Biological Molecules - Carbs And Lipids Flashcards
What is a monomer?
small identical/similar molecules condensed to make larger molecules called polymers
eg: alpha glucose, aa, nucleotides
What is a polymer?
Large molecules made by joining 3 or more identical or similar monomers together
eg: proteins, polysaccharides, polynucleotides
define condensation reactions
- joins two monomers together to form a - chemical bond
- removal of one water molecule
eg: alpha glucose -> starch
define hydrolysis reaction
- breaks a chemical bond between two monomers
- via addition of water molecule
eg: cellulose -> beta glucose
DNA -> DNA nucleotides
disaccharide/dimer equation
C6H12O6 + C6H12O6 -> C12H22O11 + H2O
Where do condensation reactions occur in monomers?
0H group of C4 and C1 H of other group
maltose
alpha glucose + alpha glucose
maltase enzyme
lactose
galactose + glucose
lactase enzyme
sucrose
glucose + fructose
sucrase enzyme
differences between monosaccharides
a-glucose: both OH (hydroxyl groups) down
galactose: both OH up
B-glucose: one OH down, one OH up
fructose: pointy top
starch
- insoluble, doesn’t affect water potential
- helical, so compact
- large so doesnt diffuse out of cells
- branched so high surface area for rapid hydrolysis to release glucose for respiration
amylose
a (1-4) glycosidic bonds
- linear chains of a glucose
- compact and helical, so good for storage
- insoluble, doesnt affect water potential
amylopectin
branched
a(1-4) and a(1-6) glycosidic bonds
large SA for rapid hydrolysis
isnoluble doesnt affect water potential
large so doesnt diffuse out of cells
glycogen
- shorter chains
- more branched
-larger SA
-humans/animal cells
-short chains so more rapidly hydrolysed into glucose for RP
-insoluble, doesnt affect water potential - large, so doesnt diffuse out of cells
a strach molecule has a spiral shape. explain why this shape is important to its function in cells.
helical so compact
insoluble, no effect in WP
large molecule so cant leave out of cell
structure of cellulose is related to its role in plant cell walls. explain how.
- long straight unbranched chains of beta glucose
- joined by many weak hydrogen bonds
- form microfibrils
- provides strength
cellulose structure
- long straight unbranched chains
- every other b glucose inverted 180 degrees
- b (1-4) glycosidic bonds
- weak Hydrogen bonds form cross links between chains
- forms microfibrils (that form cellulose fibres, stregnth and support in plant cell walls)
hydrogen bonds are important in cellulose molecules. explain why.
hydrogen bonds forms cross links between chains
-form microfibrils
- provides strength and rigidity
-weak hydrogen bonds provide strength in large numbers
reducing sugars
galactose
glucose
lactose
maltose
fructose
non reducing sugars
sucrose
describe how the student would show that a reducing sugar was present in a solution
- add equal volumes of benedicts solution
- and heat to 95 degrees celsius
- red/orange PPT is a positive result
2 different reducing sugar solutions (A+B) of same concentration have an enzyme added before benedicts. They both produced red PPT.
After 20 minutes, B had twice as much PPT as A. Suggest why.
A is a monosaccharide and B is a disaccharide
- enzyme hydrolysed glycosidic bonds in disaccharide, releasing 2 monosaccharides
- both monosaccharides are reducing sugar, PPT doubled in B
describe how the student would show a non-reducing sugar was present in a solution
- complete benedicts test and observe a negative result (blue solution)
- add acid (HCl) and heat to 95 degrees
- then neutralise with alkali
- add benedicts solution again and then heat to 95
- red/orange PPT is positive result
describe how you would produce a calibration curve for a reducing sugar of unknown concentration and use it to obtain results
- make up several known concentrations of [R sugar]
- carry out benedicts test
- use a colourimeter to measure the colour absorbance of each solution
- and plot a calibration curve:
X: known conc
Y: absorbance/transmission - find conc of unknown sample using calibration curve
standardising colourimeter method
- samples shaken before testing
- zero colourimeter before use with a control
- use same filter
- use same volume for each reading
describe how you would test a sample of food for the presence of starch
- add potassium iodide (KI) solution
- blue/black indicates starch is present
The structure of cellulose is related to its role in a plant cell wall. Explain how.
- Long, straight unbranched chains of beta glucose
- joined by many weak hydrogen bonds
- form microfibrils
- provide strength/rigidity
Give one feature of starch and explain how it enables it to act as a storage substance
- helical so compact
- insoluble so doesnt affect water potential
- large molecule so doesnt leave cell
- branched chains so rapid hydrolysis to remove glucose for respiration
Hydrogen bonds are important in cellulose. Explain why.
Holds chain together/form microfibrils
Providing strength and rigidity
Weak Hydrogen bonds provide strength in large numbers
Describe how lactose is formed and where in the cell it would be attached to a polypeptide to form a glycoprotein
Glucose and galactose
Joined by condensation reaction
Jojned by glycosidic bond
Added to polypeptide in golgi
Describe how you would test a liquid sample for the presence of a lipid vox and how you would recognise a positive result
Mix with ethanol then water
And shake
Cloudy white emulsion
Omega 3 fatty acids are unsaturated. What is an unsaturated fatty acid?
Double bond
Between carbon atoms within the hydrocarbon chain
Describe how a ester bond is formed in a phospholipid molecule
Condensation
Between a glycerol and fatty acid
What are the differences between a triglyceride and a phospholipid
Fatty acid removed
Replaced with a phosphate group
Compare and contrast the structure and properties of triglycerides and phospholipids
- both contain ester bonds
- both contain glycerol
-fatty acids on both may be saturated or unsaturated - both are insoluble in water
- both contain C, H and O but phospholipids also contain P
- triglycerides have 3 fatty acids and phospholipids have 2 fatty acids plus a phosphate group
- triglycerides are hydrophobic and phospholipids have a hydrophilic and hydrophobic region
- phospholipids form bilayer but triglycerides do not
Desctibe the biochemical tests you would use to confirm the presence of lipid, non reducing sugar and amylase in a sample
Lipid
- add ethanol then water and shake
- white milky emulsion
Non reducing sugar
- Do benedicts and get a negative result
- boil with acid and immedicately neutralise with an alkali
- heat with benedicts and red/orange precipitate
Amylase
- add bieuret reagenr and purple
- add starch
- test for reducing sugar
A student carried out the benedicts test. Suggest a method, other than usinf a colorimeter, that this student could use to measure the quantity of reducing sugar in a solution
- filter and dry the precipitate
- find mass/weight
