PRE-2016

• DESCRIBE HOW YOU WOULD TEST A PIECE OF FOOD FOR THE PRESENCE OF LIPID

  1. dissolve in ethanol then add water
  2. white emulsion shows presence of lipid
• CHOLESTROL INCREASES THE STABILITY OF PLASMA MEMBRANES. CHOLESTEROL DOES THIS BY MAKING MEMBRANES LESS FLEXIBLE. SUGGEST ONE ADVANTAGE OF THE DIFFERENT PERCENTAGE OF CHOLESTEROL IN RED BLOOD CELLS COMPARED WITH CELLS LINING THE ILEUM.

  red blood cells free in blood so cholesterol helps to maintain shape.
• E.COLI HAS NO CHOLESTEROL IN ITS CELL-SURFACE MEMBRANE. DESPITE THIS, THE CELL MAINTAINS A CONSTANT SHALE. EXPLAIN WHY.

  1. cell unable to change shape
  2. because cell has a cell wall
  3. which is rigid made of murein
• (IMAGE) EXPLAIN THE DIFFERENCE IN THE STRUCTURE OF THE STARCG MOLECULE AND THE CELLULOSE MOLECULE SHOWN IN THE DIAGRAM.

  1. starch formed of alpha glucose but cellulose formed of beta glucose
  2. position of hydrogen and hydroxyl groups on carbon 1 are inverted

  
• STARCH MOLECULES AND CELLULOSE MOLECULES HAVE DIFFERENT FUNCTIONS IN PLANT CELLS. EACH MOLECULE IS ADAPTED FOR ITS FUNCTION.

  1. insoluble
  2. doesn’t affect water potential

  OR

  1. helical
  2. compact

  OR

  1. Large molecule
  2. cannot leave cell
• EXPLAIN HOW CELLULOSE MOLECULES ARE ADAPTED FOR THEIR FUNCTION IN PLANT CELLS

  1. long and straight chains
  2. become linked together by many hydrogen bonds to form fibrils
  3. provide strength to cell wall
• DESCRIBE THE STRUCTURE OF PROTEINS

  1. Polymer of amino acids
  2. joined by peptide bonds
  3. formed by condensation
  4. primary structure is order of amino acids
  5. secondary structure is folding of polypeptide chain due to hydrogen bonding
  6. tertiary structure is 3D folding due to hydrogen and ionic or disulfide bonding
  7. quaternary structure is two or more polypeptide chains
• DESCRIBE HOW PROTEINS ARE DIGESTED IN THE HUMAN GUT

  1. Hydrolysis of peptide bonds
  2. endopeptidases break polypeptides into smaller peptide chains
  3. exopeptidases remove terminal amino acids
  4. dipeptidases hydrolyse dipeptides into amino acids
• (IMAGE) WHAT REDUCING SUGAR, OR SUGARS, WOULD YOU EXPECT TO BE PRODUCED DURING CHEWING? GIVE A REASON FOR YOUR ANSWER.

  1. Maltose
  2. salivary glands breaks down starch

  
• IN THIS MODEL OF DIGESTION IN THE HUMAN GUT, WHAT OTHER ENZYME IS REQUIRED FOR THE COMPLETE DIGESTION OF STARCH?

  maltase

  
• (IMAGE) WHAT WAS THE PURPOSE OF STEP 2 IN WHICH SAMPLES WERE MIXED WITH WATER, HYDROCHLORIC ACID AND PEPSIN?

  mimics effects of the stomach

  
• (IMAGE) IN THE CONTROL EXPERIMENTS, COOKED WHEAT WAS CHOPPED UP TO COPY THE EFFECT OF CHEWING. SUGGEST A MORE APPROPRIATE CONTROL EXPERIMENT. EXPLAIN YOUR ANSWER.

  1. add boiled saliva
  2. everything same as the experiment but salivary amylase is denatured

  
• (IMAGE) EXPLAIN WHAT THESE RESULTS SUGGEST ABOUT THE EFFECT OF CHEWING ON THE DIGESTION OF STARCH IN WHEAT.

  1. some starch already digested when chewing
  2. faster digestion of chewed starch
  3. same amount of digestion without chewing at the end

  
• (IMAGE) NAME A AND B

  A - phospholipid layer

  B - transport protein

  
• (IMAGE) C IS A PROTEIN WITH A CARBOHYDRATE ATTACHED TO IT. THIS CARBOHYDRATE IS FORMED BY JOINING MONOSACCHARIDES TOGETHER. NAME THE TYPE OF REACTION THAT JOINS MONOSACCHARIDES TOGETHER.

  condensation reaction

  
• SOME CELLS LINING THE BRONCHI SECRETE LARGE AMOUNTS OF MUCUS WHCIH CONTAINS PROTEIN. NAME ONE ORGANELLE THAT YOUD EXPECT TO FIND IN LARGE NUMBERS IN A MUCUS-SECRETING CELL AND DESCRIBE ITS ROLE IN THE PRODUCTION OF MUCUS.

  1. Golgi apparatus
  2. package proteins

  OR

  1. rough endoplasmic reticulum
  2. make polypeptide or forming peptide bonds

  OR

  1. mitochondria
  2. release energy

  OR

  1. vesicles
  2. secretion or transport of protein
• (IMAGE) SUGGEST THREE REASONS WHY IT IS MORE EFFICIENT TO ATTACH LACTASE TO THE BEADS

  1. lactase / beads can be reused
  2. no need to remove from milk
  3. allows continuous process
  4. the enzyme is more stable
  5. avoid end-product inhibition

  
• MONOSACCHARIDES AND DISACCHARIDES TASTE SWEET. THE LACTOSE-FREE MILK MADE AFTER HYDROLYSIS WITH LACTASE TASTES SWEETER THAN THE COWS MILK CONTAINING LACTOSE. SUGGEST WHY.

  1. lactose hydrolysed to galactose and glucose
  2. so more sugar molecules
  3. so more sugars produced are sweeter than lactose

  
• GIVE TWO WAYS IN WHICH THE STRUCTURE OF STARCH IS SIMILAR TO CELLULOSE.

  BOTH

  1. are polymers, polysaccharides, made up of monomers, monosaccharides
  2. Contain glucose, carbon, hydrogen, oxygen
  3. contain glycosidic bonds
  4. have 1-4 links
  5. hydrogen bonding within structure
• GIVE TWO WAYS IN WHICH THE STRUCTURE OF STARCG IS DIFFERENT FROM CELLULOSE

  STARCH

  1. Contains alpha glucose
  2. helical, compact, coiled, branched
  3. 1,6 bonds, 1,6 branching
  4. glucose same way up
  5. no hydrogen bonds between molecules
  6. no micro fibrils
• (IMAGE) EXPLAIN THE DIFFERENCE IN THE STRUCTURE OF THE STARCH MOLECULES AND THE CELLULOSE MOLECULE SHOWN IN THE DIAGRAM

  1. starch formed from alpha glucose but cellulose formed from beta glucose
  2. position of hydrogen and hydroxyl groups on carbon atom 1 inverted

  
• EXPLAIN ONE WAY IN WHICH STARCH MOLECULES ARE ADAPTED FOR THEIR FUNCTION IN PLANT CELLS.

  1. Insoluble
  2. so don’t affect water potential

  OR

  1. Helical
  2. so compact

  OR

  1. Large molecule
  2. can’t leave cell
• EXPLAIN HOW CELLULOSE MOLECULES ARE ADAPTED FOR THEIR FUNCTION IN PLANT CELLS

  1. Long and straight chains
  2. become linked together by many hydrogen bonds to form fibrils
  3. provide strength to cell wall
• IN HUMANS, THE ENZYME MALTASE BREAKS DOWN MALTOSE TO GLUCOSE. THIS TAKES PLACE AT NORMAL BODY TEMP.

  EXPLAIN WHY MALTASE:

  • ONLY BREAKS DOWN MALTOSE
  • ALLOWS THIS REACTION TO TAKE PLACE AT NORMAL BODY TEMP.

  1. Tertiary structure means
  2. active site is complementary to maltose fitting like a lock and key
  3. DESCRIPTION OF INDUCED FIT
  4. enzyme is a catalyst which lowers the activation energy
  5. by forming enzyme-substrate complex
• DESCRIBE COMPETITIVE AND NON-COMPETITIVE INHIBITORS OF AN ENZYME

  1. Inhibitors reduce binding of enzyme to substrate preventing the formation of ES-complexes

  COMPETITIVE

  1. Inhibitor similar shape to substrate
  2. binds to active site of enzyme
  3. inhibition can be overcome by more substrate

  NON-COMPETITIVE

  1. Inhibitor binds to site on enzyme other than active site
  2. which Changes the shake if the active site
  3. this can’t be overcome by adding more substrate
• (IMAGE) WHAT CONCLUSIONS CAN YOU MAKE ABOUT THE DATA?

  1. less likely children with asthma eat fish
  2. less likely that children with asthma eat oily fish
  3. little difference in children with or without asthma who eat non-oily fish

  
• DESCRIBE HOW YOU COULD USE THE EMULSION TEST TO SHOW THE PRESENCE OF OIL IN A SAMPLE OF FISH

  1. Shake with ethanol
  2. then add water
  3. white emulsion should form if oil is present
• DESCRIBE HOW YOU WOULD TEST A PIECE OF FOOD FOR THE PRESENCE OF LIPID

  1. Dissolve in alcohol then add water
  2. white emulsion shows presence of lipid
• (IMAGE) OTHER THAN ETHICAL REASONS, SUGGEST TWO REASONS WHY THEY CHOSE TO USE CATS AS MODEL ORGANISMS

  1. they're mammals so likely to have the same reactions as humans
  2. small enough to keep in a laboratory
  3. Produce enough milk to extract
  4. can use a large number

  
• (IMAGE) EXPLAIN WHY MONITORING THE PH OF THE MIXTURE COULD SHOW WHETHER THE CATS MILK CONTAINED LIPASE

  1. Hydrolysis if lipids produces fatty acids
  2. which lowers pH of mixture

  
• (IMAGE) WHAT CAN YOU CONCLUDE FROM THE FIGURE ABOUT THE IMPORTANCE OF BILE-ACTIVATED LIPASE IN BREAST MILK

  1. Bile activated lipase increases growth rate of kittens
  2. results for formula with lipase not significantly different from breast milk
  3. showing addition of bile activated lipase is the likely cause of increased growth
  4. lipase increases rate of digestion of lipids

  
• SOME SEEDS CONTAIN LIPIDS. DESCRIBE HOW YOU COULD USE THE EMULSION TEST TO SHOW THAT A SEED CONTAINS LIPIDS.

  1. Crush or grind the seeds
  2. with ethanol
  3. then add water
  4. forms cloudy emulsion
• (IMAGE) A TRIGLYCERIDE MOLECULE IS FORMED BY CONDENSATION. FROM HOW MANY MOLECULES IS THIS TRIGLYCERIDE FORMED.

  4

  
• THE STRUCTURE OF A PHOSPHOLIPID MOLECULE IS DIFFERENT FROM THAT OF A TRIGLYCERIDE. DESCRIBE HOW A PHOSPHOLIPID IS DIFFERENT.

  1. phosphate
  2. instead of one of the fatty acids
• (IMAGE) USE THE DIAGRAM TO EXPLAIN WHAT IS MEANT BY AN UNSATURATED FATTY ACID

  1. double bonds present which means it’s not saturated with hydrogen
  2. in fatty acid C

  
• OMEGA-3 FATTY ACIDS ARE UNSATURATED. WHAT IS AN UNSATURATED FATTY ACID?

  double bonds between carbon
• (IMAGE) DOES THE DATA SHOW THAT EATING OMEGA-3 FATTY ACIDS PREVENTS CORONARY HEART DISEASE? EXPLAIN YOUR ANSWER.

  1. Graph shows negative correlation
  2. but this Correlation doesn’t mean causation, only shows lower risk
  3. May be due to another factor

  
• (IMAGE) NAME BOND X

  glycosidic

  
• (IMAGE) A TRIGLYCERIDE DOES NOT CONTAIN SUCROSE OR BOND X. GIVE ONE OTHER WAY IN WHICH THE STRUCTURE OF A TRIGLYCERIDE IS DIFFERENT TO OLESTRA.

  Contains glycerol

  three fatty acids

  forms three ester bonds

  
• (IMAGE) STARTING WITH SEPARATE MOLECULES OF GLUCOSE, FRUCTOSE AND FATTY ACIDS, HOW MANY MOLECULES OF WATER WOULD BE PRODUCED WHEN ONE MOLECULE OF OLESTRA IS FORMED?

  9