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CARBOHYDRATE S & LIPIDS

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Presentation on theme: "CARBOHYDRATE S & LIPIDS"— Presentation transcript:

1 CARBOHYDRATE S & LIPIDS
Compounds of carbon, hydrogen and oxygen are used to supply and store energy. Topic 2.3 IB Biology Miss Werba

2 TOPIC 2 – MOLECULAR BIOLOGY
2.1 MOLECULES TO METABOLISM 2.2 WATER 2.3 CARBOHYDRATES & LIPIDS 2.4 PROTEINS 2.5 ENZYMES 2.6 STRUCTURE OF DNA & RNA 2.7 DNA REPLICATION, TRANSCRIPTION & TRANSLATION 2.8 CELL RESPIRATION 2.9 PHOTOSYNTHESIS J WERBA – IB BIOLOGY 2

3 THINGS TO COVER U.1 U.2 U.3 U.4 Statement Guidance
Monosaccharide monomers are linked together by condensation reactions to form disaccharides and polysaccharide polymers. The structure of starch should include amylose and amylopectin. Sucrose, lactose and maltose should be included as examples of disaccharides produced by combining monosaccharides. U.2 Fatty acids can be saturated, monounsaturated or polyunsaturated. Named examples of fatty acids are not required U.3 Unsaturated fatty acids can be cis or trans isomers. U.4 Triglycerides are formed by condensation from three fatty acids and one glycerol. J WERBA – IB BIOLOGY 3

4 THINGS TO COVER A.1 A.2 A.3 A.4 S.1 S.2 IM.1 Statement Guidance
Structure and function of cellulose and starch in plants and glycogen in humans. A.2 Scientific evidence for health risks of trans fats and saturated fatty acids. A.3 Lipids are more suitable for long-term energy storage in humans than carbohydrates. A.4 Evaluation of evidence and the methods used to obtain the evidence for health claims made about lipids. S.1 Use of molecular visualization software to compare cellulose, starch and glycogen. S.2 Determination of body mass index by calculation or use of a nomogram. IM.1 Variation in the prevalence of health problems around the world. NOS 5.2 Evaluating claims J WERBA – IB BIOLOGY 4

5 MONOSACCHARIDES U.1 J WERBA – IB BIOLOGY 5

6 MONOSACCHARIDES Glucose Galactose Fructose Examples Details
Example use in plants Example use in animals Glucose C6H12O6 hexose sugar 5C form the ring, with the 6th corner taken by oxygen Source of energy for cellular respiration Galactose same as glucose but one set of side chains are reversed Found in cereals Found in milk Fructose pentose sugar Component of fruits; Sweet so it attracts animals for seed dispersal. Component of honey; Good energy source. J WERBA – IB BIOLOGY 6

7 MONOSACCS  DISACCS Condensation reaction Water is removed
U.1 Condensation reaction Water is removed Glycosidic bond formed J WERBA – IB BIOLOGY 7

8 MONOSACCS  DISACCS U.1 glucose + glucose  maltose + water glucose + galactose  lactose + water glucose + fructose  sucrose + water J WERBA – IB BIOLOGY 8

9 DISACCHARIDES U.1 d Maltose O Lactose J WERBA – IB BIOLOGY 9

10 DISACCHARIDES Maltose Lactose Sucrose Examples Details
Example use in plants Example use in animals Maltose C12H22O11 glucose dimer Intermediate in the digestion of starch. Sweet; Lactose glucose and galactose Found in some seeds as a energy storage molecule. Found in milk to provide newborns with energy. Sucrose table sugar glucose and fructose Transport form for carbon; Provides energy for plants unable to photosynthesise. Sweet; Good energy source. J WERBA – IB BIOLOGY 10

11 MONOSACCS  POLYSACCS Condensation reaction
U.1 Condensation reaction Water is removed (no. of water = no. of monomers – 1) Glycosidic bonds formed Often very long and may be branched Examples: Starch – used as an energy storage molecule in plants Glycogen – used as an energy storage molecule in animals Cellulose – structural component of plant cell walls J WERBA – IB BIOLOGY 11

12 POLYSACCHARIDES: Cellulose
Function: structural component of plant cell walls J WERBA – IB BIOLOGY 12

13 POLYSACCHARIDES: Cellulose
Structure: Made of β-glucose molecules linked together Condensation reactions link C1 of one glucose to C4 of the next The molecule is a straight chain (not curvy) The molecule is unbranched Cellulose can adhere to itself through hydrogen bonding  forming cellulose myofibrils with a high tensile strength It’s this strength that allows the cell wall to resist turgor pressure. J WERBA – IB BIOLOGY 13

14 POLYSACCHARIDES: Starch
U.1 A.1 Function: used as an energy storage molecule in plants J WERBA – IB BIOLOGY 14

15 POLYSACCHARIDES: Starch
U.1 A.1 Structure: Made of α-glucose molecules linked together Condensation reactions link C1 of one glucose to C4 of the next The molecule is a curvy chain (not straight) Starch is hydrophilic, but are not soluble due to their large size There are two forms of starch: Amylose – contains only C1-C4 bonds and forms linear helices Amlyopectin – also contains C1-C6 bonds, which causes branching J WERBA – IB BIOLOGY 15

16 POLYSACCHARIDES: Glycogen
U.1 A.1 Function: used as an energy storage molecule in animals J WERBA – IB BIOLOGY 16

17 POLYSACCHARIDES: Glycogen
U.1 A.1 Structure: Made of repeating glucose molecules linked together Condensation reactions link C1 of one glucose to C4 of the next and potentially C1 to C6. The molecule is a compact Hydrophilic but insoluble – therefore does not upset the osmotic balance of the cells J WERBA – IB BIOLOGY 17

18 POLYSACCHARIDES: Glycogen
U.1 A.1 J WERBA – IB BIOLOGY 18

19 MOLECULAR VISUALISATION SOFTWARE
Use jmol from the biotopics website to compare the structure of cellulose, starch and glycogen: Instructions: Select the glucose molecule and identify the colours used to represent carbon, hydrogen and oxygen. Compare the glucose, sucrose and fructose molecules. Look at the amylose model and zoom out. Describe it’s shape. Zoom in on amylose. Describe the glycosidic bonds here. Are they all the same? Look at the amylopectin model and zoom in on a branchpoint. Describe the glycosidic bonds here. Investigate glycogen and cellulose in a similar way. J WERBA – IB BIOLOGY 19

20 LIPID STRUCTURE Monomer unit: fatty acid
Polymer: triglyceride (3 fatty acids & glycerol) J WERBA – IB BIOLOGY 20

21 LIPID STRUCTURE U.4 Triglycerides (3 fatty acids & glycerol) are formed by condensation reactions 3 reactions are needed ∴ 3 water molecules are produced covalent bond = ester bond J WERBA – IB BIOLOGY 21

22 FATTY ACID STRUCTURE Fatty acids can be:
saturated monounsaturated or polyunsaturated Saturation determined by the number of double bonds: saturated = 0 double bonds monounsaturated= 1 double bond polyunsaturated = 2+ double bonds J WERBA – IB BIOLOGY 22

23 FATTY ACID STRUCTURE U.2 J WERBA – IB BIOLOGY 23

24 FATTY ACID STRUCTURE Unsaturated fatty acids can be: cis or
trans isomers J WERBA – IB BIOLOGY 24

25 FATTY ACID STRUCTURE U.3 J WERBA – IB BIOLOGY 25

26 FATTY ACID STRUCTURE Cis- isomers Trans- isomers very common in nature
rare in nature – produced artificially to produce solid fats (eg. margarine) H atoms on same side H atoms on different sides double bond causes a bend double bond does not cause a bend loosely packed closely packed triglycerides with cis-isomers have low melting points triglycerides with trans-isomers have highmelting points J WERBA – IB BIOLOGY 26

27 HEALTH CLAIMS ABOUT LIPIDS
The relative amounts of different types of fatty acid in a person’s diet can be correlated with health issues. Western diets tend to be high in saturated fatty acids Linked to an increased risk of coronary heart disease (CHD) But correlation ≠ causation – could be linked to another factor The Maasai people of Kenya do not fit this correlation. Diet is high in saturated fatty acids (from milk, meat, etc) but CHD is almost unknown in this population Mediterranean diets also tend to be high in fatty acids – but cis-monounsaturated fatty acids Linked to an lower incidence of CHD J WERBA – IB BIOLOGY 27

28 HEALTH CLAIMS ABOUT LIPIDS
Genetic factors could be involved All fatty acids are high in energy Excess of fat in the diet linked to an increased risk of obesity Positive link between diets high in trans-fats and Coronary Heart Disease (CHD) Link between low levels of omega-3 saturated fatty acids and affected brain and nerve development J WERBA – IB BIOLOGY 28

29 HEALTH CLAIMS ABOUT LIPIDS
Evaluation of evidence and the methods used to obtain the evidence for health claims made about lipids. Evaluation = make an appraisal by weighing up the strengths and limitations of the evidence Evidence = from research; some more scientifically valid than others. J WERBA – IB BIOLOGY 29

30 HEALTH CLAIMS ABOUT LIPIDS
Evaluation of evidence and the methods used to obtain the evidence for health claims made about lipids. Strengths? Has a positive or negative correlation been determined? Has the standard deviation or statistical significance of the difference between the two means tested? Has the range of data been considered? J WERBA – IB BIOLOGY 30

31 HEALTH CLAIMS ABOUT LIPIDS
Evaluation of evidence and the methods used to obtain the evidence for health claims made about lipids. Limitations? What was the definition of “health” being tested? What was the sample size? Was the sample representative of the general population? Was the data gathered from human or animal trials? Was it a realistic study? What methods were used to gather the data? J WERBA – IB BIOLOGY 31

32 LIPIDS FOR LONG-TERM ENERGY STORAGE
Functions of lipids include: Structure: phospholipids in cell membranes Hormonal signalling: steroid hormones are lipids (eg. oestrogen, testosterone) Insulation: fat is an insulator; lipids also insulate neurons Protection: fats protect internal organs from damage Storage of energy: fats can be used as a long-term energy storage source J WERBA – IB BIOLOGY 32

33 LIPIDS FOR LONG-TERM ENERGY STORAGE
Carbohydrates and lipids contain a lot of chemical energy and can be used for energy storage Lipids less readily digested so they are more suitable for long-term storage These lipids are stored as fats in adipose tissue (cf. Topic 1) under the skin and around organs, including the kidneys. J WERBA – IB BIOLOGY 33

34 LIPIDS FOR LONG-TERM ENERGY STORAGE
Lipids release more energy per gram than sugars (or proteins). Lipids are therefore lighter than sugars of equal yield: Lipids add 1/6 as much to body mass as sugars Glycogen is stored with water (b/c it’s soluble), while fats are not J WERBA – IB BIOLOGY 34

35 LIPIDS FOR LONG-TERM ENERGY STORAGE
So why use glycogen at all???? Breakdown of sugars is much faster, making them ideal for short-term energy storage Sugars are soluble in water (cf. blood plasma), making them easy to transport to sites of use Glucose can be used aerobically (with O2) or anaerobically (without O2) but fats can only be utilised in aerobic conditions J WERBA – IB BIOLOGY 35

36 BODY MASS INDEX (BMI) CALCULATION
Body Mass Index (BMI) is a screening tool to identify potential weight issues. It should be used in conjunction with other measures, such as a skinfold thickness test, diet evaluations, fitness assessments and family history. J WERBA – IB BIOLOGY 36

37 BODY MASS INDEX (BMI) CALCULATION
BMI is calculated in the same way for kids and adults. Units for BMI = kg m-2 J WERBA – IB BIOLOGY 37

38 You can also use a BMI chart to assess BMI
J WERBA – IB BIOLOGY 38

39 BODY MASS INDEX (BMI) CALCULATION
An alternative method uses a nomogram. Draw a line between the weight and weight. It will cross a third axis, a BMI scale. J WERBA – IB BIOLOGY 39

40 BODY MASS INDEX (BMI) CALCULATION
Q1 A woman has a mass of 72kg and a height of 1.69m. Calculate her BMI. Deduce the body mass status of this woman using the table. Outline the relationship between height and BMI for a fixed body mass. J WERBA – IB BIOLOGY 40

41 BODY MASS INDEX (BMI) CALCULATION
A1 A woman has a mass of 72kg and a height of 1.69m. BMI = 72 ÷ (1.69)2 = 25.2 kg m-2 The woman would be classified as overweight. The taller the person, the smaller their BMI; negative correlation. J WERBA – IB BIOLOGY 41

42 BODY MASS INDEX (BMI) CALCULATION
Q2 A man has a height of 150cm and a BMI of 40. Calculate the minimum amount of body mass he must lost to reach normal body mass status. Show your working. Suggest 2 ways in which the man could reduce his body mass. J WERBA – IB BIOLOGY 42

43 BODY MASS INDEX (BMI) CALCULATION
A2 A man has a height of 150cm and a BMI of 40. BMI = mass / height = mass / mass = 40 x = 90 kg BMI = mass / height = mass / mass = 24.9 x = 56 kg Weight loss needed = 90 – 56 = 34kg Reduce calorie intake/reduce fat intake/diet; Exercise/increase activity level. J WERBA – IB BIOLOGY 43

44 CARBOHYDRATES & LIPIDS
Q3 What is the difference between galactose and lactose? Lactose is a disaccharide and galactose is a monosaccharide. Lactose is the product of anaerobic respiration in humans and galactose is the product of anaerobic respiration in yeast. Lactose is an enzyme and galactose is a hormone. Galactose is a sugar found in milk but lactose is not found in milk. J WERBA – IB BIOLOGY 44

45 CARBOHYDRATES & LIPIDS
Q4 What chemical is shown in the diagram below? Saturated fatty acid Cis-monounsaturated fatty acid Trans-monounsaturated fatty acid Polyunsaturated fatty acid J WERBA – IB BIOLOGY 45

46 CARBOHYDRATES & LIPIDS
Q5 Outline the role of condensation and hydrolysis in the relationship between fatty acids, glycerol and triglycerides.[6] J WERBA – IB BIOLOGY 46


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