Pretend you are a teacher preparing a quiz for the text I will provide. 
            The quiz must be in the same language as the input text. Each quiz question should follow the format below. 
            Ensure that you carefully recheck the questions and answers for accuracy. 
            The output should be a ready-to-use list of dictionaries for the Python eval function:

            [

                    "question": "WRITE THE QUESTION HERE",
                    "options": ["A) OPTION_1", "B) OPTION_2", "C) OPTION_3", "D) OPTION_4"],
                    "answer": INDEX_OF_CORRECT_ANSWER_IN_OPTIONS_LIST,
                    "explanation": "EXPLAIN WHY THE ANSWER IS CORRECT"

            ]

            Text:
            of the -OH and -H on 
carbon atoms 2, 3 and 4 of glucose. 
•
The most important epimers of glucose are mannose and galactose. 
Mannose is formed by epimerization at carbon 2 whereas, galactose by 
epimerization at carbon 4. 
H 
C        O      
H        C       OH  
HO     C      H 
H        C     OH 
H        C      OH 
           CH2OH 
 
  D-GLUCOSE 
 
H 
C        O      
HO     C        H  
HO     C      H 
H        C     OH 
H        C      OH 
           CH2OH 
 
D-MANNOSE 
 
CARBON-2 EPIMERS 
CARBON-4 EPIMERS 
H 
C        O      
H        C       OH  
HO     C      H 
H        C     OH 
H        C      OH 
           CH2OH 
D-GLUCOSE 
 
H 
C        O      
H        C      OH  
HO     C        H 
HO     C        H 
H        C      OH 
           CH2OH 
D-GALACTOSE 
 
(5) Aldoses and Ketoses: 
•
Depending on the presence of aldehyde or keto group in the sugar. 
•
Fructose contains ketone group in position 2, whereas there is an 
aldehyde group in position 1 of glucose. 
Benedict’s Test 
Benedict’s 
reagent 
undergoes 
a 
complex colour change when it is 
reduced 
The intensity of the colour change is 
proportional to the concentration of 
reducing sugar present 
The colour change sequence is: 
– Blue… 
– green… 
– yellow… 
– orange… 
– brick red 
The carbonyl group - monosaccharides 
The carbonyl group is ―free‖ in 
the straight chain form. 
But not free in the ring form. 
BUT remember – the ring form 
and the straight chain form are 
interchangeable. 
So all monosaccharides are 
reducing sugars. 
All 
monosaccharides 
reduce 
Benedict’s reagent. 
Optical activity 
•Optical activity is the ability of a compound to rotate plane polarized light either 
to the left [-] (levorotatory) or to the right [+] (dextrorotatory). 
•The plane- polarized light (PPL) can be obtained by passing a beam of ordinary 
light through a NiKol’s prism (CaCO3). 
•The naturally occuring form of glucose is D (+) isomer. 
•The compounds having asymmetric carbon atoms can rotate the beam of plane 
polarized light and are said to be optically active. 
•DL-mixture is a mixture containing  equal amounts of D and L isomers,  the 
resulting solution has no optical activity since the activities of each isomer cancel 
one another. Such structure is called racemic or DL-mixture. 
•All monosaccharides are optically active except dihydroxyacetone because they 
contain asymmetric carbon atoms. 
Monosaccharide Derivatives 
Deoxy sugars:  
are sugars in which a hydrogen atom replaces one or more of the -OH 
groups in a monosaccharide. 
D-ribose and its derivative D-2-deoxyribose (deoxy = minus one 
oxygen atom) are found in various coenzymes and in DNA. 
DISACCHARIDES
 
These are sugars formed of 2 monosaccharide units linked by a glycosidic 
linkage. 
The physiologically important disaccharides are maltose, sucrose and 
lactose. 
 
•
Reducing: Maltose, Lactose –with free aldehyde or keto group. 
•
Non-reducing: Sucrose, Trehalose –no free aldehyde or keto group. 
 
The 

            Quiz: