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AS Human Biology Unit 3


Text questions

P38/Q1 a)
b)
c)
monomer = fat
polymer = starch
macromolecule = protein

P39/Q2   6 + 6 = 12
 
P40/Q3   C12H22O11
 
P41/Q4   All starches are polymers of alpha glucose with glycosidic bonds between C1 and C4 of adjacent glucose molecules. Starches from different plants will differ due to:
  • Different number and position of side - branches
  • Different number and position of hydrogen bonds linking various parts of the chain together
P42/Q5   Food intake is generally higher for males than for females. Starch rich foodstuffs form the basis of the most common diets hence starch intake is higher in males.

P43/Q6   Protein perform many functions, eg. structural, functional (enzymes). There are many different types of proteins designed to perform structural functions and there are many different proteins which serve as enzymes controlling various metabolic processed within the cell.

P44/Q7   8

P44/Q8   N-terminus a free amino group (NH2), i.e. not engaged in the formation of a peptide bond.
C-terminus a free carboxyl group (COOH), not engaged in the formation of a peptide bond.

P45/Q9   (b) & (d)

P48/Q10   A vaccine (inactive influenza viruses) stimulates some of the WBC to produce influenza antibodies. These antibodies will defend the person if he/she is exposed to living influenza viruses. An antibody is a protein. The structure of this protein matches some protein features found on the surface of the virus. «1968»  strains of influenza virus have different protein features than «1985» strains. Therefore «1968» antibodies will not match «1985» strains of the virus. Therefore «1968» antibody will not defend a person exposed to the «1985» virus.

P49/Q11   Fatty acid with at least two double bonds (C = C) in the hydrocarbon chain.

P50/Q12   Triglyceride: 1 molecule of glycerol plus 3 molecules of fatty acids. Molecule of a triglyceride is non-polar and it does not mix with water (hydrophobic).

Phospholipid: 1 molecule of glycerol + 2 molecules of fatty acids + 1 phosphate group. Molecules of fatty acids are non-polar (hydrophobic). Phosphate group is polar (hydrophilic).

P51/Q13   Monounsaturated.

P52/Q14   The distance moved by substance is a fraction of the distance moved by solvent front.


Assignment

P54/Q1   Figures are given per 100g of milk because it is a standard way of providing this information
("nutritional value"), which was accepted by the industry. This standard enables comparisons to be made easily. It also allows a quick expression of values as %.

P54/Q2 a)

b)
Carbon is oxidised: C + O2 CO2.

Ash represents mineral salt/ions content in milk.

P54/Q3   ? EMPTY

P54/Q4 a) 10.8kg - 100%
34.4kg - ?

? = (34.4 x 100) / (10.8) = 318.5%

Very high protein content synthesis of new proteins for tissue/body building.
Very high fat content synthesis of seal's own fat/adipose tissue.
High ash content ions (Ca2+) used in bone building.

P55/Q5 a)

b)
They have no double bonds in their hydrocarbon chain.

Total concentration of saturated and unsaturated  fatty acids in mg per g of milk produced.
  Vegan Control
Saturated 325 497
Unsaturated 657 466

  c) The concentration of saturated fatty acids (SFA) in the milk produced by vegans is lower than the  concentration of SFA in the milk produced by the control group. The reason for the difference is the diet. Vegans eat exclusively non-animal foodstuffs (vegetarian diet). Fats produced by plants have a much higher concentration of unsaturated fatty acids than fats of animal origin, which are eaten by the control group, (hence higher saturated fatty acid content in the milk of the control group). The concentration of unsaturated fatty acids (UFA) in the milk of vegans is higher than that of the control group because of the reason given above.
 


Examinations

P56/Q1 a) Glycerol 3C.

  b) (i) The smallest molecule of an amino acid is

It contains 2C.

Instead of H there can be a hydrocarbon chain consisting of anything between 1C and 9C. Hence in total there could be anything from 2 up to  9 + 2 = 11 atoms of carbon.

(ii) The number of hexose (6C) molecules of glucose in a molecule of starch is large but not exactly defined.
 

  c) During the reaction molecules of water are removed and as result a new (glycosidic) bond between adjacent glucose molecules can be formed, joining these two molecules together. Water molecule is removed to condense two (glucose) molecules into one larger molecule.
 
P56/Q2 a) A strip of chromatography paper - pencil line ruled to show the origin - a spot of the substance is build up gradually on the origin by transferring small drop of a solution of the substance at a  time, drying it up, transferring another one, and so on - the paper is fastened with a drawing pin to a bung and placed in a boiling tube with some  amount of solvent inside - care needs to be taken to ensure that the spot is suspended above the level of the solvent - the completed chromatogram is examined and the distances  A(moved by substance) and B (moved by solvent front) are measured to enable calculation of the Rf value.
 
  b) (i) A certain amount of fatty acids remains on the chromatography paper before the maximum height ("distance moved by substance") is reached.

(ii) The level of radioactivity measured in this case is determined by the number of radioactive atoms of C. Hence, the larger the molecule of the fatty acid, the higher number of radioactive C atoms in its hydro-carbon chain and, consequently, the higher total level of the molecule's radioactivity.
 

 

 

 

 

 

 
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