to 24.58 °C.

Calculate the calorific value of sucrose (in kJ mol−1) given the specific heat capacity of water in Table 2 of the Data Booklet. [3]

05MS3 C1. (c) The energy content of a vegetable oil was determined using a calorimeter. A 5.00 g sample of the oil was completely combusted in a calorimeter containing 1 000 g of water at an initial temperature of 18.0 oC. On complete combustion of the oil, the temperature of the water rose to 65.3 oC.

Calculate the calorific value of the oil in kJ g−1 [4]

C.2 Proteins (3h)

2-aminopropanoic acid shows the essential structure shared by all 2-amino acids: H2N-CH(CH3)-COOH The –CH3 group can be changed for one of 19 other groups to make the 20 amino acids encoded in our genes.

The –NH2 group of one amino acid can join to the –COOH group of another. This condensation reaction forms a peptide bond and a water molecule.

To use either of these techniques the peptide bonds in the proteins must first be hydrolysed to release individual amino acids. Include the use of Rf values in paper chromatography. Given isoelectric points, students should be able to determine a suitable pH to achieve good separation in electrophoresis.

The amino acids of a protein can be identified if the protein is hydrolysed. Hydrolysis usually involves refluxing the protein with acid or alkali to give the separate amino acids. Chromatography separates the amino acids according to whether they prefer to dissolve in a solvent, or remain attached to a solid. Careful choice of solid and solvent allow amino acids to be separated as the solvent moves over the solid.

The ratio of the movement of the amino acid, to the movement of the solvent, is called the Rf value.

Example: Using paper as the solid phase and a 8:1:1 mix of ethanol:water:ammonia as the solvent, the Rf values of Aspartic acid, lysine and leucine were found to be 0.13, 0.35 and 0.84 respectively. Draw a chromatogram showing the expected pattern of spots for the three pure amino acids and a mixture of all three.

Electrophoresis uses an electric field to separate the amino acids. Positively-charged amino acids will tend towards the cathode and negatively charged amino acids will move towards the anode.

Each amino acid has an isoelectric point, given in table 20 of your data book. This is the pH which gives the amino acid an overall neutral charge. Above this pH, the amino acid is negative (deprotonated) and below this pH the amino acid is positive (protonated). For example, at pH 5.4:

Charge Structure Movement Asparagine

Cysteine

Glutamine

• Note how neutral amino acids form zwitterions.

The sequence of amino acids in a protein is called its primary structure. The sequence is given from the free amine end (N-terminal) to the free acid end (C-terminal). The backbone of the protein can form two regular patterns due to hydrogen bonding. These are called secondary structures:

A single chain can form an -helix. Hydrogen bonds form between the C=O of one amino acid and the H-N of the amino acid four residues further down the chain.

When two chains run alongside one another, the C=O groups of one chain and N-H groups of the other chain can be aligned and hydrogen bonds form. This forms a -sheet.

Tertiary structure is the folding of the protein into its final shape. The main driving force is for oily amino acids to be buried within the structure and polar amino acids to be exposed on the surface. Other forces involved are hydrogen bonds, salt bridges (ionic bonds between acidic and basic amino acids) and disulphide bridges (covalent bonds between cysteine residues).

A fictional peptide showing two disulphide bridges, a salt bridge, and inter-residue hydrogen bonding.

Some proteins may be formed from multiple polypeptide chains. The assembly of two or more chains to form one protein is called quaternary structure. Like tertiary structure, the bonding may involve hydrophobic, ionic, covalent and hydrogen bonding.

These are structure, biological catalysts (enzymes) and energy sources. Proteins are involved in holding biological structures together (e.g. collagen, keratin) Enzymes are biological catalysts which control metabolism. They are all proteins. Protein can be used as a source of energy but is generally only used if fat and carbohydrate is not available.

Questions

06MS3 C1. (a) (i) Deduce the structure of one of the dipeptides that can be formed when the two aminoacids below react together.[2]

(ii) State the name given to this type of reaction and identify the other product of the reaction. [2] (b) Describe how a mixture of aminoacids can be analysed using electrophoresis. [4] (c) (i) Explain what is meant by the primary structure of proteins. [1] (ii) Explain, with reference to hydrogen bonding, why the α-helix and β-sheet secondary structures of proteins are different. [2] (iii) Identify three types of interactions responsible for the tertiary structure of proteins. [2]

C.3 Carbohydrates (2.5h)

Monosaccharides contain a carbonyl group (C=O) and at least two -OH groups, and have the empirical formula CH2O. Monosaccharides are organic molecules with three or more carbon atoms in a straight chain. Each carbon has a single –OH group, except one which has a C=O group. Aldose sugars have an aldehyde group (i.e. C=O is on carbon number 1). Ketose sugars have their C=O in the middle of the chain. Fructose:

Glucose:

Note that glucose and galactose are identical except that one chiral centre is different: Galactose:

Having four chiral centres, glucose has seven isomers which differ in the orientation at one or two of the chiral centres. Galactose is just one these. Glucose has an enantiomer (L-glucose) which differs at all four chiral centres. You need to know the straight-chain formula of glucose.

Monosaccharides in solution form a ring structure involving the C=O group:

α-glucose straight-chain version of glucose β-glucose You need to know these formulae of glucose. Galactose undergoes a similar reaction:

The –OH group on carbon 1 of -glucose can link to the –OH group on carbon 6 of fructose. H2O is lost and a bridging –O- links the two molecules. This disaccharide is called sucrose. If the –OH group on carbon 1 of -galactose links to the –OH group on carbon 4 of glucose, a similar reaction occurs resulting in the disaccharide lactose.

A chain of -glucose molecules linked carbon 1 to carbon 4 is called amylose. It is part of starch and glycogen.

Amylose chains can be linked together by connecting the carbon-1 at one end to the carbon-6 of another chain. This branched polymer is called amylopectin and it is the second component of starch and glycogen.

These are energy sources, energy reserves (eg glycogen) and precursors for other biologically important molecules.

Glucose is broken down into water and carbon dioxide to yeld metabolic energy. Other carbohydrates are converted into glucose and metabolized in a similar manner. Starch and glycogen are used as stores of glucose in plants and animals respectively. Carbohydrates are used in other roles but are usually modified beforehand e.g. as glycoproteins, glycolipids, hormones.

Questions

03NS3 C2. (a) Draw the straight chain structure of glucose. [1] (b) The structure of α-glucose is shown below.

Outline the structural difference between α-glucose and β-glucose. [1] (c) Glucose molecules can condense to form starch which can exist in two forms, amylose and amylopectin. Describe the structural differences between the two forms. [2] 05NS3 C1. (a) The equilibria which exist in an aqueous solution of glucose is shown in the structures below.

(i) Identify the α and β forms of glucose [2] (ii) State, with a reason, whether or not the two ring forms of glucose are enantiomers. [1] (iii) In structure B identify, by stating the numbers, the carbon atoms which are not chiral. [1] (b) The structure of lactose, a disaccharide formed from glucose and galactose, is shown in the Data Booklet. Draw the ring structure of galactose and state whether it is an α or β isomer. [2] (c) State one major function of polysaccharides such as starch and glycogen. [1]

C.4 Lipids (2.5h)

Lipids generally do not dissolve in water. There are 3 different kinds of lipids in the human body:

- Triglycerides

- Phospholipids

- Steroids

Tricglycerides(Oils)

Fats and oils (collectively known as lipids) have a variety of chemical structures, but they are all insoluble in water.

The most common lipid structure is the triacylglyceride (TAG) which is an ester. Three long-chained acids form ester bonds to a single molecule of glycerol (propane-1,2,3-triol).

Saturated fatty acids have no C=C double bonds. Unsaturated fatty acids have one or more C=C double bonds. Polyunsaturates have more than one C=C.

Saturated fatty acids tend to have higher melting points. The intermolecular bonds must be greater. This is due to the more efficient van der Waals' forces that are possible when the flexible saturated chains lie alongside one another – the contact surface is maximised. In unsaturated fatty acids the double bonds reduce the flexibility of the molecules, and this prevents the molecules aligning so efficiently.

Margarine manufacturers use unsaturated vegetable oils and hydrogenate them until they become saturated enough to have the right semi-solid consistency at room temperature.

The number of C=C bonds can be determined from the number of moles of I2 which add to one mole of fat.

Like Br2 in the test for alkenes, I2 will add to the C=C double bonds in unsaturated fatty acids. Unlike Br2 it only reacts with C=C bonds.

The iodine number is the mass of iodine which reacts with 100 g of a lipid. For saturated fatty acids such as palmitic acid (C15H31COOH) the iodine number is zero. For unsaturated fatty acids like hexadecadienoic acid C15H27COOH the iodine number is calculated as follows:

The number of double bonds is calculated by comparing the fatty acid to the appropriate saturated fatty acid. For every two hydrogens missing, there must be one double bond. In this case, compare C15H27COOH with C15H31COOH: There are four missing hydrogens and therefore two C=C bonds.

Example

Calculate the molar mass of the fatty acid and the iodine: C15H27COOH 252 g mol-1 I2 254 g mol-1 Every C=C will react with one iodine, so: one mole of C15H27COOH will react with two moles of I2 252 g of C15H27COOH will react with 508 g of I2 100 g of C15H27COOH will react with 202 g of I2

Hydrolysing TAGs into glycerol and fatty acids is achieved by refluxing with sodium hydroxide. The fatty acid salts which result acts as soaps. The ionic salt end is water-soluble but the long alkane chain is oily. The molecule can therefore bridge between water and oily impurities. C.4.5 List the major functions of fats in the body.

These are energy sources, insulation and cell membranes.

Fats are used as a source of energy. They are more slowly broken down than carbohydrates so they are not as useful in an emergency as glycogen. On the other hand, fats do not accumulate water so they are more efficient energy stores, pound for pound, than glycogen. Fat is a good thermal insulator (blubber).

Phospholipids

Made of three structures:

-Glycerol(backbone) -2 fatty acid chains -phosphate chain.

It is amphipathic meaning the head, glycerol is polar but the chains are not.

Phosopholipids form bilayers.

The bilayers form cell membranes

Steroids

3 Cyclohexane rings and 1 cyclopentane ring bonded together

Examples of steroids are: Vitamins, hormones and cholesterol.

Questions

03NS3 C3. Linoleic acid (C17H31COOH, Mr = 280) and stearic acid (C17H35COOH, Mr = 284) both contain eighteen carbon atoms and have similar molar masses. (a) Explain why the melting point of linoleic acid is considerably lower than the melting point of stearic acid. [3] (b) Determine the maximum mass of iodine (I2 Mr = 254) that can add to (i) 100 g of stearic acid: [1] (ii) 100 g of linoleic acid: [2] (c) (i) Draw the simplified structural formula of a fat containing one stearic acid and two linoleic acid residues. [1] (ii) Give the formulas of the products formed when this fat is hydrolyzed by sodium hydroxide. [1]

05MS3 C1. (a) A brand of vegetable fat consists of 88 % unsaturated fats and 12 % saturated fats. State the major structural difference between unsaturated and saturated fats. [1] (b) Linoleic acid, CH3(CH2)4CH=CHCH2CH=CH(CH2)7COOH, and palmitic acid, CH3(CH2)14COOH, are components of vegetable fat. Explain why palmitic acid has the higher melting point. [3] (d) List two functions of fats in the human body. [2]

06MS3 C2. (a) The formula of oleic acid is CH3(CH2)7CH=CH(CH3)7COOH. Using R to represent CH3(CH2)7CH=CH(CH3)7, show the structure of the triglyceride formed from this acid. [1] (b) Explain why some triglycerides that are liquid at room temperature become solids when they are completely hydrogenated. [3]

05NS3 C2. (a) The general formula for saturated fatty acids is CnH2nO2. The molecular formula of linoleic acid is C18H32O2. (i) Determine the number of carbon to carbon double bonds in linoleic acid. [1] (ii) Iodine number is defined as the number of grams of iodine that adds to 100 g of a fat or an oil in an addition reaction. Determine the iodine number of linoleic acid. [2] (b) (i) State one structural similarity between fats and oils. [1] (ii) Explain, by referring to their structures, why fats are solid at room temperature, but oils are liquid. [3] (c) When a fat is reacted with aqueous sodium hydroxide, soap and one other product are formed. (i) State the condition required for this reaction. [1] (ii) Draw the structural formula of the other product. [1]

B.5 Micronutrients and Macronutrients

B.5.1 - Outline the difference between micronutrients and macronutrients

• Micronutrients are substances required in very small amounts (mg or ug) and that mainly function as a co-factor of enzymes (<0.005% body weight). Examples include vitamins and trace minerals (Fe, Cu, F, Zn, I, Se, Mn, Mo, Cr, Co and B).

  Macronutrients are chemical substances that are required in relatively large amount (>.005% body weight). Examples include proteins, fats, carbohydrates, and minerals (Na, Mg, K, Ca, P, S and Cl)

B.5.2 - Compare the structures of retinol (vitamin A), calciferol (vitamin D) and ascorbic acid (vitamin C)

• Vitamin C - Has hydroxyl groups (4), Ester within in the ring, polar because hydroxyl groups make it easy to dissolve in water.

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• Vitamin A - Only 1 hydroxyl group, long polyene chain. Has small polar parts, but overall non-polar (therefore doesn't dissolve in water, fat soluable)

  Vitamin D - Only one hydroxyl group. Some alkene groups, small parts are polar areas but overall no big difference. Overall non-polar due to the carbon chains and cyclocarbons.

B.5.3 - Deduce whether a vitamin is water or fat soluable from its structure Water soluble molecules have a high proportion of electronegative atoms in their structures. Fat soluble vitamins are relatively non-polar and consist mainly of hydrocarbon.

• Water-soluable are vitamins B and C

  Fat-soluable are vitamins A, D, E, K

B.5.4 - Discuss the causes and effects of nutrient deficiencies in different countires and suggest solutions. Micronutrient defiencies include:

• Iron - anemia

  Iodine - goitre

  Retinol (vitamin A) - Xerophythalmia, night blindness

  Niacin (Vitamin B3) - Pellagra

  Thiamin (Vitamin B1) - Beriberi

  Ascorbic Acid (Vitamin C) - Scurvy

  Calciferol (Vitamin D) - Rickets

Macronutrient definciencies

• Protein - Marasmus and kwashiorkor

Solutions include

• Providing food rations that are composed of fresh and vitamin -and mineral- rich foods

  Adding nutrients missing in commonly consumed foods

  Genetic modification of foods

  Providing nutrition supplements

  Providing selenium supplements to people eating foods grown in selenium-poor soil

Questions

04MS3 C1. (a) The structures of three important vitamins are shown in Table 22 of the Data Booklet. State the name of each one and deduce whether each is water-soluble or fat-soluble, explaining your choices by reference to their structures. [5]

(b) Identify the metal ion needed for the maintenance of healthy bones and state the name of the vitamin needed for its uptake. [2]
(c) State the name of the vitamin responsible for maintaining healthy eyesight and the name of the functional group which is most common in this vitamin. [2]
(d) Identify one major function of vitamin C in the human body and state the name of the most common disease caused by deficiency of this vitamin. [2]
(e) Fresh fruits and vegetables are good sources of vitamin C. Explain why some meals made from these foods may contain little vitamin C. [2]

05NS3 C3. (a) State the name of a disease which results from the deficiency of each of the following vitamins. [2] vitamin A vitamin C vitamin D

(b) A person consumes an excess of both vitamin A and C. State, with a reason, which one is more likely to be stored in the body and which is more likely to be excreted. [2]

06MS3 C3. The structures of vitamins A and C are shown in Table 22 of the Data Booklet. State, with a reason, whether each is fat soluble or water soluble. [3]

C.6 Hormones (2.5h)

Questions

04MS3 C2. The structures of two sex hormones, progesterone and testosterone, are shown in Table 22 of the Data Booklet. (a) State the names of two functional groups that are present in both hormones. [2]

(b) Identify which of the two hormones is the female sex hormone and where in the human body it is produced. [2]

(c) Outline the mode of action of oral contraceptives. [3]

05MS3 C2. (a) By referring to Table 22 of the Data Booklet, identify one vitamin that is water soluble and one vitamin that is fat soluble. Explain the differences in solubility in terms of their structures and intermolecular forces. [4] (b) Vitamins C and D are vital in a balanced diet. State one major function of each of these vitamins and state a disease that results from the deficiency of each one. [4]

B.7 Enzymes

Made of amino acid chains

B.8 Nucleic Acids

DNA RNA