SPM Biology 2018


2. Diagram 2.1 shows the food which contains carbohydrates.

Rajah 2.1 menunjukkan makanan yang mengandungi karbohidrat.

(a) (i) Based on Diagram 2.1, name the type of carbohydrates in rice and honey.

Berdasarkan Rajah 2.1, namakan jenis karbohidrat dalam nasi dan madu.

(a) (ii) Explain why diabetic patient are advised not to consume excessive amount of rice in their daily diet.

Terangkan mengapa pesakit diabetes dinasihatkan untuk tidak mengambil nasi secara berlebihan dalam diet harian mereka.

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(b) Diagram 2.2 shows the formation of molecule V in milk. Based on the Diagram 2.2, describe:

Rajah 2.2 menunjukkan pembentukan molekul V dalam susu. Berdasarkan Rajah 2.2, huraikan:

(i) The formation of molecule V.

Pembentukan molekul V.

(ii) The breakdown of molecule V.

Penguraian molekul V.

(c) When sucrose solution is heated with Benedict’s solution, the blue solution remains unchanged. Explain why Benedict’s test gives negative results on sucrose.

Apabila larutan sukrosa dipanaskan dengan larutan Benedict, warna biru larutan itu tidak berubah. Terangkan mengapa ujian Benedicr memberi keputusan negative pada sukrosa.

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Answer ( 1 )


    2. a) i)

    Rice: glucose

    Honey: fructose


    ii) Diabetic patients have a high blood glucose level and either produce low levels of insulin or their bodies are not sensitive to the insulin produced. Thus, excess glucose cannot be efficiently converted to glycogen resulting in glucose build up within the bloodstream. So diabetic patients need to reduce their blood glucose levels and they can do this by avoiding the consumption of excessive amounts of rice which contains glucose.


    b) i) Molecule V is lactose. To form molecule V, glucose and galactose undergoes a condensation process to form lactose and this process removes one water molecule.

    ii) The breakdown of molecule V involves a hydrolysis process by adding water to V. Lactose with the addition of water produces glucose and galactose.


    c) Benedict’s solution reduces from a blue solution (copper II sulphate) to a brick red precipitate (copper I oxide) in the presence of reducing sugars. The Benedict’s test gives a negative result on sucrose because sucrose is a non-reducing sugar, the reaction doesn’t take place. In order to get a positive result sucrose needs to be hydrolysed with dilute hydrochloric acid to produce fructose and glucose. Monosaccharides are reducing sugars so a positive result can then be obtained.





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