Homework 3                     answers to Homework 2

Refer to the figure below for question 1-3:
     1.  is this molecule 
           a) a monosaccharide     b) disaccharide
           c) a fat                       d) a nucleotide
           e) none of these is correct
     2.  the covalent link between these molecules is 
          a) a peptide bond     b) hydrogen bond
          c) glycosidic bond    e) none of these
     3.  the covalent kink between these molecules is
          a) alpha     b) beta    c) Esther Williams
          d) ionic      e) none of these
Refer tot he figure below for questions 4-5
      4.  The molecule above is classified as a?
           a) sugar   b) nucleotide   c) amino acid
           d) fat (lipid)        e) none of these
      5.  The classification of the above
           molecule is as it is because the 
           molecule is insoluble in an aqueous
           environment.    a) true     b) false
refer to the figure below for questions 6-7

      6.  The type of weak electrostatic bonds
           depicted in A.  is ?
           a) ionic bond      
           b) van der Walls force
           c) hydrophobic attraction
           d) hydrophilic  attraction
           e) none of the above 
      7.  The approximate energy required to break
           electrostatic bond in the figure above is?
           a) greater than 100 Kc/mole
           b) over 7,300 Kc/mole
           c) between 1 and 5 Kc/mole
           d) none of the above 
 refer to figure below for question 8-9

     8.  The molecule above is
          a) a nucleotide       b) an amino acid
          c) a fatty acid       d) a sugar
          e) none of these 
     9.  Is this molecule?
          a) saturated          b) unsaturated
          c) neither saturated or unsaturated


  Answers to homework 2

2:  1 (d),  2(b),  3(b),  4(b),  5(c),  6(d),  7(a),  8(d)  and

   9)  You could examine the filtered sap solution
   under an electron microscope to determine if there
   was a cellular agent present, i.e., very small bacteria,
   or is a non-cellular agent as a virus was present.

   You could also try to cell culture the filtrate.  If it was
   bacterial, you might be able to grow them in the absence
   of other cells, but if it were viral, you might have to
   culture it on host cells.

   You could chemically analyze the filtrate for a genome,
   DNA or RNA and measure the amount of genetic material
   present.  If the infective agent was cellular, it would be
   expected to have a larger genome than if it were viral.